\magnification=\magstep1
\centerline{MONT 104N -- Modeling the Environment}
\centerline{Solutions for Sample Midterm Exam Problems}
\centerline{October 22, 2012}

\bigskip
\noindent
{\it Sample Mathematical Questions}
\bigskip
\noindent
A.  (Refer to the tables on pages 6 - 8; something like those
would be supplied for your use on a question like this.)
\medskip
\item{1.}  A committee on rail freight transport recommends
increasing the mass that can be carried in a trailer
from 42 to 46 tonnes.  How much are these masses in kilograms?
In pounds?
\bigskip
\item{} {\it Solution:}  One tonne is 1000 kilograms.  So 
the recommendation is that the mass increases from 42000 kg to 
46000 kg.  Similarly, from the table on page 8, we see
1 lb = .4536 kg, so one 1 kg = 1/.4536 lb = 2.2046 lb.
This gives the increase as from 92593.2 lb to 101411.6 lb.  
\bigskip 
\item{2.}  About 24 million liters of rainwater are 
discharged by a storm drain into a river.  The
concentration of toxic arsenic in the rainwater
averages 170 micrograms of arsenic per liter.
How many kilograms of arsenic are discharged into
the river each year.
\bigskip
\item{} {\it Solution:}  Since one microgram is $10^{-6}$ gram,
the amount of arsenic discharged is
$$170 \times 10^{-6} {{\rm g}\over
{\rm liter}} \times 24 \times 10^{6}\ {\rm liter} \times {10^{-3}} {{\rm kg} \over {\rm g}} = 4.08\ {\rