% ElectricityBill/contents.tex
%
\chapter{\mychaptername}
\label{\here}
\tocnotetoo{
We use an electricity bill as a hook on which to hang an introduction
to functions in general and linear functions in particular, in algebra
and in Excel. Then we apply what we've learned to study taxes ---
sales, income and social security. You'll also find here a 
general discussion of energy and power.
}

\begin{goals}

\begin{goal}{directproportion}Understand direct proportion as a
linear equation with intercept 0.
\end{goal}

\begin{goal}{linearfunctions}
Study situations governed by linear equations.
\end{goal}

\begin{goal}{slopeintercept}
Stress the meaning and units of the slope and intercept.
\end{goal}

\begin{goal}{functions}
View functions as tables, graphs and formulas.
\end{goal}

\begin{goal}{linearexcel}
Construct flexible spreadsheets to model linear equations.
\end{goal}

\begin{goal}{incometax}
Understand the piecewise linear income tax computations.
\end{goal}

\begin{goal}{energyandpower}
Sort out the confusing distinction between energy and power.
\end{goal}

\end{goals}

\qrsection[directproportion]{Rates}
\teachertag{}
\begin{teacher}
In a course devoted more to real life applications than to modeling,
you can move directly to the section on taxes. There's no explicit
need there for the slope-intercept description of a linear model.
\end{teacher}

We began Chapter~\ref{Units} with a discussion of the relationship
\begin{center}
distance = rate $\times$ time.
\end{center}
or, with units,
\begin{equation}\label{eq:rtd}
\text{miles traveled = (rate, in miles/hour)} \times \text{(hours on
the road) }.
\end{equation}

There we worked with particular numbers; now we want to
look at that relationship a little more generally.
If we work with a fixed rate (say 60 miles/hour), then we can find the
distance traveled whenever we know the time. To say that with some
algebra, write  $D$ for distance and $T$ for time. Then
%
\begin{equation*}
D = 60 \times T,
\end{equation*}
or, more generally,
\begin{equation*}
D = r \times T,
\end{equation*}
where $r$ is the rate of travel, in miles/hour. 

That formula says that distance traveled is
\emindex{proportional} to travel time.
The rate in miles per hour is
the \emindex{proportionality constant}.
If you drive for twice as long you go twice as far. 
If you drive ten times as long you go ten times as far. 

In \sref*{mpg} we introduced the units  (gallons per hundred miles) as
a useful way to measure automobile fuel economy: the amount of gas you
use is proportional to the distance you drive.
Let $F$ be the amount of gasoline you use, in gallons, to drive $D$ hundred
miles. Then
%
\begin{equation*}
F = r \times D
\end{equation*}
%
where the proportionality constant $r$ is the fuel
use rate, with units gallons per hundred miles. If you drive 10
times as far you use 10 times as much gas.
You don't use any gas at all just sitting in the driveway
(unless you're idling to warm up the car.

The proportionality constant is always a rate: it 
appears with units. In these examples the units are miles per hour
and gallons per 100 miles. 

The unit pricing discussion in \sref*{unitprices} provides more
examples of proportionality.

Finally, sales tax is computed as a proportion. If you spend twice as
much you pay twice as much tax. The tax rate is the proportionality
constant. If it's 5\% then you pay (five dollars of tax) per
(hundred dollars of purchase).

\qrsection[tamworth]{Reading your electricity bill}

The more electricity you use at home, the more you pay. But the
relationship isn't quite proportional. You don't pay twice as much to
use twice as much. Figure~\ref{tamworthbill} shows a simple sample
electricity bill from \url{www.squashedfrogs.co.uk};
there's a local copy at 
\link{how\_to\_read\_electricity\_bills.doc}.

\figfile{TamworthElectricBillcropped.pdf}
\begin{figure}[ht]
\centering
\framebox{
%  \includegraphics[height=60mm]{\thefigurefilename}
  \includegraphics[height=60mm]{\here/TamworthElectricBillcropped.pdf}
}
\caption{Tamworth electricity bill}
\figsource{dead url: \url{www.squashedfrogs.co.uk/resources/2005/10/how\_to\_read\_electricity\_bills.doc}}
\figcomment{Permission requested from www.squashedfrogs.co.uk/contact.cfm}
\label{tamworthbill}
\end{figure}
\figfile{}

This bill explains itself. We'll study it before we look
at a real one. It comes from England, so the costs are expressed
in pounds and pence rather than dollars and cents, and it comes once a
quarter (every three months) rather than once a month, but you can ignore
that while you read it --- from the bottom up.

The last line is the total bill, computed as
\begin{center}
\framebox{
\texttt{Cost of electricity + fixed charge} 
} .
\end{center}
Checking the arithmetic:
\begin{equation*}
\mbox{\textsterling} 34.62 + \mbox{\textsterling} 9.49.
= \mbox{\textsterling} 44.11 .
\end{equation*}
The third line from the bottom explains the \textsterling 34.62:
\begin{center}
\framebox{
\texttt{Number of units $\times$ cost per unit} 
} .
\end{center}
That's our old friend proportionality.
The previous line gives the proportionality constant:
the cost per unit as 7.35 pence per unit. Later on in
the document we're told that a unit is just a kilowatt-hour,
abbreviated ``kWh''.
(It's too bad the bill talks about ``units'' instead
of just ``kWh'' since for us ``units'' has a more general meaning.)

Thre are 100 pence in a pound, so the cost per unit is
%
\begin{equation*}
0.0735 \frac{\text{\textsterling}}{\mbox{kWh}}.
\end{equation*}
%
(The computation would have been much
more complicated before February 15, 1971 --- the day England converted
rom pounds/shillings/pence to decimal currency --- 
\url{news.bbc.co.uk/onthisday/hi/dates/stories/february/15/newsid\_2543000/2543665.stm}.)

In the current quarter this customer used 471 kWh of electricity ---
the difference between the meter reading before and after the quarter.

Here's all the arithmetic, with units:
\begin{equation*}
\mbox{\textsterling} 44.11 = 
0.0735 \frac{\mbox{\textsterling}}
{\mbox{\cancel{kWh}}} 
\times 471 {\mbox{\ \cancel{kWh}}}
+
\mbox{\textsterling} 9.49.
\end{equation*}

That English bill is easy to read. Figure~\ref{nstarbill} shows a real
one that's a little more complex, from NStar, in Boston.

\figfile{NStarElectricBill.jpg}
\begin{figure}[ht]
\centering
\framebox{
  \includegraphics[height=60mm]{\thefigurefilename}
}
\caption{NStar electricity bill (2007)}
\figsource{Scan of one of the authors' bills - need to replace! Change
  to Eversource. Look   for image on line, or scan at higher resolution.}
\figcomment{Permission granted.}
\label{nstarbill}
\end{figure}
\figfile{}

We can identify the same two components. The 
fixed charge is the \$6.43 labelled ``Customer Charge.''
It's the part of the \$145.26 total  that does not 
depend on the amount of electricity used --- in this case, 813
kwh. The six lines on the bill that do depend on that contribute
\begin{align*}\
(0.04432 + 0.01039 + 0.00468 + 0.00050 + 0.00250 + 0.10838) \times
813 
&     =  0.17077 \times 813   \\
&     =  138.83601 
\end{align*}
to the total bill, which is 
\begin{equation*}
\$145.26 = 
0.17077 \frac{\$}
{\mbox{\cancel{kwh}}} 
\times 813 {\mbox{\ \cancel{kwh}}}
+
\$6.43.
\end{equation*}
\noindent
(If you check the calculation you will discover that the electric
company rounded \$138.83601 down to \$138.83 rather than up to the
nearest penny. We should be grateful for small favors.)

\qrsection[linearfunction]
{Linear functions}\index{function}
\index{linear function}

So far \CommonSense{} has called for hardly any algebra.
Now a little bit will come in handy.

Suppose you buy your electricity from NStar as in the example above 
and want to study how your
bill changes when you use different amounts of electricity.
The monthly \$6.43 Customer Charge does not change.
The rest of your bill is proportional to
the amount of electricity. The proportionality constant is
0.17077 \$/kwh in the sample bill. We will assume that it does
not change, although in fact it does change slightly when the electric
company changes its rates.

If in a given month you use $E$ kwh of electricity your total bill $B$ 
can be computed with the formula
\begin{equation*}
B\$ = 
0.17077 \frac{\$}
{\mbox{\cancel{kWh}}} 
\times E {\mbox{\ \cancel{kWh}}}
+
\$6.43 .
\end{equation*}

That formula captures how the dollar amount of your electricity
bill depends on the amount of electricity you use, measured in kWh.
The first term captures the part that's
proportional to the amount of electricity used. The second term (the amount \$6.43) is
fixed. It represents the electric company's fixed costs: things like 
generating the bill and mailing it to you and  maintaining the power
lines on the street in front of your house. Those are expenses they
must cover even if you're on vacation and have turned off all the
appliances.

You \myindex{probably} encountered a similar formula once
in an algebra class --- it may look more
familiar without the units
\begin{equation*}
B = 0.17077 \times E + 6.43 .
\end{equation*}

It may look even more familiar if we call the variables by the
traditional names $x$ (for the independent variable) and $y$ (for the
dependent) instead of $E$ and $B$:
\begin{equation*}
y = 0.17077 x + 6.43.
\end{equation*}
This is a \emindex{linear function}, which standard algebra texts write in
\emph{slope-intercept} form
\begin{equation*}
y = m x + b.
\end{equation*}
In this example the {\emindex{slope} $m$ is 
$0.17077 \frac{\$}{\mbox{kwh}}$ and the \emindex{intercept} $b$ is \$6.43. 
For the English bill the slope is
$0.0735 \frac{\mbox{\textsterling}}{\mbox{kwh}}$ and the 
intercept is \textsterling{}9.49.%
\teachertag{}
\begin{teacher}
We usually treat the slope and intercept as given data, since that's
how they appear in the world. Computing the slope as $\Delta y /
\Delta x$ belongs in an algebra class that's leading to calculus, but
not here. 
\end{teacher}

There are many everyday examples where a
linear equation captures what's happening: a total is computed by
adding a varying part that's a proportion 
to a constant. The proportionality constant is the slope.

\begin{itemize}

\item 
The most familiar examples are the ones where the intercept is 0:
all the ones in \sref*{directproportion}.

\item When renting a truck, the amount you pay is 
\begin{center}
(rate in dollars/mile) $\times$ (miles driven) + (fixed charge) .
\end{center}

\item Your monthly cell phone bill might be
\begin{center}
(rate in dollars per minute) $\times$ (number of minutes) +
(fixed fee) .
\end{center}

(A real cell phone bill will \myindex{probably} be more complicated,
perhaps with separate charges for phone minutes, text messages and
data transfer, perhaps with some of each kind of use built in to the
fixed fee.) 

\item If you work as a salesperson and your commission is 15\%  of
total sales 
your total wages are
\begin{center}
0.15 $\times$ (total sales) + (your base salary) .
\end{center}
\end{itemize}

The pattern is
\begin{center}
total = (rate) $\times$ (amount of some quantity) + (fixed constant).
\end{center}

In each case the slope is the rate and the intercept is the fixed
constant.  The units of a slope are always those of a rate.
In the truck example, the slope is the rate in dollars per mile; in
the cell phone example, the units of the slope are dollars per minute; 
in the salesperson example, the slope is 0.15 dollars of commission per
dollar of total sales. 

Think of the intercept as an initial or starting
value; it's what happens when the input is zero.  It has proper units
too --- in each of these examples that unit is dollars.
If you rent a truck but don't drive it anywhere, you still pay the fixed
charge. If you make no cell phone calls you still pay the
fixed fee. If you don't sell anything in a month, your commission is
\$0 but you still earn your base salary.   


\qrsection[linearexcel]{Linear functions in Excel}

In \sref*{tamworth} we saw that the amount you pay for
electricity in a month is a linear function of the amount you use.
In this section we'll use Excel to calculate electricity bills and to
draw a picture of the results.

Figure~\ref{fig:TamworthElectricScreenShot1} is a screen shot of the 
Excel spreadsheet \link{TamworthElectric.xlsx}. 
We put the slope 0.735 in
cell \cell{C4}, with its units  {\textsterling}/kWh in cell \cell{D4}.  We put the intercept  9.49 in cell
\cell{C5} and the units  (\textsterling) in cell \cell{D5}.

\figfile{TamworthElectricScreenShot1.png}
\begin{figure}[ht]
\centering
  \includegraphics[height=60mm]{\thefigurefilename}
\caption{Tamworth electricity bill}
\figsource{Excel screen shot}
\label{fig:TamworthElectricScreenShot1}
\end{figure}
\figfile{}

Then we entered column labels in cells \cell{B7:C8} and
a few values in rows \cell{9} through 
\cell{13}  in column \cell{B}. Finally, we asked Excel to calculate
the electricity bills in column \cell{C}. To do that, we started with 
the formula
\displayexcel{
%	=C\$4*B9+C\$5
	=C4*B9+C5
}
\noindent in cell \cell{C9}. The \excel{=} sign tells Excel to
multiply the numbers in cells \cell{C4} and \cell{B9} and add the
number in \cell{C5}. The result is \excel{9.49}, as expected.

The next step is to copy that formula from \cell*{C9} to
\cell*{C10}. With any luck Excel will guess what we want to do, change
\excel{B9} to  
\excel{B10}, compute
\displayexcel{
%	=C\$4*B9+C\$5
	=C4*B10+C5
}
\noindent
and display \excel{44.11}. But that's not what happens! Excel
shows \excel{4469.79} instead! If you look at the contents of
\cell*{C10} you will find
\displayexcel{
	=C5*B10+C6
}
so Excel added 1 to the row numbers for cells \cell{C4} and \cell{C5}
as well as to \cell{B9}. 
There's nothing in \cell*{C6}. Excel treats
that as a zero and adds it to $471 \times 9.49$ to get $4469.79$.

That's not what we want. Changing \cell{B9}
to \cell{B10} is right, but we want Excel to leave the
references to cells \cell{C5} and \cell{C6} alone. The trick
that makes that happen is to put a \cell{\$} in front of the \cell{5}
and the \cell{6}. This is not something you could have figured
out. There's no particular reason why this trick should work. You
can't figure it out. Just remember it. The right formula to use in
\cell*{C9} is 
%
\displayexcel*{
	=C{\textbf\$}4*B9+C\textbf{\$}5
}
\noindent
When we copy that formula from row \cell{9} to rows
\cell{10:13} we get Figure~\ref{fig:TamworthElectricScreenShot1}.

Figure~\ref{fig:TamworthElectricScreenShot2} is a screen shot
of the same spreadsheet --- after we asked Excel to show the
formulas for each cell instead of the values. 

\figfile{TamworthElectricScreenShot2.png}
\begin{figure}[ht]
\centering
  \includegraphics[height=60mm]{\thefigurefilename}
\caption{Tamworth electricity bill --- formulas}
\figsource{Excel screen shot}
\figcomment{Need to recapture at higher resolution.}
\label{fig:TamworthElectricScreenShot2}
\end{figure}
\figfile{}

In Chapter~\ref{IncomeDistribution} we learned how to use Excel to
draw bar charts and histograms so that we could visualize data
organized into categories. The $x$-axis displayed category names, with
corresponding values on the $y$-axis. That won't work for the data in
Figure~\ref{fig:TamworthElectricScreenShot1}, since there both the
$x$- and $y$-axes have numerical values. Instead, after selecting
cells \cell{B9:C13} we must ask Excel for a chart of
type\index{chart type} \excel{XY(Scatter)}\index{scatter
  chart}. Figure~\ref{fig:linearGraph} shows the result. 

The graph is a straight line --- 
that's why the function is called ``linear''. The slope tells us how
steep the line is and the intercept tells us where it crosses the
vertical axis --- in this case at the value 
\textsterling{}9.49, the total bill when you use no electricity at
all.
\teachertag{}
\begin{teacher}
Since the entries in the Tamworth Electricity Bill table are out of
order, Excel has drawn some of the segments in the graph twice. You can
see that if you look carefully. You might want to point this out, or not.
\end{teacher}

\figfile{TamworthElectricBillExcel.png}
\begin{figure}[ht]
\centering
  \includegraphics[height=80mm]{\thefigurefilename}
\caption{Tamworth electricity bill}
\figsource{REDRAW hires. Chart from Excel spreadsheet we built.}
\label{fig:linearGraph}
\end{figure}
\figfile{}

Excel will let you change the type of a chart once it's built. If you
change the chart in Figure~\ref{fig:linearGraph} to a 
\excel{Line Chart}\index{line chart} Excel 
will use the data in the column \cell{B} as category labels rather
than as the numbers of kilowatt-hours. 
It will space them evenly along the $x$-axis, whatever their
values, and draw the nonsense you see in Figure~\ref{nonsense}.

\figfile{TamworthElectricBillLineChart.png}
\begin{figure}[ht]
\centering
  \includegraphics[height=80mm]{\thefigurefilename}
\caption{How NOT to draw a line}
\label{nonsense}
\figsource{REDRAW hires. Chart from Excel spreadsheet we built.}
\end{figure}
\figfile{}

If you change the chart type to scatter you get two
scatters, one for each column. You can get 
Figure~\ref{fig:linearGraph} only if you start with a scatter plot.
If you select the two columns of data and build a line chart first,
things are even worse. 
Excel thinks each row is a category for which you have two
pieces of data. It labels the categories 1, 2, \ldots and shows a line 
for each.
Try this and see what happens.

\qrsection[comparelinear]{Which truck to rent?}

Table~\ref{table:TruckRental} shows the cost of renting a truck for
one day in Boston in March of 2015. Four companies offer equivalent
models. Which one should you choose?

\begin{center}
\begin{tabular}{cS[table-format=2.0]S[table-format=2.2]S[table-format=2.2]S[table-format=2.2]}
\toprule
           & {Watertown} & {U-Haul} & {Budget} & {Enterprise} \\
\midrule
fixed cost & 79 & 29.95 & 29.95 & 59.99 \\
\$/mile & 0 & 1.39 & 0.99 & 0.59 \\
\bottomrule
\end{tabular}
\captionof{table}{One day truck rental costs}
\label{table:TruckRental}
\end{center}

It's clear from the numbers that for a very few miles Budget will be
cheapest, since it's tied with U-Haul for the the lowest fixed cost
and charges less per mile. For a really long move Watertown will be
best since there is no mileage charge. 
The Excel chart in
Figure~\ref{fig:TruckRental} tells the whole story. Budget is cheapest
up to about 50 miles. For longer trips, choose Watertown. It  never
makes financial sense to rent from U-Haul or Enterprise.

The figure is a good reminder of the meaning of slope and intercept
for straight line graphs. A line's intercept is where it crosses the
vertical axis. In this example intercepts represent the fixed
costs. The units of the intercept are dollars --- the units on the
$y$-axis. A line's slope measures how steep it is.
That's particularly visible when you compare U-Haul
and Budget, which have the same intercept but different slopes ---
1.39 dollars/mile and 0.99 dollars/mile. The units of the slope are
always (units on $y$-axis)/(units on $x$-axis),
The line for Watertown is horizontal since its slope is 0 dollars/mile. 

\figfile{TruckRentalCropped.pdf}
\begin{figure}[ht]
\centering
  \includegraphics[height=3in]{\thefigurefilename}
\caption{Comparing truck rental costs}
\figsource{Chart from an Excel spreadsheet we built.}
\label{fig:TruckRentalScreenshot}
\end{figure}
\figfile{}

The spreadsheet with that figure is at
\link{TruckRental.xlsx}. Figure~\ref{fig:TruckRentalScreenshot}
shows how we arranged the formulas in the spreadsheet to compute the
total cost for each company.
Column \cell{A} lists the mileages we're considering, from
0 in \cell*{A11} to 100. The formula \excel{=A11+10} in
\cell*{A12} fills column \cell{A} when we copy it to cells
\cell{A13:A21}. The formula in \cell*{B11} is
\displayexcel*{
=B\$7+\$A11*B\$8
}
It uses three \excel{\$} signs to keep Excel from adjusting
references for rows \cell{7} and \cell{8} and for column
\cell{A}. That allowed us to copy it to all of the range
\cell{B11:E21}.

\figfile{TruckRentalscreenshot.png}
\begin{figure}[ht]
\centering
  \includegraphics[height=2in]{\thefigurefilename}
\caption{Computing truck rental costs}
\figsource{Screen shot from Excel spreadsheet we built.}
\label{fig:TruckRentalscreenshot}
\end{figure}

The problem we've just solved is typical of situations where you have
to decide among options, some with small startup cost but a
high ongoing rate, others the reverse. Here are some examples:

\begin{itemize}
\item Insulate your house (high initial investment compared to doing
nothing) in order to pay less for heat in the winter (lower rate for
use).

\item Buy a hybrid instead of a conventional car (higher initial cost,
lower rate of fuel consumption).

\item Buy energy efficient light bulbs (more expensive to start with,
but they use less electricity to run),

\item Select a phone plan with unlimited text messaging (more
expensive than pay-as-you-text, but the slope (\$ per text) is zero).
\end{itemize}

Each of these can be thought of as looking at linear equations to
see where their graphs cross. You can do that by building a table of
values  or by drawing the graphs (in Excel, or with pencil
and paper) or by writing down the equations and solving them with
remembered algebra, or by \myindex{guess-and-check}. 

But you can't rely on just this mathematics to make a decision. There
are always other important things to think about. How long does the
more expensive purchase last? Can you afford the initial high payment?
If so, what else might you rather do with that money? Do you need to
take depreciation or inflation into account?


\qrsection[energypower]{Energy and power}

How much electrical energy does a 100 watt light bulb use? That
depends on how long it's on. When it's switched off, it doesn't
use any at all. If it's on for two hours it must use twice as much as
it does in one hour. 
Figure~\ref{fig:TamworthPower} (from the second page of the Tamworth
bill) shows the 
proportion lurking there.

\begin{figure}[ht]
\centering
  \includegraphics[height=20mm]{\here/TamworthPowercropped.pdf}
\caption{How much electrical energy?}
\figsource{From the Tamworth .doc file.}
\label{fig:TamworthPower}
\end{figure}

The second line in that figure displays the units for the
quantities in the first line. Time is measured in hours,
of course. Power is measured in kilowatts. Energy is
measured in kilowatt-hours: the product of the units for power
and for time.

That tells us right away that energy and power are not the
same thing. Comparing the figure to Equation~\ref{eq:rtd}, you can see
that energy is like distance --- a thing that's consumed or
traveled, while power is like speed --- the rate at
which energy is used or distance covered.%
\teachertag{}
\begin{teacher}
We're undecided about how much time to spend on power and energy.
It takes a lot of teaching time (and energy) to
convey the distinction convincingly in class. Perhaps those resources
are better spent on other parts of the curriculum. But the topic is
important and ultimately interesting, because students care about
climate change and alternative power sources. We have included
problems that explore the issue further.

If you do choose to spend more time on the difference between power
and energy, and the confusion because the name ``watt-hour'' contains
a unit of time, consider discussing the light-year, which is a measure
of distance, not time. So ``light-years ago'' is never right. The
knot --- one nautical mile per hour --- is also a rate, like power, that
doesn't mention time. 
\end{teacher}

If you turn a 100 watt light bulb on for 2 hours the
formula in Figure~\ref{fig:TamworthPower} tells you how much
electrical energy you use:
\begin{align}
100 \text{ watts} \times 2 \text{ hours } 
& = 200 \text{ watt-hours} \label{eq:energypower1}\\
& = 0.2 \text{ kilowatt-hours} \label{eq:energypower2}
\end{align}
(\ref{eq:energypower1}) is just multiplication.
(\ref{eq:energypower2}) changes watt-hours to kilowatt-hours.

How much does it cost to leave a 40 watt bulb on all the time in
your basement for a year? There are about 9,000 hours in a year. That
is 9 kilo-hours, so you'll use about
%
\begin{equation*}
40 \text{ watts} \times 9 \text{ kilo-hours} = 360 \text{ kilowatt-hours}.
\end{equation*}
If you pay \$0.10 per kWh for electricity it will cost you about \$36 per
year to guarantee that you don't fall down the basement steps in the dark.

The electrical energy that flows through the wires in your house to
your appliances \myindex{probably} comes to you from a power plant, which
might be burning coal or natural gas or extracting the energy from nuclear
fuel. (You might have a wind turbine in your neighborhood, or a hot
spring, or solar panels on your roof, but these are unlikely power
sources for most people.)\index{solar energy}
So power plants produce energy, not power. The power of a power
plant is the rate at which it can produce energy, so
 ``energy plant'' would be a better name than 
``power plant''.

The website for Chicago's Cook Nuclear Plant says that
\begin{qwrap}
\begin{quotation}
\firstline{The 1,048 net megawatt (MW) Unit 1 and 1,107 net MW Unit 2}
  combined produce enough electricity for more than one and one half
  million average homes.%
%\webref{www.cookinfo.com/cookplant.htm}
\webref{www.cookinfo.com/cookplant.htm}
\end{quotation}
\sourceinfo[420]{www.cookinfo.com/cookplant.htm}
\end{qwrap}

Let's check this. The combined total power is 2,155 megawatts.  This
is the rate at which that plant produces electrical energy when
it is running at full power. (When it's not running it's still just as
powerful, but not producing any energy.) When it's running, how many
average homes could it produce electricity for?

If the Cook plant ran all year (about 9,000 hours) it would produce
\begin{align*}
2,155 \text{ megawatts} \times 9,000 \text{ hours}
& \approx 18,000,000 \text{ megawatt-hours} \\
& = 18,000,000,000 \text{ kilowatt-hours}
\end{align*}
of electrical energy.
Googling ``average household electricity usage'' finds
\begin{quotation}
 6,000 kWh per household per year for 3 residents average per
household
\end{quotation}
from
\url{www.physics.uci.edu/~silverma/actions/HouseholdEnergy.html}.
The source is a physics professor's web site, so it's \myindex{probably}
reliable. 
At 6,000 kWh per household per year Cook could power 3 million
homes. The quotation claims half that, so it's clearly in the right
ballpark. The 6,000 kWh per household per year is a southern
California average --- households in northern Illinois might well use
more electricity.

Energy comes in many forms besides electric. The Cook plant
converts the energy in its nuclear fuel to electricity. 
Driving a car uses the energy stored in the gasoline.
Running a \myindex{marathon} uses the energy in the food you eat.
Each form of energy has its own units. We've seen that electrical
energy is measured in kilowatt-hours.
If you cook on a gas stove, the energy in the gas is measured in
\emph{therms}.\index{therm} The energy in the oil that heats  
your house is measured in \emindex{British Thermal Units}
\index{BTU} or BTUs. The energy in the
food you eat is measured in \emph{calories}\index{calorie}. Physicists
measure energy in \emph{ergs} or \emph{joules}; you rarely see those
units in everyday life. You can look up conversion factors for these
units --- for example, the energy in a barrel (42 gallons) of oil is
about 5.8 million BTU, which is
equivalent to 1700 kilowatt-hours. So it would take about a fifth of a
barrel to keep that 40 watt light bulb burning for a year.

Converting among the units for energy is just like converting among
the units for length (meters, feet, yard, miles, \ldots). You can use
a table, an online calculator like the one at the National Institute
of Standards and Technology
(\url{physics.nist.gov/cuu/Constants/energy.html}) or the
Google calculator.

Possibly the most interesting energy conversion is the one that
Einstein discovered in 1905: mass and energy are the same thing,
measured in different units. The conversion factor is the square of
the speed of light --- hence the famous equation
\begin{equation*}
e = mc^2.
\end{equation*}

To see that at work, look again at the yearly energy output of the
Cook plant. The Google calculator tells us that
\gc*{
18 000 000 000 kilowatt hours = $6.48 \times 10^{16}$ joules
}
The National Institute of Standards and Technology website says
that corresponds to a mass of about 0.72 kg, which is 720
grams. That means just about 1.6 pounds of matter must be converted to
energy to power millions of Chicago homes for a year.
The Google calculator does not know Einstein's equation, so
it wouldn't convert kwh to grams directly!

\qrsection[taxes]{Taxes: sales, social security, income}

Taxes are a part of life (the only other certainty is death),
so it's only common sense to learn how they work. In \sref*{salestax}
and \sref*{directproportion} we studied sales taxes. They are
collected by cities and states and are computed
as a percentage of the purchase price. In this section we'll explain
two important federal taxes that depend on your income, not on how you
spend it. 

The first of these is the FICA tax.%
\index{FICA| see {social security}}
``FICA'' is the acronym for the ``Federal Insurance Contributions
Act''. Those taxes pay for \myindex{social security} and
\myindex{medicare}.
In 2014 the tax rate was 6.2\% for social security and 1.45\% for
medicare. The social security tax is collected only on
the first \$117,000 of your earnings. Up to that income level the
combined rate is 7.65\%.
(The actual rules are a little more
complicated. First, the tax applies only to wages. Other income (like
stock dividends or interest) are not subject to this tax. Second, the
real rates are twice the quoted amounts, but  your employer is
required to pay half. If you're self-employed you pay it all.
)

Here are some sample computations.
\begin{itemize*}
\item If you earn \$1,000 your FICA tax is $0.0765 \times \$1,000 =
\$76.50$.
\item If you earn \$50,000 your FICA tax is $0.0765 \times \$50,000 =
\$3,825$.
\item If you earn \$117,000 your FICA tax is $0.0745 \times \$117,000
= \$8,950.50$.
\item If you earn \$500,000 your FICA tax is 
%
\begin{equation*}
\text{social security + medicare} = 0.062 \times \$117,000 + 0.0145
\times \$500,000
= \$14,504.
\end{equation*}
%
\end{itemize*}

Once you make more than \$117,000
the amount of social security tax remains constant; the medicare tax
continues at the rate of 1.45\%.
That means the percentage of your earnings collected for FICA taxes
decreases as your earnings increase. That percentage is
called the \emindex{effective tax rate}. 
Up to \$117,000 the effective rate is 7.45\%.
For \$500,000 the effective rate is just
$\$14,504/\$500,000 = 2.9\%$. For higher incomes, the effective rate
is even smaller.
Because they decreases as earnings increase, social security taxes are
\emph{regressive}\index{regressive tax}. Once you reach the social
security maximum your effective tax rate
decreases. So the more you earn, the smaller a percentage you pay.

Figure~\ref{fig:socialsecurity} is a
screenshot of the spreadsheet 
\link{SocialSecurityTax.xlsx} showing a data table and a chart
illustrating social security tax computations. The formula bar shows
how Excel computes the tax on \$500,000.

\figfile{SocialSecurityTax.png}
\begin{figure}[ht]
\centering
  \includegraphics[height=60mm]{\thefigurefilename}
\caption{FICA (Social security and medicare) tax}
\figsource{REDRAW hires. Excel spreadsheet screen shot}
\label{fig:socialsecurity}
\end{figure}
\figfile[}

Income tax is a little more complicated. It's a \emph{progressive
graduated tax}\index{progressive tax}\index{graduated tax}.
When you make more money you not only pay more tax, some of your
income may be taxed at a higher rate. 
Table~\ref{table:2014taxtable} shows the 2014 \emph{tax brackets}
for single taxpayers. 

\begin{table}[ht]
\centering
\begin{tabular}{r@{\ --\ }lS[table-format=2.1]}
\toprule
\multicolumn{2}{c}{Bracket (\$)} & {Marginal Tax Rate (\%)} \\ 
\midrule
0  &  9,075 & 10 \\
9,075  &  36,900 & 15 \\
36,900  &  89,350  & 25 \\
89,350  &  186,350 & 28 \\
186,350  &  405,100  & 33 \\
405,100 & 406,750 & 35  \\
\multicolumn{2}{c}{406,750+} & 39.6 \\
\bottomrule
\end{tabular}
\caption{2014 single taxpayer brackets and rates}
\tablesource{Public information.}
\label{table:2014taxtable}
\end{table}

That table tells you that the first \$9,075 of your income is taxed at
10\%. If you make exactly that much, you pay \$907.50 in
tax. If you make more, the extra income is taxed at a higher rate ---
you have moved to a higher tax bracket\index{tax bracket}.
For example, if you make between \$9,075 and \$36,900 you will pay 15\% of
the amount you earn over \$9,075. If you earn more than
\$36,900 you start paying at a 25\% rate on the extra.

Let's try an example.  If you earn \$50,000, then your total tax is
\begin{align}\label{eq:fiftyK}
\text{total tax} & =  0.10 \times \$9,075
+ 0.15 \times (\$36,900 - \$9,075) 
+ 0.25 \times (\$50,000 - \$36,900) \notag \\
& =  \$907.50 + \$4,173.75 + \$3,275.00  \\
& =  \$8,356.25 \ . \notag
\end{align}

Note carefully that when you are in a higher tax bracket the higher
rate applies only to the extra income. The taxpayer in this
example is in the 25\% bracket, but that rate applies only to her
earnings in that bracket.
Figure~\ref{fig:2014taxtable} explains this rule in another way.
\teachertag{}
\begin{teacher}
In the tax rate brouhaha in the 2008 election we recall reading a
story about a dentist who complained that he would need to be careful
not to let his income exceed \$250,000 --- where candidate Obama drew a
no-new-taxes line --- lest his overall tax rate increase. If you find
the story let us know and we'll turn it into an exercise.
\end{teacher}

\figfile{2014taxbrackets.png}
\begin{figure}[ht]
\centering
  \includegraphics[height=60mm]{\thefigurefilename}
\caption{2014 single taxpayer tax calculation}
\figsource{CHECK SIZE www.forbes.com/sites/kellyphillipserb/2013/10/31/irs-announces-2014-tax-brackets-standard-deduction-amounts-and-more/}
\label{fig:2014taxtable}
\end{figure}
\figfile{}

The graphs in Figure~\ref{fig:taxgraphs}
show that the dependence of tax on income is \emindex{piecewise linear}
--- built from pieces of straight lines that become steeper as income
increases.  You can find these graphs in the Excel spreadsheet
\link{GraduatedTax.xlsx}. 

\figfile{GraduatedTaxcropped.pdf}
\begin{figure}[ht]
\centering
  \includegraphics[height=60mm]{\thefigurefilename}
\caption{2014 single taxpayer tax liability}
\figsource{Charts from an Excel spreadsheet we built.}
\label{fig:taxgraphs}
\end{figure}
\figfile{}

Figure~\ref{fig:effectivetax} shows the \myindex{effective tax rate}
--- the percentage of your income you pay in Federal income tax. If you
earn less than  \$9,075 you are in the lowest bracket and your
effective tax rate is the rate in that bracket: 10\%. If you earn
\$50,000 you are in the third bracket. You pay 10\% of the first
\$9,075, 15\% of the part in the second bracket, and 25\% of the rest.
The arithmetic in (\ref{eq:fiftyK}) shows that your total tax
is \$8,356.25. Your effective tax rate is 
$\$8,356.25 / \$50,000 = 16.71\%$. 
This is a weighted average of the three bracket rates
10\%, 15\% and 25\%, with weights the amount of income taxed at each rate.
Even if you make \$500,000, which puts you in the top 39.6\% bracket, your
effective tax rate is still only about 31\%, because you pay at a
lower rate on the first part of your income.

\figfile{EffectiveTaxcropped.pdf}
\begin{figure}[ht]
\centering
  \includegraphics[height=60mm]{\thefigurefilename}
\caption{2014 effective tax rates}
\figsource{Chart from an Excel spreadsheet we built.}
\label{fig:effectivetax}
\end{figure}
\figfile{}

\exstart

\begin{exx}{\untested\sref{tamworth}\gref{linearfunctions}
\gref{slopeintercept}}Your electricity bill.

Verify the computations on your current electricity bill, either with
a calculator or by modifying the Tamworth bill spreadsheet at
\link{TamworthElectric.xlsx}.

If you don't have a current electricity bill, check the website for
your local electric company, which \myindex{probably} provides a
sample bill you can use instead.

\end{exx}

\begin{exx}{\untested\sref{tamworth}\gref{adjustforinflation}
\gref{linearfunctions}}
Electricity costs now and then and here and there.

Compare residential electricity cost in Boston in 2007
(the date of the NStar bill in Figure~\ref{nstarbill}) to the cost
where you live today.  Your answer should take inflation into account.

If you have a current electricity bill, use it.
If you don't, try to find the fixed monthly cost and the cost of
electricity in \$/kWh from your local electric company. Perhaps their
website has that information. 

\end{exx}

\begin{exx}{\routine\hassolution\sref{tamworth}\gref{linearfunctions}
\gref{slopeintercept}}
How much electricity does it use \ldots ?

The document \link{how\_to\_read\_electricity\_bills.doc} containing
the Tamworth bill has several questions about how much electricity
various appliances use. Answer them.

\begin{sol}
\begin{enumerate}

\item Electricity consumed
\begin{abcd}

\item 	100W lamp on for 2 hours.
\begin{equation*}
100 \text{W} \times 2 \text{ hours} \times \frac{\text{kilo}}{1000} = 
0.2 \text{ kilowatt-hours}.
\end{equation*}

\item 500W TV on for 5 hours.

\begin{equation*}
500 \text{W} \times 5 \text{ hours} \times \frac{\text{kilo}}{1000} = 
2.5 \text{ kilowatt-hours}.
\end{equation*}

\item 2kW kettle on for half an hour.

1 kilowatt-hour. 

\item 10W electric blanket on for 15 minutes.
\begin{equation*}
10 \text{W} \times 0.25 \text{ hours} \times \frac{\text{kilo}}{1000} = 
0.0025 \text{ kilowatt-hours}.
\end{equation*}

\end{abcd}

\item Comparisons

\begin{abcd}
\item 2kW heater for 2 hours or 3 kW heater on for 3 hours.

Clearly 4kW hours is less than 9 kW hours. 

\item 900W toaster for 15 minutes or 2kW grill for 10 minutes.

It's easier to do this one if I leave the times in minutes.

$0.9 \times 15 = 1.35$ kW-minutes is less than
$2 \times 10 = 20$  kW-minutes.

\item 100W radio for 2 hours or 500W radio for 45 minutes.

I'll do this one in minutes too, and use watts rather than kilowatts.

$100 \times 120 = 12,000$ watt-minutes is less than 
$500 \times 45 = 22,500$ watt-minutes.


\item Which appliance would be cheaper to use in each case?

In each case the first one is cheaper - the amount of electricity is
about half.

\end{abcd}

\end{enumerate}
\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{tamworth}
\gref{directproportion}\gref{energyandpower}}
Direct current.

In an article from \theTimes{} on November 17, 2011 headlined
``From Edison's Trunk, Direct Current Gets Another Look''
\index{Edison, Thomas Alva}
you can read that

\begin{qwrap}
\begin{quotation}
In a data center redesigned to use more direct current, monthly
utility bills can be cut by 10 to 20 percent, according to Trent
Waterhouse, vice president of marketing for power electronics at
General Electric. Verizon Communications, a G.E. customer, expects to
save 1 billion kilowatt-hours a year from a nationwide retrofit of its
data centers, which translates to roughly enough to power 77,000
homes.%
\webref{www.nytimes.com/2011/11/18/business/energy-environment/direct-current-technology-gets-another-look.html}
\end{quotation}
\sourceinfo{www.nytimes.com/2011/11/18/business/energy-environment/direct-current-technology-gets-another-look.html}
\end{qwrap}

\begin{abcd}
\item How many watt-hours is a billion kilowatt-hours?

\item About how much electricity is Verizon using now, given that they hope
to save a billion kilowatt-hours per year?

\item Verify the estimate that those billion kWh are ``roughly enough
to power 77,000 homes''. (You won't find anything useful in the
original article. How else will you search?)

\end{abcd}

\begin{sol}
\begin{abcd}
\item How many watt-hours is a billion kilowatt-hours?

Since ``kilo'' means ``multiply by 1000'', a billion kilowatt-hours is
(a billion times a thousand) watt-hours. That's a trillion
watt-hours, or $10^{12}$ watt-hours, or a Terawatt-hour. If you have
to write the zeroes, it's 1,000,000,000,000 watt-hours.

\item About how much electricity is Verizon using now, given that they hope
to save a billion kilowatt-hours per year?

If the billion kWh saved were 10\% of what they are using now, that
would mean they were using 10 billion kWh now. If the billion kWh were
20\% of current use, that would make current use 5 billion kWh. So
they are using between 5 and 10 billion kWh now.

\item Verify the estimate that those billion kWh are ``roughly enough
to power 77,000 homes''. (You won't find anything useful in the
original article. How else will you search?)

The website 
\url{www.eia.gov/tools/faqs/faq.cfm?id=97&t=3}
says that

\begin{quotation}
In 2010, the average annual electricity consumption for a
U.S. residential utility customer was 11,496 kWh, an average of 958
kilowatthours (kWh) per month. Tennessee had the highest annual
consumption at 16,716 kWh and Maine the lowest at 6,252 kWh. 
\end{quotation}

Using the figures in the original quotation
\begin{equation*}
\frac{ 1 \text{ billion kWh}}{77,000 \text{ homes}}
= 12,987.013 
\frac{\text{kWh}}{\text{home}}
\end{equation*}
which is close enough to the government's figure of 11,496 kWh/home.

\end{abcd}
\end{sol}

\end{exx}

\begin{exx}{\untested\sref{linearfunction}\gref{linearfunctions}}
How hot was it?

Figure~\ref{fig:weather} shows a weather forecast for Hamilton,
Ontario, Canada. The temperature there is displayed in degrees Celsius,
marked ${}^{\circ}\text{C}$. 

\figfile{HamiltonWeather.png}
\begin{figure}[ht]
\centering
\framebox{
  \includegraphics[height=60mm]{\thefigurefilename}
}
\caption{Weather in Canada}
\figsource{RECAPTURE hires Screen capture from
www.theweathernetwork.com/weather/caon0289/}
\label{fig:weather}
\end{figure}
\figfile{}

\begin{abcd}
\item Look up the linear relationship for converting temperatures
measured on the Celsius scale to temperatures on the Fahrenheit scale.%

\item Use the formula you found to calculate the 
temperature in degrees Fahrenheit  at the Hamilton airport 
on Friday, July 27, 2012 at 14:00 EDT.

\item Check your answer using the Google calculator.

\item What does ``Wind: E 17 km/h'' in the figure mean?

\item What does ``Pressure: 101.12 kPa'' in the figure mean?

\end{abcd}
\end{exx}

\begin{exx}{\hassolution\sref{linearfunction}\gref{linearfunctions}
\gref{slopeintercept}\gref{functions}} 
The \myindex{Jollity Building}.

In \exref{liebling} there's an implicit linear model
for Mr. Ormont's weekly income. Write the linear equation for that
model. Clearly identify the independent and dependent variables and
the units for the slope and the intercept.

\begin{sol}
The clue to finding the linear equation is the phrase
``fifty dollars a week and 2 percent of the total rents.'' I'll let
$W$ stand for Mr. Ormont's weekly income and $R$ for the weekly rent.
Then 
\begin{equation*}	
	W = \$50 + 0.02R.
\end{equation*}
The constant amount (the intercept), is \$50. He gets that much even if
the building is empty. The slope is 0.02. Its units are
\begin{equation*}
\frac{\text{dollars for Mr. Ormont}}
{\text{dollar of weekly rent}}.
\end{equation*}
\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{linearfunction}
\gref{linearfunctions}\gref{slopeintercept}}
Newton trees .

In an article in the April 2012 issue of the \myindex{Newton
Conservators} newsletter you can read that

\begin{qwrap}
\begin{quotation}
\firstline{In the early 1970s there were approximately 40,000 trees}
lining the streets of Newton. Today, that number is about 26,000 --- a
35\% loss. The current annual rate of decline is about 650 trees per
year. At this rate, if unchecked, public street trees would diminish to 
approximately 10,000 within a generation (25 years), and in 40 years,
public street trees would no longer be part of the Newton landscape. 
\webref{%
www.newtonconservators.org/newsletters/apr12.pdf
}
\end{quotation}
\sourceinfo{Newton Conservators' newsletter,
www.newtonconservators.org/newsletters/apr12.pdf 
}
\sourcecomment{I am sure I can get permission for this. I know the
conservators and the arborist --- eb}
\sourcewc{825}
\end{qwrap}

\begin{teacher}
Exponential decay might be a better model than linear for this
exercise on the Newton trees. Consider returning to this exercise when
you reach that topic later in the text. 
\end{teacher}

\begin{abcd}
\item Check the arithmetic that leads to the claimed ``35\% loss''.

\item Check the arithmetic that leads to a ``current annual rate of
650 trees per year''.

\item Check the predictions in the last sentence. Are they 
likely\index{probability}  to
come to pass?

\item Write the equation  for the linear model implicit in this
quotation (use years since 2012 as the independent variable).
Identify the slope and the intercept, with their units.%

\end{abcd}

\begin{sol}

\begin{abcd}
\item Check the arithmetic that leads to the claimed ``35\% loss''.

The 1+ trick tells me $40K \ times 0.65 = 26K$.

\item Check the arithmetic that leads to a `` current annual rate of
650 trees per year''.

If I take ``the early 1970's'' to be 1972 then the interval is 40 years.
\begin{equation*}
\frac{14,000 \text{ trees}}{40 \text{ years}} =
350 \frac{\text{trees}}{\text{years}}.
\end{equation*}
That doesn't match the figure in the article. Perhaps the rate of loss
is larger in more recent years.

\item Check the predictions in the last sentence. Are they 
likely to come to pass?

Using the rate of 650 trees per year, the loss in 25 years would be
$25 \times 650 \approx 16,000$ trees. Subtracting that from the
current 26,000 trees would indeed leave just 10,000.

In another 15 years the loss would be just about all the 10,000
remaining trees.

\item Write the equation  for the linear model implicit in this
quotation (use years since 2012 as the independent variable).
Identify the slope and the intercept, with their units.

The independent variable is $Y$, years since 2012. The dependent
variable is $T$, the number of trees. The equation is
\begin{equation*}
T = 26,000 - 650Y.
\end{equation*}
The intercept is 26,000 trees, the slope is -650 trees/year.
\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\routine\hassolution\artificial\sref{linearexcel}
\gref{linearfunctions}\gref{slopeintercept}\gref{linearexcel}}
Apple time.

Kuipers Family Farm, in Maple Park, IL, charged \$9 for admission to
the apple orchard for the 2013  apple season.  This includes a
$\frac{1}{4}$ peck bag of apples and a hayride to the orchard.
Visitors can pick additional peck bags of apples for \$15 each. (A 
bag of apples in the grocery store usually contains $\frac12$ peck.)  
 
\begin{abcd} 	
\item  If you pick that additional peck of apples, how much do you pay?
\item  If you pick two additional pecks of apples, how much do you pay?
\item  If you just enjoy the free cider, your $\frac14$ peck of apples
  and the sunshine, what do you pay? 
\item  Write a linear function that shows how the total cost of the
farm visit depends on the how many pecks of apples you pick. Identify
the slope and intercept, with their units. 
\item Build an Excel spreadsheet to compute this function and use it
to check the values you worked out by hand above. Include a chart in
your spreadsheet.
\item Use your spreadsheet to calculate your apple cost (in
  \$/peck) when you pick no extra pecks, then when you pick 1, 2 or
  10 extra pecks.
\end{abcd}

\begin{sol}

I found the function first. If $C$ is the total cost of the visit when
you pick $A$ pecks of apples then $C$ depends on $A$ this way:
%
\begin{equation*}
	C = 15 A + 9.00.
\end{equation*}
%
That's a linear function with slope 15~\$/peck and intercept
\$9.00.

I used the function to find the  following answers.

\begin{abcd} 	
\item If you pick that additional peck of apples, how much do you pay?

\$24.00

\item   If you pick two additional pecks of apples, how much do you pay?

\$39.00

\item   If you just enjoy the free cider, your $\frac14$ peck of
  apples and the sunshine, what do you pay? 


\$9.00. Didn't need the formula for this one.

\item  Write a linear function that shows how the total cost of the
farm visit depends on the amount of apples picked. Identify
the slope and intercept, with their units.

Did that first.

\item Build an Excel spreadsheet to compute this function and use it
to check the values you worked out by hand above.

See \slink{U-PickApplesSolution.xlsx}

\item Use your spreadsheet to calculate your apple cost (in
  \$/peck) when you pick no extra pecks, then when you pick 1, 2 or
  10 extra pecks.

The number of pecks of apples you go home with is $A + \frac14$. I
used the spreadsheet to compute $C/(A+ \frac14)$ for $A=0, 1, 2$ and
$10$ and found the cost per peck to be $\$36.00, \$19.20, \$17.33$ and
$\$15.51$. I'm not surprised that the more you pick the closer the
cost per peck is to $\$15$.

\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\hassolution}{\artificial\sref{comparelinear}
\gref{linearfunctions}\gref{linearexcel}} 
Comparing telephone calling plans.\index{telephone bill} 

A cell phone company has introduced a pay-as-you-go price structure,
with three possibilities.

\begin{center}
\begin{tabular}{ccr}
\toprule
	Plan 1 &  \$10 a month & 10 cents per minute \\
	Plan 2 &  \$15 a month  & 7.5 cents per minute \\
	Plan 3 &  \$30 a month & 5 cents per minute \\
\bottomrule
\end{tabular}
\end{center}

\begin{abcd}

\item For each plan, find a linear function that describes
how the total cost for one month depends on the number of minutes
used. 

\item Construct a table in Excel showing the total cost for
one month for each of the three plans. Organize your data this way:

\begin{itemize}
\item Create a sequence of cells in column A for the various possible
numbers of minutes. Label that column. Start with 0 minutes. What's a
good step to use? What's a reasonable place to stop?

\item Use columns \cell{B}, \cell{C} and \cell{D} for each of the
  three plans. The fixed charge  and charge per call should be in
  cells in those columns too, so you can use the same formula
  everywhere in the data table. (That will call for clever use of
  the \excel{\$} to keep Excel from changing row numbers and column
  letters when you don't want it to.)  
\end{itemize}

\item Use Excel to draw one chart showing how the monthly bill
  ($y$-axis) depends on the number of minutes you use the phone
  ($x$-axis) for all three plans.  

\item Write a paragraph explaining to your friend how she should go about
choosing the plan that's best for her. 
\end{abcd}

\begin{hint}
Finding the places
where the lines in your Excel chart cross is the key to the last part
of the problem.
\end{hint}


\begin{sol}

\begin{abcd}

\item For each plan, find a linear function that describes
how the total cost for one month depends on the number of minutes
used. 

Let $B$ represent the monthly cost, in dollars, and $M$ the
number of minutes used. Then the three linear functions are
\begin{align*}
	B & =  0.10 M + 10, \\
	B & =  0.075 M + 15, \\
	B & =  0.05 M + 30
\end{align*}

\item Use Excel to draw one chart showing how the monthly bill
  ($y$-axis) depends on the number of minutes you use the phone
  ($x$-axis) for all three plans.  

See the spreadsheet at \slink{PhoneCostsSolution.xlsx}.

\item 
If you talk for less than 200 minutes, choose Plan 1. If you talk for
more than 600, choose Plan 3. For anything in between, choose Plan
2. You can check my work by looking at the spreadsheet --- the
conclusion is plain from both the table and the chart.

\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{comparelinear}\gref{linearfunctions}
\gref{linearexcel}}
Prepaid phones\index{prepaid phones}.

Until summer 2012, if you wanted an \myindex{iPhone} you needed to
lock into a two-year contract.  Then some mobile companies started
selling the iPhone and letting you choose your own plan, with no contract.
You can use the ideas from this chapter to make a quantitative
comparison (we'll let you decide what other factors, such as paying a
lot up front, matter in your decision).

Virgin Mobile began selling a 16GB iPhone 4S for
\$649.99, with no plan or contract.  They offered a \$55 per month
``unlimited" data plan (in fact it was not unlimited, as once you go
past 2.5GB of data they slowed the phone speed down considerably).  If
you purchased the phone through Sprint, it cost just \$149.99.
However, you had to sign up for a two-year contract.  The least
expensive option was \$79.99 with what they called unlimited data. 

\begin{abcd*}
\item  For each plan, find a linear function that shows how the total cost of
the phone depends on the number of months you have it.

\item  Build a spreadsheet in Excel using your functions and fill in
  the cost of the two different options over several months. 

\item  Graph the data in your spreadsheet.  

\item  Write a short statement comparing the two plans.  Clearly the
  Virgin Mobile plan is more expensive at first.  When does it become
  the less expensive plan?  If you had to choose one of the plans,
  which one would you choose and why?  What other factors would you
  consider? 
\end{abcd*}

\begin{sol}
The answers to all the questions are in the spreadsheet at
\slink{PrepaidPhoneSolution.xlsx}.
\end{sol}
\end{exx}


\begin{exx}{\hassolution\sref{comparelinear}\gref{slopeintercept}
\gref{linearexcel}}
Hybrid payback.\index{hybrid}

The ``Best \& Worst Cars 2011'' issue of \emph{Consumer Reports}
provides the following data for new Toyota Camrys:

\begin{center}
\begin{tabular}{ccc}
 & conventional & hybrid \\
\midrule
cost & \$19,720 & \$26,575 \\
fuel economy & 26 MPG & 34 MPG \\
\bottomrule
\end{tabular}
\end{center}

Assume gasoline costs \$3.50/gallon.

\begin{abcd}
\item Questions about the conventional Camry. 

\begin{enumerate}[(i)]

\item
Once you own the car, how much does it cost to run, in 
dollars per mile? Does your answer make sense?

\item
Calculate the total cost (purchase plus gasoline) to drive the
conventional Camry 10,000 miles.

\item
Write the linear equation that computes the total cost $C$ of
driving the conventional Camry $M$ miles.

\item
Identify the slope and the intercept of this equation, with
their units. 
\end{enumerate}

\item Open the spreadsheet \link{ConventionalvsHybrid.xlsx}. Enter the
numerical data from the table and the cost of
gasoline in the appropriate cells. What formula should you enter in cell
\cell{C15} to check your answer to part (i)? 

\item Copy your formula to cells \cell{B14:D29} to fill in the
table. Where must you add \$ signs to keep Excel from
changing row and column references?

\suspend{abcd}

Create a properly formatted and labelled chart in Excel showing how
the cost of driving each car depends on the number of miles
driven. Use your graph along with the table to answer the following
questions.

\resume{abcd}

\item If you drive 120,000 miles will you recover in gas savings the
extra initial cost of the hybrid? Write a complete sentence or two and
use appropriate precision for the numbers you use to make your
argument.

\item When will you recover the extra initial cost in gas savings 
if the government (re)instates a \$3,000 tax rebate for hybrid
purchases? 

\item With the original initial costs, how much would the price of
gasoline have to be in order for the breakeven point to occur at
30,000 miles?

\item Restore all the inputs to their original values. Arrange 
  your spreadsheet so that it will print on one page, with the chart
  below the data table. (Use \excel{Print Preview}.) 

\end{abcd}

(If you're thinking of 
There's a lot more that goes into the cost of driving one car or
another (or any car at all) than just these simple computations using
initial cost and miles driven.)

\begin{sol}

\begin{abcd}
\item Questions about the conventional Camry. 

\begin{enumerate}[(i)]

\item Once you own the car, how much does it cost to run, in 
dollars per mile? Does your answer make sense?

The cost in dollars per mile is
\begin{equation*}
3.50 \frac{\$}{\text{gallon}} \times 
\frac{ 1 \text{ gallon}}{26 \text{ miles}}
= 0.1346 \frac{\$}{\text{mile}}.
\end{equation*}

That's about 13 cents per mile, which makes sense to me.

\item Calculate the total cost (purchase plus gasoline) to drive the
conventional Camry 10,000 miles.

\begin{equation*}
\$19,720 + 
0.1346 \frac{\$}{\text{mile}} \times
10,000 \text{ miles} 
=\$21,066.1538 \approx \$21,000.
\end{equation*}

\item Write the linear equation that computes the total cost $C$ of
driving the conventional Camry $M$ miles if gasoline costs
\$3.50/gallon.
\begin{equation*}
	C = 19,720 + 0.1346\ M.
\end{equation*}

\item Identify the slope and the intercept of this equation, with
their units. 

The slope is 0.1346 \emph{dollars per mile}, the intercept
is 19,720 \emph{dollars}.
\end{enumerate}

\item Open the spreadsheet \link{ConventionalvsHybrid.xlsx}. Enter the
data from the table on page 1 and the cost of gasoline in the
appropriate cells. What formula should you enter in cell
\cell{C15} to check your answer to part (i) ? 

\displayexcel{
=C9+C6*B15/C10 
}

\item Copy your formula to cells \cell{B14:D29} to fill in the
table. Indicate here where you put \$ signs in to keep Excel from
changing row and column references.

\displayexcel{
=C\$9+\$C\$6*\$B15/C\$10 
}

Note that I have no \cell{\$} before two of the \excel{C} references so this
formula will do the right thing when I move it to column \cell{D} for
the hybrid.

\suspend{abcd}

Create a properly formatted and labelled chart in Excel showing how
the cost of driving each car depends on the number of miles
driven. Use your graph along with the table to answer the following
questions.

See \slink{ConventionalvsHybridSolution.xlsx}. 

\resume{abcd}

\item If you drive 120,000 miles will you recover in gas savings the
extra initial cost of the hybrid? Write a complete sentence or two and
use appropriate precision for the numbers you use to make your
argument.

No. The total cost for driving the hybrid is about \$39K. That's about
\$3,000 more than the total cost for driving the conventional Camry.

\item When will you recover the extra initial cost in gas savings 
if the government (re)instates a \$3,000 tax rebate for hybrid
purchases? 

The breakeven point is at about 120,000 miles.
According to Excel the costs for the conventional and hybrid are
\$35,873.85 and \$35,927.94 respectively.  These are only about \$50
apart, which is much too much precision for a prediction that far
away.

\item With the original initial costs, how much would the price of
gasoline have to be in order for the breakeven point to occur at
30,000 miles?

By trial and error, changing the gasoline price in the spreadsheet, I
found out that gas would have to cost about \$23 per gallon for
this to happen. 

\end{abcd}

\end{sol}

\end{exx}


\begin{exx}{\hassolution\sref{comparelinear}\gref{slopeintercept}
\gref{linearexcel}}
Contract or not?

Table~\ref{table:cellphoneplans} provides data that appeared in a
story in \theGlobe{} on June 14, 2012 
headlined ``Pay full price for \myindex{iPhone}, avoid contract''.%
\webref{bostonglobe.com/business/2012/06/14/bgcom-techlab/AskcWIPBv1qccvmqIx7DmK/story.html}

\begin{table}[ht]
\centering
\begin{tabular}{lS[table-format=3.2]S[table-format=2.0]S[table-format=4.2]}
\toprule
Plan & {Phone cost (\$)} & {Monthly charge (\$)} & {Two year cost (\$)} \\
\midrule
Cricket Wireless & 499.99 & 55 & 1,819.99 \\
Virgin Mobile &  649 & 30 & 1,369 \\
\bottomrule
\end{tabular}
\caption{Comparing cell phone plans}
\label{table:cellphoneplans}
\end{table}

%\begin{figure}[ht]
%\centering
%\includegraphics[height=20mm]{\here/cellphoneplansCropped.jpg}
%\caption{Comparing cell phone plans}
%\figsource{Scanned image from \theGlobe, June 14 2012, illustrating
%\url{bostonglobe.com/business/2012/06/14/bgcom-techlab/AskcWIPBv1qccvmqIx7DmK/story.html}}
%\figcomment{Couldn't find the graphic on the web. Can be redrawn.}
%\label{fig:cellphoneplans}
%\end{figure}
%
\begin{abcd}

\item How much would it cost (in total) to buy the Cricket phone and
use it for two months?

\item Write an equation for the total cost to buy and use the
Virgin Mobile phone for $M$ months.

\item Identify the slope and the intercept of your equation, with
proper units for each. 

\item Create an Excel spreadsheet and use Excel formulas to complete a
table like this:

\begin{center}
\begin{tabular}{|c|c|c|}
\hline
Months & Cricket & Virgin \\
\hline
0 & & \\
1 & & \\
\dots & & \\
24 & & \\
\hline
\end{tabular} .
\end{center}

\item Check that your spreadsheet produces the answers in the table
for 24 months.

\item Create a properly labelled and formatted chart displaying the
data in your table. 

\item When (in terms of months of use) would it be better to choose
the Cricket phone?

\item Suppose the monthly charge for the Cricket phone was just \$45
while that for the Virgin phone increased to \$35/month. Answer the
previous question with this new data.

\end{abcd}

\begin{sol}
\begin{abcd}

\item How much would it cost (in total) to buy the Cricket phone and
use it for two months?

Cricket phone two months: \$609.99.

\item Write an equation for the total cost to buy and use the
Virgin Mobile phone for $M$ months.
\begin{equation*}
\text{Virgin Cost} = 650 + 30M			
\end{equation*}

\item Identify the slope and the intercept of your equation, with
proper units for each. 

The slope is 30 dollars/month; the intercept is 650 dollars.

\item Create an Excel spreadsheet and use Excel formulas to complete
  the table.

See \slink{cellphonesolution.xlsx}.

\item Check that your spreadsheet produces the answers in the graphic
for 24 months.

It does.

\item Create a properly labelled and formatted chart displaying the
data in your table. 

See \slink{cellphoneplansolution.xlsx}.

\item When (in terms of months of use) would it be better to choose
the Cricket phone?

Cricket is cheaper up to 6 months. Then Virgin is the better buy.

\item Suppose the monthly charge for the Cricket phone was just \$45
while that for the Virgin phone increased to \$35/month. Answer the
previous question with this new data.

Change values in cells \cell{C9} and \cell{D9}. Then graphs cross at
14 months.

\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{energypower}\gref{linearfunctions}
\gref{directproportion}\gref{energyandpower}}
Regenerative braking.
\index{regenerative braking}

When you apply the brakes in a Toyota Prius 
\index{Prius} the car uses some of
the energy of the forward motion to recharge the battery.
Figure~\ref{fig:prius}
shows how the dashboard
displays a little car icon each time
that recharging has collected 50 watt-hours. The figure shows nine 
of those icons.
The image is from 
\url{nudges.wordpress.com/2008/05/11/does-the-prius-fuel-gauge-change-driving-habits/}. The text there is interesting too.

\figfile{prius-dashboard-5.jpg}
\begin{figure}[ht]
\centering
  \includegraphics[height=60mm]{\thefigurefilename}
\caption{The Prius dashboard}
\figsource{\url{nudges.wordpress.com/2008/05/11/does-the-prius-fuel-gauge-change-driving-habits/}}
\figcomment{RECAPTURE hires? We have permission to use this.}
\label{fig:prius}
\end{figure}
\figfile{}

\begin{abcd*}
\item Estimate the energy equivalent of each icon in gallons of gasoline.
\item Estimate the dollar value of that gasoline.
\item Compare your estimate to the dollar value of 50 watt-hours of
electricity in your house (what it would cost to keep a 100 watt bulb
on for half an hour).
\item Discuss the value of the display.
\end{abcd*}

\begin{hint}
For part (a) 
you will have to look up the conversion factors among various forms of
energy in order to convert watt-hours to gallons of gasoline. Part (c)
is about psychology, not quantitative reasoning. Your answer might
begin ``The display is valuable because \ldots On the other hand \ldots''
\end{hint}

\begin{sol}

\begin{abcd*}
\item Estimate the energy equivalent of each icon in gallons of
gasoline.

I tried asking the Google calculator for
\gc{
50 watt-hours in gallons of gasoline
}
hoping to find a quick solution. Google didn't know the conversion
factors, but it did suggest I look at
\url{%
www.uwsp.edu/CNR/wcee/keep/Mod1/Whatis/energyresourcetables.htm
}. There I discovered that
1 kilowatt-hour equals 3413 Btu (British Thermal Units) 
and 
1 gallon of gasoline equals 125,000 Btu.
Then
\begin{equation*}
50 \text{ watt-hours} \times
\frac{3413 \text{ Btu}}{1000 \text{ watt-hours}} \times
\frac{ 1 \text{ gal gasoline}}{125,000 \text{ Btu}}
= 0.0013652 \text{ gal},
\end{equation*}
so each icon is equivalent to just over one tenth of one percent of a
gallon of gasoline. 

\item Estimate the dollar value of that energy.

I'll use \$3 per gallon for the price of gasoline. Then one tenth of
one percent of a gallon will cost about 0.3 cents. Call it half a
penny.

That was a much smaller number than I expected

\item Compare your estimate to the dollar value of 50 watt-hours of
electricity in your house (what it would cost to keep a 100 watt bulb
on for half an hour).

Electricity in my house costs about 10 cents per
kilowatt-hour. So there too 50 watt-hours does indeed cost half a penny!
So I think my answer to the previous part of the problem is probably
right, even though it's surprising.

\item Discuss the value of the display.

Regenerative braking doesn't save much gasoline money at all ---
just pennies for even moderately long trips. So the value of the gas
is negligible. But the display might be valuable. I know from driving
my daughter's Prius on occasion and watching others drive that the
display of the energy efficiency encourages people to drive more
slowly and use less gas. 
\end{abcd*}

\end{sol}
%From: David McCandless <dm@davidmccandless.com>
%To: eb@cs.umb.edu
%Subject: Fwd: using images and data from your web site
%Date: Tue, 31 Aug 2010 10:34:37 +0100
%
%go for it Ethan
%
%Begin forwarded message:
%
%> From: "Rusu, Meredith" <Meredith.Rusu@HARPERCOLLINS.com>
%> Date: 21 July 2010 14:42:53 GMT+01:00
%> To: 'David McCandless' <dm@davidmccandless.com>
%> Subject: FW: using images and data from your web site
%>
%> Hey David.  Received this request for images used your website and
%> wanted to pass along.  Thanks!
%
%> Meredith
%>
%> Meredith Rusu
%> Assistant Publicist
%> Harper Perennial | It Books | Collins Design | Avon
%> HarperCollinsPublishers
%> 10 East 53rd Street
%> New York, NY 10022-5299
%> 212-207-7301
%> meredith.rusu@harpercollins.com
%>
%> -----Original Message-----
%> From: Ethan Bolker [mailto:eb@cs.umb.edu]
%> Sent: Wednesday, July 21, 2010 9:34 AM
%> To: Rusu, Meredith
%> Cc: eb@cs.umb.edu; maura.mast@umb.edu
%> Subject: using images and data from your web site
%>
%> We are writing a college text for teaching quantitative reasoning.
%> What's your policy on including a few images from
%> www.informationisbeautiful.net
%> , with attribution, of course?
%>
%> At the moment we have a draft for our students. I don't yet have a
%> commercial publisher.
%>
%> Thanks
%> Ethan Bolker


\end{exx}

\begin{exx}{\hassolution\sref{energypower}
\gref{directproportion}\gref{energyandpower}}
Computers don't sleep soundly.

The website 
\url{michaelbluejay.com/electricity/computers.html} gives
information about how much energy a computer uses while asleep, in
standby mode, or in use.  The iMac G5, for example, uses 97 watts
while ``doing nothing'', compared to 3.5 watts while asleep.   
\begin{abcd}

\item  What do you think ``doing nothing'' means?

\item  If this type of computer is doing nothing all day (24 hours),
how much electricity does it use?  Express your answer in
kilowatt hours.

\item  Now suppose the computer goes to sleep after 15 minutes of
doing nothing. How much electricity does it use in an idle day?

\item If a kilowatt hour of electricity costs 20 cents, how much
money is saved in a day because the computer is smart enough to go to
sleep?

\end{abcd}

\begin{sol}

\begin{abcd}
\item  What do you think ``doing nothing'' means?

According to the quote, ``doing nothing'' can't really be doing
nothing, since it uses 97 watts. That's probably running the screen
saver program, and checking from time to time for email or updates to
various websites --- things the computer does when it's turned on even
if no one is typing at the keyboard or using the mouse.

\item  If this type of computer is doing nothing all day (24 hours),
how much electricity does it use?  Express your answer in
kilowatt\-hours.

Consuming electricity at a \emph{rate} of 97 watts for a day uses
97 \emph{watt-days}. That's
$24*97 =2,328$ watt-hours or 2.3kwh of electricity.

\item  Now suppose the computer goes to sleep after 15 minutes of
doing nothing. How much electricity does it use in an idle day?

A lot less: $0.25*97 + 23.75*3.5= 107.375$ watt hours, which is about
0.11 kwh.

\item If a kilowatt hour of electricity costs 20 cents, how much
money is saved in a day because the computer is smart enough to go to
sleep?

The sleep saves about 2.2 kwh of electricity, worth 44 cents.

\end{abcd}

\end{sol}
\end{exx}


\begin{exx}{\hassolution\complex\sref{energypower}
\gref{directproportion}\gref{energyandpower}}
How Much Water Does Pasta Really Need?

On February 24, 2009 \thetimes{} published an article
by Harold McGee addressing that question.
\index{pasta}

There you could read:

\begin{qwrap}
\begin{quotation}

\firstline{After some experiments, I've found that we can indeed make}
  pasta in just a few cups of water and save a good deal of
  energy. Not that much in your kitchen or mine ---
 just the amount  needed to keep a burner on high for a few more
 minutes. But Americans cook something like a billion pounds of pasta
 a year, so those minutes could add up.   

My rough figuring indicates an energy savings at the stove top of
several trillion B.T.U.s. At the power plant, that would mean saving
250,000 to 500,000 barrels of oil, or \$10 million to \$20 million at
current prices. Significant numbers, though these days they sound like
small drops in a very large pot.%
\webref{%
www.nytimes.com/2009/02/25/dining/25curi.html
}
\end{quotation}
\sourceinfo[1209]{www.nytimes.com/2009/02/25/dining/25curi.html}
\end{qwrap}

\begin{abcd}

\item
Verify the author's conversion of ``several trillion B.T.U.s'' to
barrels of oil and then to dollars.

\item Does McGee's estimate of a billion pounds of pasta per year make
  sense?

\item How much water do Americans use cooking pasta? How much would
  they use if they followed McGee's advice?

\item Does not boiling the extra water really save the amount of energy
  McGee claims?
\end{abcd}


\begin{sol}

\begin{abcd}

\item
Verify the author's conversion of ``several trillion B.T.U.s'' to
barrels of oil and then to dollars.

The web site
\url{www.unitjuggler.com/convert-energy-from-boe-to-MMBtu.html}
says that the energy in one barrel of oil is 5.55 million BTU.
If I take ``several trillion BTU'' to be ``2 trillion BTU'' then
that's equivalent to (2 trillion)/(5.5 million) = 360 thousand barrels
of oil. That's just about what McGee says. 

On October 24, 2014 oil cost about \$80/barrel. 250,000 barrels
would cost about \$20 million. That's double McGee's estimate, so when
he wrote in 2009 oil must have cost about \$40/barrel. The web site
\url{www.ioga.com/Special/crudeoil_Hist.htm} confirms that
price for early 2009.

\item Does McGee's estimate of a billion pounds of pasta per year make
  sense?

That would be about 
%
\begin{equation*}
\frac{1 \text{ billion pounds}}{100 \text{ million households}}
= 
10 \frac{\text{pounds}}{\text{household}} .
\end{equation*}
%
I guess I can believe that order of magnitude --- it's pasta for dinner
about once a month on average. Some households never cook it; those
that do probably have it much more often than that.

\item How much water do Americans use cooking pasta? How much would
  they use if they followed McGee's advice?

If the usual method is to use two quarts of water for a pound, this much
pasta would take two billion quarts of water. McGee suggests two cups 
instead of two quarts. Since there are four quarts in a cup, doing
pasta McGee's way would take only one fourth as much water, or half a
billion quarts.

\item Does not boiling the extra water really save the amount of energy
  McGee claims?

To answer this I looked up how much energy it takes to boil
water. Several web sites told me it was about 350 kilojoules per
quart. Google told me that was about 330 BTU per quart. To boil the
extra billion and a half quarts of water would take about 500 trillion
BTU. That's more than the ``several trillion'' McGee claims. I wonder
where I went wrong --- if I did.


\end{abcd}

\end{sol}
\end{exx}


\begin{exx}[solarenergy1]{\untested\sref{energypower}\gref{energyandpower}}
  Solar energy.
\index{solar energy}

On May 1, 2013, \emph{The Arizona Republic} reported the opening of a
new power plant near Phoenix.  The article said, 

\begin{qwrap}
\begin{quotation}
\firstline{The \$500 million Arlington Valley Solar Energy II project}
  will be among the largest solar plants in the world when finished this
year. \ldots 

The project adds to Arizona's large list of solar-power plants. If it
were complete today, it would be among the 10 largest in the world. It
will have 127 megawatts of capacity when finished. One megawatt is
enough electricity to supply about 250 Arizona homes at once, when the
sun is shining on the solar panels. 
\webref{%
www.azcentral.com/business/arizonaeconomy/articles/20130501new-arlington-valley-solar-site-packs-power.html
}
\end{quotation}
\sourceinfo{www.azcentral.com/business/arizonaeconomy/articles/20130501new-arlington-valley-solar-site-packs-power.html}
\end{qwrap}

\begin{abcd}
\item  Compare the estimated power needs of a home in this article, 
(1/250)th of a megawatt, to the estimate in
\exref{solarenergy2}.
\item
Estimate the power needed to supply the electricity needs of your
home.

\end{abcd}

\end{exx}


\begin{exx}[solarenergy2]{\untested\complex\sref{energypower}
\gref{directproportion}\gref{energyandpower}}
Solar power at \myindex{Wellesley College}.

The sign on a solar panel array at Wellesley College reads:
\begin{center}
\framebox{
\begin{minipage}{5in}
\begin{center}
Solar Photovoltaic Array \\
This 10-kilowatt Solar PV Array is composed \\
of 48 panels, each 210 watts. It will generate approximately \\
13,000 kilowatt hours of electricity per year, enough to \\
power 2 homes, 32 metal halide street lights \\
or 85 LED street lights for an entire year. \\
For real time electrical output please go to: \\
\url{www.sunwatchmeter.com/home/day/wellesley-college} \\
PLEASE KEEP OFF THE PANELS
\end{center}
\end{minipage}
}
\end{center}

You can see a picture at
\url{www.theswellesleyreport.com/2010/09/wellesley-college-saves-the-planet/solar-panel/}

\begin{abcd}
\item Check the consistency of some of the numbers.

\item How many hours of sunshine per year do the designers expect the
installation to see?

\item Visit the website and write about it. What do the various graphs
and meters represent? What is happening there now? What has happened
recently?  
\end{abcd}

\end{exx}


\begin{exx}{\complex\sref{energypower}\gref{energyandpower}}
Wind power.
\index{wind power}

From \emph{The Los Angeles Times}, March 1, 2009:

\begin{qwrap}
\begin{quotation}
\firstline{The U. S. last year surpassed Germany as the world's No.}
1 wind-powered nation, with more than 25,000 megawatts in
place. Wind could supply 20\% of America's electricity needs by
2030, up from less than 1\% now, according to a recent Energy
Department report.%
\webref{articles.latimes.com/2009/mar/01/business/fi-wind-bootcamp1/}
\end{quotation}
\sourceinfo{articles.latimes.com/2009/mar/01/business/fi-wind-bootcamp1/}
\end{qwrap}

\begin{abcd}

\item What do these data say when you calculate wind power
in megawatts per person, or as a percentage of the total power
available? 

\item Explain why it might or might not be true
to say that when this article appeared the U.S. now produced more
wind energy than Germany?
\end{abcd}

\end{exx}

\begin{exx}{\untested\sref{energypower}\gref{energyandpower}
\gref{directproportion}}
World \myindex{solar power}.

In a posting on their website on July 31, 2013, the Earth Policy
Institute reported that

\begin{qwrap}
\begin{quotation}
\firstline{The world installed 31,100 megawatts of solar photovoltaics}
  (PV) in 2012 --- an all-time annual high that pushed global PV
capacity above 100,000 megawatts.  There is now enough PV operating to
meet the household electricity needs of nearly 70 million people at
the European level of use.%
\webref{%
www.earth-policy.org/indicators/C47/solar_power_2013
}
\end{quotation}
\sourceinfo{www.earth-policy.org/indicators/C47/solar_power_2013}
\end{qwrap}

\begin{abcd}
\item  Use data in the quote to estimate average household electricity usage.
\item  How does the ``European'' level of use compare to the usage in
  the United States?    
You may want to use your estimates from other problems, including
\exref{solarenergy2}. 
\end{abcd}
\end{exx}

\begin{exx}{\hassolution\complex\sref{energypower}\gref{directproportion}\gref{energyandpower}}
Chilling out by the quarry.

On August 16, 2010 \theGlobe{} described a local business's plan to cool
its corporate facility with water from a nearby quarry rather than
with conventional air conditioning.

\begin{quotation}
[Director of facilities] Dondero estimated that the cooling system,
which eliminates the need 
for any type of refrigerant in the building, saves about \$75,000 a
year, reduces annual water use by one million gallons, and cuts yearly
energy use by about 300,000 kilowatt hours --- enough to power about 30
homes.%
\webref{%
www.boston.com/business/technology/articles/2010/08/16/chilling_out_by_the_quarry/
}
\end{quotation}

\teachertag{}
\begin{teacher}
The article on the quarry water cooling system also asserts that
\begin{quotation}
The system ... cost only about \$700,000 more than a traditional
cooling system, meaning Biogen Idec should get a return on its
investment in eight to 10 years. 
\end{quotation}
Discussing payback time might be interesting --- or too difficult.
\end{teacher}

\begin{abcd*}
\item What rate in dollars per kWh is Dondero using to support his
assertion that this change will save \$75,000 a year?

\item Is the claim that 300,000 kilowatt hours would power 30 homes
for a year reasonable?
\end{abcd*}

\begin{sol}

\begin{abcd*}
\item What rate in dollars per kwh is Dondero using to support his
assertion that this change will save \$75,000 a year?

This is easy if I just work with the units. I want dollars per kWh so
I divide
\begin{equation*}
\frac{\$75,000}{300,000 \text{ kWh}}
=
0.25
\frac{\$}{\text{kWh}}
\end{equation*}.

I didn't even need a calculator to find that nice round number.

\item Is the claim that 300,000 kilowatt hours would power 30 homes
for a year reasonable?
\end{abcd*}

There are lots of web sites that begin to answer this question. I
wanted one I thought might be authoritative, so I started at the
U.S. Energy Information Administration FAQ at
\url{www.eia.gov/tools/faqs/index.cfm}. There the answer to the
question ``How much electricity does an American home use?'' is

\begin{quotation}
In 2008, the average annual electricity consumption for a
U.S. residential utility customer was 11,040 kWh
\end{quotation}

The newspaper article assumes 10,000 kWh per household per year for 30
homes. That's the right order of magnitude.


\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{energypower}\gref{directproportion}\gref{energyandpower}} Energy savings at \myindex{MIT}.

On March 26, 2011 Jon Coifman wrote in \theGlobe{} that

\begin{quotation}
In just 36 months, [MIT and NStar] plan to cut the university's energy
use 15 percent --- enough to power 4,500 Massachusetts homes for a year.
\webref{%
www.boston.com/bostonglobe/editorial_opinion/oped/articles/2011/03/26/an\_energy\_program\_too\_efficient\_for\_its\_own\_good/
}
\end{quotation}

\begin{abcd}

\item If all of MIT's energy use were devoted to powering
Massachusetts homes, how many homes would that be?

\item Compare your answer in part (a) to the number of homes in
Cambridge.

\item 
Estimate MIT's
total annual energy use, in BTUs. 

\item Convert your answer to the previous question from BTUs to watt-hours.

\item 
The cool web site
\url{cogen.mit.edu/powermit/} reported that on April 4, 2011
MIT used about 550 megawatthours of energy.
Use this information to estimate the total amount of energy MIT uses
in a year. Is your answer consistent with the estimate from the previous
problem?

\end{abcd}

\begin{sol}

\begin{abcd}

\item If all of MIT's energy use were devoted to powering
Massachusetts homes, how many homes would that be?

It follows from the article that MIT uses enough energy in a year
to power $4,500/0.15 = 30,000$ homes. 

\item Compare your answer in part (a) to the number of homes in
Cambridge.

Cambridge has about 100,000 people, so about 30,000 homes. That means
MIT uses as much energy as all the rest of the city (domestically).

Wikipedia says Cambridge has about 40,000 homes, but my estimate is
the same order of magnitude, so the conclusion remains the same.

\item Estimate MIT's
total annual energy use, in BTUs. 

Wikipedia says annual household energy use in the
Northeast is about 180,000,000 BTU, half of which is electricity. 
(\url{en.wikipedia.org/wiki/File:US\_household\_energy\_usage.png}).

\begin{equation*}
30,000 \text{ households} 
\times
180,000,000 \frac{\text{BTU}}{\text{household}}
= 5,400,000,000,000 {\text{ BTU}}
= 5.4 \text{ TeraBTU}.
\end{equation*}

\item Convert your answer to the previous question from BTUs to watt-hours.
Write your answer using the appropriate metric prefix, not with lots
of zeroes. 

1 kilowatt-hour = 3412.1416 BTU, so
\begin{equation*}
5.4 \times 10^{12} \text{ BTU} \times 
\frac{ 1 \text{ kilowatt-hour}}{3412.1416 \text{ BTU}}
= 1.58 \text{ Terawatt-hours}.
\end{equation*}

\item 
The cool web site
\url{cogen.mit.edu/powermit/} reported that on April 4, 2011, 
MIT used about 550 megawatthours of energy.

Use this information to estimate the total amount of energy MIT uses
in a year. Is your answer consistent with the answer to the previous
problem?

550 megawatthours per day would come to about 200,000 megawatthours
per year. That's 2 TeraWattHours, which is the same order of magnitude
as the 1.58 TWH answer in the previous part of the problem.

The MIT cogen web site seems to be reporting just about electricity,
so I should probably take the answer to the previous part of the
problem as 0.75 TWH (half is electric energy). Even then the answers
are in the same ballpark.
\end{abcd}

\end{sol}


\end{exx}


\begin{exx}{\hassolution\sref{energypower}\gref{directproportion}
\gref{energyandpower}} 
The AP gets the units wrong.

On May 13, 2011 the \emph{Norwich Bulletin} reported an item from the
Associated Press that said

\begin{qwrap}
\begin{quotation}
\firstline{Connecticut Governor Daniel Malloy recently signed off on a}
  deal to tax electricity generators one quarter of one cent per
  kilowatt hour, or 25 cents per \$100.
\webref{%
www.norwichbulletin.com/news/x832282338/Attorney-General-Connecticut-electricity-tax-could-cost-Mass-26M
}
\end{quotation}
\sourceinfo{www.norwichbulletin.com/news/x832282338/Attorney-General-Connecticut-electricity-tax-could-cost-Mass-26M}
\end{qwrap}

\begin{abcd}

\item What is wrong with the units in this quotation?

\item Estimate the percentage change in the cost of electricity that
would result from a one-quarter of one cent increase per
kilowatt-hour.

\item What do you think the Associated Press intended to say?

\end{abcd}


\begin{hint}
What did electricity cost in the spring of 2011?
\end{hint}

\begin{sol}

\begin{abcd}

\item What is wrong with the units in this quotation?

The author says that some number of cents per kilowatt-hour is
equivalent to some number of cents per dollar. That makes no sense at
all.

\item Estimate the percentage change in the cost of electricity that
would result from a one-quarter of one cent increase per
kilowatt-hour.

In the spring of 2011 a kilowatt-hour costs about 10 cents in
Massachusetts.  I need to find out what percentage 0.25 cents (a
quarter of a cent) is of 10 cents. That's easy: $0.25/10 = 0.025 =
2.5\%$.

\item What do you think the Associated Press intended to say?

I think the author meant to say that a quarter of a cent per
kilowatt-hour is 25 cents per hundred kilowatt-hours, not per hundred
dollars.
\end{abcd}

\end{sol}
%
\end{exx}



\begin{exx}{\untested\sref{taxes}\gref{directproportion}
\gref{linearexcel}\gref{incometax}}
Your total federal tax bill.

Modify the graph in Figure~\ref{fig:effectivetax} to show how total
tax and the effective tax rate for (income tax + social security)
depends on income.

\end{exx}


\begin{exx}{\untested\complex\sref{taxes}\gref{directproportion}
\gref{incometax}}
President Obama's income tax. 
\index{Obama, Barack}

According to the White House web site

\begin{qwrap}
\begin{quotation}
\firstline{[The President] and the First Lady filed their income tax}
  returns jointly and reported adjusted gross income of \$481,098. The
  Obamas paid \$98,169 in total tax. 

The President and First Lady also reported donating \$59,251 --- or about
12.3 percent of their adjusted gross income --- to 32 different
charities. The largest reported gift to charity was \$8,751 to the
Fisher House Foundation. The President's effective federal income tax
rate is 20.4 percent. \ldots  The President and First Lady also
released their Illinois income tax return and reported paying \$23,328
in state income tax.%
\webref{%
www.whitehouse.gov/blog/2014/04/11/president-obama-and-vice-president-biden-s-2013-tax-returns}
\end{quotation}
\sourceinfo{www.whitehouse.gov/blog/2014/04/11/president-obama-and-vice-president-biden-s-2013-tax-returns}
\end{qwrap}


Warning: some of the questions that follow are complex. They may not
have simple numerical answers, and may depend on information that's
hard to find. Sometimes you will only be able to find partial answers.
Make sure what you write makes sense. You can write about what kind of
information is missing.

\begin{abcd}

\item What tax bracket was the president in?

\item Use the Married Filing Jointly brackets and rates in 
the spreadsheet at 
\link{Federalindividualratehistory.xlsx} 
to compute the Obamas' 2012 Federal income tax bill. If your result
does not match the reported figure, what might explain the difference?

\item If the Obamas had not made those charitable contributions the money
would be added to their net income. 

\begin{abcd*}
\item What would their taxes have been? 
\item What fraction of the contribution was (essentially) made by the
government?
\item What fraction of the Obamas' income did they contribute to charity? 
\item Did they tithe? 
\item How does their contribution compare to the national average?
\end{abcd*}

\end{abcd}

\end{exx}

\begin{exx}{\hassolution\sref{taxes}\gref{directproportion}\gref{incometax}}
Using the tax table.

Use Table~\ref{table:2014taxtable} to answer the following questions.

\begin{abcd}

\item Compute the tax due in 2014 on a net taxable income of
\$80K. Show your work.

\item Compute the \myindex{effective tax rate} for that income.

\item Check your answers with those in the spreadsheet

\item Compute the effective tax rate a second time, as a 
weighted average of the rates in the various brackets, 
using as weights the amount of income subject to
tax at each rate. You should get the same answer.


\link{GraduatedTax.xlsx}.

\end{abcd}

\begin{sol}

Note: this answer is to the version of the exercise assigned in Spring
2013. It needs to be updated since the exercise has changed.
\begin{abcd}


\item The tax on an income of \$40K is
\begin{equation*}
10\% \times \$8025 + 15\% \times (\$32550-\$8025) + 25\% \times
(\$40000-\$32550)
=
\$6343.65.
\end{equation*}

\item The effective tax rate is \$6343.65/\$40000 = 0.1586 = 15.86\%.

\item The effective tax rate computed as a weighted average of the
three percentages is
\begin{equation*}
\frac{ \$8025 \times 10\% + (\$32550-\$8025) \times 15\%+ 
(\$40000-\$32550) \times 25\%}
{\$40,000}
= 0.1586 = 15.86\%,
\end{equation*}
which matches, as it should.


\item The answers match those I get by temporarily
replacing the \$10,000 entry in cell B42 by \$40,000.

\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{taxes}\gref{directproportion}\gref{incometax}}
Taxes and inflation.
\index{inflation}

The spreadsheet \link{Federalindividualratehistory.xlsx} contains a
complete history of income tax brackets and rates from the inception
of the income tax in 1913 through 2013, in both dollars current in
each year and adjusted for inflation (2012 dollars).

\begin{abcd}

\item Compute the tax due in 2003 for a single taxpayer with a net
taxable income of \$30K. What is her effective tax rate?

\item Suppose that taxpayer received raises each year that kept up with
inflation. Use an inflation calculator to calculate her net taxable
income in 2012.

\item Use the 2012 tax tables to compute her tax in 2012. What is her
effective tax rate?

\item 
Compare her 2003 tax and effective tax rate with her 2012 tax
and effective tax rate, taking inflation into account. Has her tax
gone up, or down, or stayed the same?

\item Compare your answer to the previous problem with the answer you
  get using the inflation adjusted data in the spreadsheet.
\end{abcd}

\begin{sol}

Note: These are answers to the original version of the question, as
assigned in Spring 2013. Redo these answers next time ...

\begin{abcd}

\item The tax in 2003 on an income of \$30K is
\begin{equation*}
10\% \times \$7000 + 15\% \times (\$28400-\$7000) + 25\% \times
(\$30000-\$28400)
= \$4310.
\end{equation*}

%.10 * 7000 + .15*(28400-7000) + .25 * (30000-28400)

Her effective tax rate is 
$\$4310/\$30000 = 0.143666667 \approx 14.46\%$.

\item 
The inflation calculator at
\url{data.bls.gov/cgi-bin/cpicalc.pl} says that \$30,000 in
2003 has the same buying power as \$32,869.57 in 2006. 

\item The tax in 2006 on an income of \$32,869.57 is
\begin{equation*}
10\% \times \$7550 + 15\% \times (\$30,650-\$7550) + 25\% \times
(\$32,869.57-\$30,650
= \$4774.90.
\end{equation*}

Her effective tax rate is 
$\$4774.90/\$32869.57 = 0.145267872 \approx 14.53\%$.

%.10 * 7550 + .15*(30650-7550) + .25 * (32869.57-30650)

\item 
Her tax has gone up by \$465, or 9.7\%. Her income has gone up by
$\$32869.57/\$30000 = 1.09565233$, which is a 9.57\% increase. That's
a little bit less than the percentage increase in her taxes, which is
why her effective tax rate increased by just 0.07 \emph{percentage
points}. That's not a lot. The IRS did a pretty good job adjusting the
brackets for inflation.

\end{abcd}

\end{sol}

\end{exx}


\begin{exx}{\hassolution\sref{linearfunction}\gref{linearfunctions}
\gref{slopeintercept}}
\myindex{Pandora} growing fast!

On June 10, 2011, CNN Money reported that 

\begin{qwrap}
\begin{quotation}
\firstline{[Pandora], (t)he Oakland, Calif. company, is growing fast,}
averaging a new user every second.  As of last April, Pandora had
90 million registered users \ldots %
\webref{%
money.cnn.com/2011/06/10/technology/pandora_ipo/index.htm
}
\end{quotation}
\sourceinfo{money.cnn.com/2011/06/10/technology/pandora_ipo/index.htm}
\end{qwrap}

\begin{abcd}
\item Write the linear equation that models this scenario, using
seconds since the article appeared for the independent
variable. What are the slope and the intercept, with their units?

\item Rewrite your equation using millions of users for the dependent
variable and years for the independent variable. What are the slope
and intercept?

\item How long could Pandora continue to grow at that rate?

\item Did it continue to grow at that rate?

\end{abcd}

\begin{sol}


\begin{abcd}
\item Write the linear equation that models this scenario, using
seconds since the article appeared for the independent
variable. What are the slope and the intercept, with their units?

Let $U$ be the number of Pandora users and $S$ the number of seconds
since the article appeared on June 14, 2011. Then the equation
suggested would be
\begin{equation*}
U = 90,000,000 + S
\end{equation*}
with a $y$-intercept of 90 million users and a slope of 1 user per
second.

\item Rewrite your equation using millions of users for the dependent
variable and years for the independent variable. What are the slope
and intercept?

There are about 30,000,000 seconds in a year, so the equation
becomes
\begin{equation*}
U = 90  +  30Y
\end{equation*}
when $U$ is measured in millions of users and $Y$ is measured in
years. The intercept is 90 (million users) and the slope is 
30 (million users per year).

\item How long could Pandora continue to grow at that rate?

Well at that rate in two years (June of 2013) about half the
population of the United States would be Pandora users. I don't think
that could happen.

\item Did it continue to grow at that rate?

I found this from the \emph{Los Angeles Times} on April 5, 2012

\begin{qwrap}
\begin{quotation}
\firstline{The Oakland company said Thursday that the number of active}
listeners (that is, people who have used the service at least once in
the past 30 days) grew to 51 million in March, up 59\% from a year earlier. 
\webref{
latimesblogs.latimes.com/entertainmentnewsbuzz/2012/04/pandora-users.html
}
\end{quotation}
\sourceinfo[333]{latimesblogs.latimes.com/entertainmentnewsbuzz/2012/04/pandora-users.html}
\end{qwrap}

That seems to say there were just about 51/1.59 = 32 million users in
the spring of 2011, not 90 million. Something doesn't match. I have no
idea why.	
\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\untested\gref{energyandpower}}Bicycle power in
  \myindex{Times Square}.

From an Associated Press story in \theGlobe{} on December 30, 2013.

\begin{qwrap}
\begin{quotation}
\firstline{Over the weekend, six Citibikes from the city's bike-share}
  program were installed in Times Square and connected to 12-volt deep
  cycle batteries. New Yorkers and tourists will generate power by
pedaling. That will help illuminate the famed ball.

Each bike will generate an average of 75 watts an hour. It takes 
50,000 watts to light up the ball's LEDs.
\end{quotation}
\sourceinfo{www.bostonglobe.com/news/nation/2013/12/30/justice-sotomayor-lead-times-square-ball-drop/Nac0Vg3INmXYcLcLIYSz4J/story.html}
\end{qwrap}

Unfortunately, the Associated Press reporter is quite confused about
the difference between energy and power. The ``generate \ldots 75
watts an hour'' in the second paragraph makes no sense. We
think what he or she is trying to say is that while someone is
actually pedaling it each bike could power a 75 watt light bulb. All
six bikes together could light up just 450 watts worth of LEDs.

\begin{abcd}
\item How many bikes would have to be pedaled simultaneously to light
  up all the ball's LEDs?

\item 
Since there are only six bikes, people pedaling during the day will
  store the energy they generate in batteries, which will then
  be used to light the ball. Suppose the lights need to be on for ten
  seconds while the ball drops at midnight.

How many hours of pedaling will it take to generate (and save) the
electrical energy needed?
\end{abcd}

\end{exx}

\begin{exx}{\hassolution} Flying twice as far.

A curious traveler asked this question on \myindex{stack exchange}:

\begin{qwrap}
\begin{quotation}
\firstline{A flight from Los Angeles to Albuquerque is about 2 hours}
  but is $\approx 670.2$ miles.

A flight from San Jose to Chicago is 4 hours but is $\approx 1859.0$ miles.

Can anyone explain why the travel time from San Jose to Chicago is not
longer and closer to 5.75 hours? 

If the distance increase by 2, shouldn't the time increase by a factor
of 2 as well?%
\webref{travel.stackexchange.com/questions/26083/explain-travel-times-and-distances-on-flight}
\end{quotation}
\sourceinfo{travel.stackexchange.com/questions/26083/explain-travel-times-and-distances-on-flight}
\end{qwrap}

\begin{abcd}

\item Write a linear model for this question. Takeoff and landing will
  take a fixed amount of time. Actual travel in the air will take time
  proportional to the distance traveled. Think about which of the
  variables (time and distance) is the independent variable, and
  identify the slope and intercept with their units.

\item Use the data in the quotation to estimate the two constants in
  your linear model.

\item Compare your answer to those at the link to the quotation.
\end{abcd}

\begin{sol}

\begin{abcd}
\item 
Write a linear model for this question. Takeoff and landing will
  take a fixed amount of time. Actual travel in the air will take time
  proportional to the distance traveled. Think about which of the
  variables (time and distance) is the independent variable, and
  identify the slope and intercept with their units.

In this problem the \emph{dependent} variable $y$ is the
travel time, in hours. The independent variable $x$ is
the distance flown, in miles. The intercept $b$ is the time in
minutes for takeoff and landing. The slope is $m$. Its units are
minutes per mile. That's the tricky part of this question --- we're
used to thinking of speed in units time/distance, not distance/time.

The linear model is then the familiar $y = mx + b$.

\item Use the data in the quotation to estimate the two constants in
  your linear model.

I know from the data (after a little rounding) that
%
\begin{equation*}
4 = m \times 1860 + b
\end{equation*}
%
and
%
\begin{equation*}
2 = m \times 670 + b
\end{equation*}

Subtracting tells me that 
\begin{equation*}
2 = (1860 - 670)m = 1190m \approx 1200m
\end{equation*}
%
so $m$ is about (2/1200) hours per mile, or 600 miles per hour. That's
a pretty good approximation for airline cruising speed. Putting $m =
1/600$ into the first equation tells me that 
%
\begin{equation*}
4 = 1860/600 + b \approx 3 + b
\end{equation*}
%
so $b \approx 1$ hour --- half an hour each for takeoff and landing.

\item Compare your answer to those at the link to the quotation.
\end{abcd}

The link say it's reasonable to estimate the cruising speed at about
500 miles per hour and the takeoff and landing time at half an
hour. Those numbers fit the data too.

\end{sol}
\end{exx}


\begin{ReviewExercises}

\begin{rexx}
If you drive at a rate of 50 miles per hour for 3 hours, how far have
you driven?  Identify each piece of this proportion:  the quantities
being measured and the proportionality constant, with the appropriate
units.  
\end{rexx}

\begin{rexx}You may remember from geometry that the circumference of a
circle is directly proportional to the diameter of that circle.  The
relationship is 
%
\begin{equation*}
c = \pi d, 
\end{equation*}
where $c$ represents the circumference and $d$ represents
the diameter and $\pi \approx 3.14$ is the proportionality constant.
If the diameter of a circle is doubled, how does the circumference
change? 
\end{rexx}

\begin{rexx}
The cost of potatoes is proportional to the weight (in pounds) you buy.
If potatoes cost \$0.69 per pound, what is the cost for 3
pounds of potatoes?  
\end{rexx}

\begin{rexx} 
The conversion from {\textsterling} to U.S. \$ is 1.53
\$/{\textsterling}. How much is {\textsterling}200
worth in U.S.\$? 
\end{rexx}

\begin{rexx}
Suppose that $y$ is directly proportional to $x$.  When $x=16$, then
$y=4$. 
\begin{abcd}
\item What is the proportionality constant?

\item  If $x=32$, what is $y$? 

\item If $y=32$, what is $x$?
\end{abcd}
\end{rexx}

\begin{rexx}
Identify the slope and intercept in each of the following.  When
appropriate, state the units. 
\begin{abcd*}
\item $y = 2.5x + 6$
\item $y = -5x +20$
\item $y = 300 + 40x$
\item $Q = 0.004E - 300$
\item  He earns is \$9.25 per hour.
\item  To rent a car for one day, the cost is \$25 plus \$0.15 per mile.
\item  My new phone cost \$25, plus a monthly charge of \$15.
\item  The conversion from {\textsterling} to U.S. \$ is 1.53
\$/{\textsterling}.
\item  The salesperson worked only on commission,
earning 20\% of the total amount sold. 
\end{abcd*}
\end{rexx}

\begin{rexx}
Solve each problem.
\begin{abcd*}  
\item If $y = 2.5x + 6$ and $x=4$, what is $y$?  
\item  If $y = -5x +20$ and $y=0$, what is $x$?
\item  If $y = 300 + 40x$ and $x=-10$, what is $y$?
\item If his salary is \$9.25 per hour and he works 5 hours, how much
does he earn? 
\item  If the conversion from U.S. dollars to pounds sterling is 1.80
\$/{\textsterling}, how much money would you get by changing \$100 to
{\textsterling}?   
\item  If my new phone cost \$25, and I pay a monthly charge of \$15,
what is my total cost after 10 months?  When does my total 
cost reach \$250?
\end{abcd*}
\end{rexx}

%\begin{rexx} 
%Verify that there are about 9000 hours in one year.  Use that to
%estimate the number of hours in a decade.  Approximately how many
%hours are in a century? 
%\end{rexx}
%

\end{ReviewExercises}

\setexercisecounter{}

\begin{ExtraExercises}


\begin{exx}{\needsquestions}
The more you earn the less you pay.

Figure~\ref{fig:FederalTaxRateByInocome2007} shows the effective
Federal tax rate by household income for the year 2007. The rate is
lower for the wealthier households because much of their income is
taxed as capital gains rather than as ordinary income.

\figfile{FederalTaxRateByIncome2007.jpg}
\begin{figure}
\centering
\includegraphics[width=3in]{\thefigurefilename}
\caption{Federal tax rate by income, 2007}
\figsource{tcf.org/blog/detail/10-reasons-to-eliminate-the-tax-break-for-capital-gains}
\figcomment{Needs permission. Resolution too low?}
\label{fig:FederalTaxRateByIncome2007}
\end{figure}

\end{exx}

\begin{exx}{\untested\sref{energypower}\gref{directproportion}
\gref{energyandpower}}
Not so deep sleep.

\begin{teacher}
This is on of several possible exercises on the same theme --- if you
combine them and assign them and let us know what happened we'll
rewrite them. 
\end{teacher}

Many electronic devices use a small amount of
electricity even 
when they are turned off.  A tv, for example, is always in standby
mode unless it is unplugged, and so is drawing a small amount of
electricity.  Similarly, computers, modems, cell phones and other
devices draw electricity even when turned off.  This is known as
``leaking electricity".  An article in 1998 estimated that this
accounts for about 45 billion kilowatt-hours of electricity consumed
in the United States each year, or about 5\% of the total electricity
use by individuals.  (See \url{www.aceee.org/pubs/a981.htm}) 
\begin{abcd}
\item  Choose one electric item in your home and research how much
electricity it uses when it is off.  You may be able to find this in
the documentation for the device, or on the web.  Make an estimate if
you can't find the exact specifications. 

\item  Suppose you unplugged this device at night, when you are done
using it.  Estimate how much electricity this will save and how much
money you will save in a year.
\end{abcd}

\end{exx}


\begin{exx}{\untested\needsquestions\sref{energypower}
\gref{directproportion}\gref{energyandpower}}
TV set top boxes.

\theTimes{} reported on June 26, 2011 that
\begin{quotation}
Those little boxes that usher cable signals and digital recording
capacity into televisions have become the single largest electricity
drain in many American homes, with some typical home entertainment
configurations eating more power than a new refrigerator and even some
central air-conditioning systems. 

A new study has found that some home entertainment systems eat more
energy than refrigerators or central air-conditioning systems. 

There are 160 million so-called set-top boxes in the United States,
one for every two people, and that number is rising. Many homes now
have one or more basic cable boxes as well as add-on DVRs, or digital
video recorders, which use 40 percent more power than the set-top
box. 

One high-definition DVR and one high-definition cable box use an
average of 446 kilowatt hours a year, about 10 percent more than a
21-cubic-foot energy-efficient refrigerator, a recent study found. 

These set-top boxes are energy hogs mostly because their drives,
tuners and other components are generally running full tilt, or nearly
so, 24 hours a day, even when not in active use. The recent study, by
the Natural Resources Defense Council, concluded that the boxes
consumed \$3 billion in electricity per year in the United States --- and
that 66 percent of that power is wasted when no one is watching and
shows are not being recorded. That is more power than the state of
Maryland uses over 12 months 
\webref{
www.nytimes.com/2011/06/26/us/26cable.htm 
}
\end{quotation}

The graphic in Figure~\ref{fig:TVSetTopBoxEnergyUse} accompanied the
article.

\figfile{TVSetTopBoxEnergyUse.jpg}
\begin{figure}[ht]
\centering
  \includegraphics[height=60mm]{\thefigurefilename}
\caption{Energy use comparisons}
\figsource{\theTimes,
6/26/2011. \url{www.nytimes.com/2011/06/26/us/26cable.htm }}
\figcomment{Need permission from NYTimes, or redraw, or do without.}
\label{fig:TVSetTopBoxEnergyUse}
\end{figure}

Hiawatha Bray's \headline{Stop power leaks; smile at savings}%
\webref{%
www.boston.com/business/technology/articles/2009/01/15/stop_power_leaks_smile_at_savings/}
from \theGlobe{} on January 15, 2009  refers you to 
\url{standby.lbl.gov/standby.html}, a site maintained by the
Lawrence Berkeley National Laboratory, at which you can find out lots
of useful stuff about standby power consumption.
\teachertag{}
\end{exx}

\begin{exx}{\needsquestions}
Romney's effective tax rate.
\index{Romney, Mitt}
\index{effective tax rate}

 \index{probably} 

An Associated Press article headlined
in the \emph{Houston Chronicle} on January
18, 2012 about the tax returns of then Presidential candidate
Mitt Romney said the following:

\begin{qwrap}
\begin{quotation}
\firstline{Speaking to reporters after a campaign stop in South}
  Carolina,  Romney said most of his income comes from investments,
  not regular wages and salary.  The tax rate on investment income is
  15 percent, much lower than the 35 percent rate applied  to wages
  for those in the highest tax bracket. 

``What's the effective rate I've been paying? It's probably closer to
  the 15 percent rate than anything,'' Romney said.  
\webref{%
www.houstonchronicle.com/news/politics/article/GOP-front-runner-Romney-backpedals-on-releasing-2586571.php}
\end{quotation}
\sourceinfo{www.houstonchronicle.com/news/politics/article/GOP-front-runner-Romney-backpedals-on-releasing-2586571.php}
\end{qwrap}

\begin{abcd}
\item  The median net income of an American family in 2012 was about \$50,000.
  What is the effective tax rate  for someone with that income?
\item  How does Romney's tax rate of ``probably'' 15\% compare to the
  ``average'' American? 
\end{abcd}

\end{exx}

\begin{exx}{\untested\complex\needsquestions} 
Does virtual save energy?

\begin{qwrap}
\begin{quotation}
\firstline{MOST PEOPLE take for granted the Earth-friendly nature of}
  electronic communication. Paperless, ink-free, no shipping supplies,
  no gas for transportation: the environmental benefits of virtual
  communication are obvious. But the reality is more complicated, at
  least according to a growing number of concerned technology experts
  and scientists. Vast stockpiles of digital data waste energy, too. 

Everyday emails aren't to blame. But large photo and video
attachments, cluttered inboxes, and massive email forwards may
be. Some analysts estimate that emailing a 4.7-megabyte attachment ---
the equivalent of four large digital photos --- can use as much energy
as it takes to boil about 17 kettles of water. The problem is
magnified when large emails are forwarded to many people and left in
inboxes undeleted. As long as emails remain in your inbox, the data
they create is physically stored somewhere. 

And that's where the problems arise: The total amount of digital
storage worldwide is approaching 1 zettabyte, or 1 million times the
contents of the Earth's largest library. Currently, that information
is archived on equipment with a mass equivalent to 20 percent of
Manhattan. Global data storage is expected to reach 35 zettabytes by
2020, which means more equipment, land, and energy. The information
industry already accounts for approximately 2 percent of global carbon
dioxide emissions. That's the same amount as the airline industry
blasts into the atmosphere. Coupled with the rapid increase in stored
data, it's an unsustainable scenario. 

Technology firms must create systems that store data with less energy,
and governments should provide incentives for them to do so. Just as
important, consumers must demand products that save energy, and use
websites like Flickr and MediaFire that allow them to share large
files without emailing. Better still, they could consider keeping some
of those embarrassing photos and home videos to themselves.%
\webref{%
www.boston.com/bostonglobe/editorial_opinion/editorials/articles/2010/09/07/dont_forward_those_photos/
}
\end{quotation}
\sourceinfo{www.boston.com/bostonglobe/editorial_opinion/editorials/articles/2010/09/07/dont_forward_those_photos/}
\end{qwrap}

\end{exx}

\begin{exx}{\untested\needsquestions}
Green gas?

Robert Bryce's op-ed in \theTimes{} on
June 7, 2011
(\url{www.nytimes.com/2011/06/08/opinion/08bryce.html})
has lots of interesting numbers about the costs in steel and land area
for solar and wind electricity generation.

\end{exx}

\begin{exx}{\untested\needsquestions} 
Every little bit counts.

On March 5, 2012 \theGlobe{} reported on the
Ocean Renewable Power Company's plans to install tidal powered
generators in Maine:
\index{tidal power}

\begin{qwrap}
\begin{quotation}
\firstline{The first unit capable of powering 20 to 25 homes will be}
hooked up to 
the grid this summer, and four more units will be installed next year
at a total cost of \$21 million \ldots

Eventually, Ocean Renewable hopes to install more units to bring its
electrical output to 4 megawatts.
\webref{
www.bostonglobe.com/business/2012/03/05/maine-company-ready-install-tidal-power-unit/daJ3ivfrUNUnHejoOt2lqJ/story.html
}
\end{quotation}
\sourcewc{240}
\sourceinfo{
www.bostonglobe.com/business/2012/03/05/maine-company-ready-install-tidal-power-unit/daJ3ivfrUNUnHejoOt2lqJ/story.html
}
\end{qwrap}
\end{exx}

\begin{exx}{\needsquestions}
Power, pollution and the internet.

On September 23, 2012 \theTimes{} printed an article with that
headline. It's full of numbers that make for interesting explorations.
More will be forthcoming:

\begin{quotation}
This is the first article in a series about the physical structures
that make up the cloud, and their impact on our environment. 
\end{quotation}

\begin{qwrap}
\begin{quotation}
\firstline{Nationwide, data centers used about 76 billion}
kilowatt-hours in 2010, 
or roughly 2 percent of all electricity used in the country that year,
\end{quotation}
\sourceinfo{www.nytimes.com/2012/09/23/technology/data-centers-waste-vast-amounts-of-energy-belying-industry-image.html}
\end{qwrap}

Here's one of the comments (many are very interesting).

\begin{quotation}
This article is weakened by hysterical reporting that relies on
meaningless unscaled scare statistics rather than a balanced
presentation of the issues. 

Buried deep in the article, one finds a more meaningful statistic:
data centers use only 2 percent of the electricity used in the
country, hardly a huge figure, given their importance in today's
economy. 

You also fail to consider the cost of alternatives. How much does it
cost to sort and deliver a first class letter? To print and distribute
a paperback book? To answer a phone bank call? To drive to a movie
theater? It is likely that the data center is saving energy, not
costing it. 

No doubt energy efficiency can be improved, and the Times is to be 
lauded for pointing that out. But no one who has ever worked with this
kind of technology can suppose that facilities of this kind can be
built without backup generators, batteries, and air conditioning: so
crucial are these servers to today's economy that when a major service
goes down for an hour, it is front page news. And in the absence of
context, the implication that data centers are disproportionately
responsible for wasting energy is questionable indeed. 
\end{quotation}
\end{exx}


\begin{exx}{\needsquestions\sref{energypower}\gref{energyandpower}}
Energy in an atomic bomb.
\index{atomic bomb}

\figfile{abombEnergyandPower.jpg}
\begin{figure}[ht]
\centering
  \includegraphics[height=60mm]{\thefigurefilename}
\caption{Energy and power from an atomic bomb}
\figsource{www.guardian.co.uk/commentisfree/2012/nov/28/ap-iran-nuclear-bomb}
\figcomment{The graph is from the AP, and it was all over the web}
\label{fig:abombEnergyandPower}
\end{figure}
\figfile{}

This graph is good for discussing the difference between energy and
power (the power energy curve is the accumulation of the power curve)
but as evidence about the Iranian nuclear program it's quite suspect
--- 
see \url{www.guardian.co.uk/commentisfree/2012/nov/28/ap-iran-nuclear-bomb}
\end{exx}


\begin{exx}{\hassolution\sref{driveandtext}\gref{percentagestrategies}
\gref{slopeintercept}}
Trucks test brakes.

Figure~\ref{fig:truckstestbrakes} shows some road signs you might see
at the top of a hill on a highway, warning that a steep grade lies
ahead.

\figfile{3630126-road-signs-to-represent-percentages-and-gradients.jpg}
\begin{figure}[ht]
\centering
\includegraphics[width=60mm]{\thefigurefilename}
\caption{Steep grade}
\figsource{Image from web page 
\url{www.123rf.com/photo_3630126_road-signs-to-represent-percentages-and-gradients.html}}
\figcomment{We will probably want to extract one of two of
these. Could redraw them or get them from here or somewhere free if
these aren't.}
\label{fig:truckstestbrakes}
\end{figure}
\figfile{}

\begin{abcd}
\item If the grade is two miles long and the grade is 10\%, how many
feet lower is the bottom of the hill than the top?

\item What could a grade of 100\% mean?
\end{abcd}

\begin{sol}
\begin{abcd}
\item If the grade is two miles long and the grade is 10\%, how many
feet lower is the bottom of the hill than the top?

Two miles is about 10,000 feet. 10\% of that is 1,000 feet. It's how
much lower you'll be at the end of the grade. I've never seen a 10\%
grade on a highway!

\item What could a grade of 100\% mean?

Fortunately the question asks what it \emph{ could} mean, not what it
\emph{ does} mean.

I can interpret it as a change of one foot in height for
every foot of travel on the road, or as one foot of travel for every
foot of horizontal progress, or as the edge of a cliff. 

If this book had a chapter on geometry this discussion would go there.


\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\untested\sref{comparelinear}
\gref{linearfunctions}\gref{slopeintercept}}
Disappearing ink.
\index{ink}

On July 5, 2013, an article in the Jefferson City, MO \emph{News
  Tribune} noted that 

\begin{qwrap}
\begin{quotation}
\firstline{Printers are getting less expensive all the time but}
  keeping them fully stocked with ink seems to be getting more
  expensive.  No sooner do you load new ink cartridges than you get
  messages warning you that the ink is running low. 
\webref{
www.newstribune.com/news/2013/jul/05/todays-inkjet-printers-give-new-meaning-disappeari/
}
\end{quotation}
\sourceinfo{www.newstribune.com/news/2013/jul/05/todays-inkjet-printers-give-new-meaning-disappeari/}
\end{qwrap}

The linear equation that describes this situation is
\begin{center}
total cost = (cost of printer) + (ink cost per page) $\times$ (pages
printed) .
\end{center}
\begin{abcd*}

\item Estimate the slope and intercept for this equation.

\item What fraction of the cost is the initial purchase price?

\end{abcd*}

\begin{hint}
You may want to think in terms of hundreds of pages rather than pages.

The answer to the second question depends on how many pages you print over the
lifetime of the printer.
\end{hint}

\end{exx}

\begin{exx}{\hassolution\artificial\sref{comparelinear}\gref{linearfunctions}\gref{slopeintercept}} 
Compact fluorescent bulbs.

Consumers are being encouraged to replace ordinary light bulbs
with compact fluorescent bulbs.\index{compact fluorescent bulb}
(CFLs). Soon they will be required to.

(This exercise should be updated to discuss LED bulbs too, and to use
real rather than invented numbers.)

A CFL uses less energy than an ordinary
incandescent bulb that produces the same amount of light, but it costs
more to buy. This table provides data with which you can
compare the two. 

\begin{center}
\begin{tabular}{ccc}
\toprule
bulb & initial cost & power \\
\midrule
ordinary & \$2.00 & 100 watts \\
CFL & \$9.00 & 25 watts \\
\bottomrule
\end{tabular}
\end{center}

Suppose electricity costs \$0.20 per kwh.

You can use pencil and paper, a calculator, or Excel to do the
arithmetic.

\begin{abcd}

\item Write a linear equation with which you can calculate the total
cost $C$ of using the ordinary bulb for $H$ hours.

\item What is the slope of that equation (with its units)?

\item What is the intercept of that equation (with its units)?

\item How much would it cost to buy the ordinary bulb and use it for
1000 hours?

\item Write a linear equation with which you can calculate the total
cost $C$ of using the CFL for $H$ hours.

\item How much would it cost to buy the CFL and use it for
1000 hours?

\item How long would you have to use the CFL to make it worth having
paid the higher purchase price?
\end{abcd}

\item Are the five numbers given in this exercise reasonable?

\item What does ``incandescent'' mean? Why are incandescent light
bulbs called that?

\begin{sol}


\begin{abcd}
\item Write a linear equation with which you can calculate the total
cost $C$ of using the ordinary bulb for $H$ hours.

\begin{equation*}
C = 2.00 + 0.1\times 0.20 H
\end{equation*}
\item What is the slope of that equation (with its units)?

The slope is 0.02 \$/hour, or 2 cents/hour.

\item What is the intercept of that equation (with its units)?

The intercept is \$2.00.

\item How much would it cost to buy the ordinary bulb and use it for
1000 hours?
\begin{equation*}
C = \$2.00 + 0.02 \frac{\$}{\text{hour}}\times 1000 \text{ hours}
= \$22.00.
\end{equation*}
\item Write a linear equation with which you can calculate the total
cost $C$ of using the CFL for $H$ hours.

\begin{equation*}
C = 9.00 + 0.025\times 0.20 H
\end{equation*}

\item How much would it cost to buy the CFL and use it for
1000 hours?

\begin{equation*}
C = \$9.00 + 0.005 \frac{\$}{\text{hour}}\times 1000 \text{ hours	}
= \$14.00.
\end{equation*}

\item How long would you have to use the CFL to make it worth having
paid the higher purchase price?

From my previous work, the answer will be less than 1000 hours. I can
find the exact time by algebra or by trial and error. It turns out to
be 467 or about 500 hours. That makes sense --- the CFL costs a penny
and a half less per hour to run, which means it will take about 500
hours to save the \$7 difference in initial cost.

\item Are the five numbers given in this exercise reasonable?

Waiting for student input.

\item What does ``incandescent'' mean? Why are incandescent light
bulbs called that?

``Incandescent'' means 
``giving off light because it's hot''. That's just how ordinary old
fashioned light bulbs work. There's a thin filament (wire) inside that
heats up and glows.

\end{abcd}

\end{sol}

\end{exx}


\begin{exx}{\untested\hassolution\sref{energypower}\gref{energyandpower}}
Not flying to London.

In \theTimes{} on April 25, 2011 you could read that

\begin{qwrap}
\begin{quotation}
``The killer application is collaboration; that is what people want,''
Dr. DeFanti said. ``You can save so much energy by not flying to London
that it will run a rack of computers for a year.''%
\webref{www.nytimes.com/2011/04/26/science/26planetarium.htm}
\end{quotation}
\sourceinfo{www.nytimes.com/2011/04/26/science/26planetarium.htm}
\end{qwrap}

Estimate the energy costs of flying to London and running a rack of
computers for a year to see if they are of the same order of
magnitude.

\begin{hint}
If you're a physicist you can make these estimates with your common
knowledge. If you're not, you can put together reliable information
from the web. Try searching for the energy cost of flying an airplane
and the energy cost of running a computer.
\end{hint}
\begin{sol}
From \url{www.inference.phy.cam.ac.uk/withouthotair/c5/page_35.shtml}:

\begin{quotation}
Imagine that you make one intercontinental trip per year by plane. How
much energy does that cost?

A Boeing 747-400 with 240 000 litres of fuel carries 416 passengers
about 8,800 miles (14,000 km). And fuel's calorific value is 10 kWh
per litre. (We learned that in Chapter 3.) So the energy cost of one
full-distance roundtrip on such a plane, if divided equally among the
passengers, is 
\begin{equation*}
\frac{2 \times 240,000 \text{ litre}}{416 \text{ passengers}}
 	\times 10 \text{ kWh/litre} \approx  12,000 \text{ kWh per passenger}.
\end{equation*}
If you make one such trip per year, then your average energy consumption
per day is 
\begin{equation*}
\frac{12,000 \text{ kWh}}{365 {\text{ days}}}
\approx  33 \text{ kWh/day}.
\end{equation*}
\end{quotation}

The round trip airline distance from Boston to London is about 6,500
miles, so that trip will cost somewhat less than the trip above, 
figure 24 kwh/day. 

At (say) 100 watts to run a computer (that's the right order of
magnitude) you use about 2,400 watt-hours or 2.4 kwh in a day. So 24
kwh will power ten computers for a day. That's a pretty small
rack, but the order of magnitude is right.
\end{sol}
\end{exx}

\end{ExtraExercises}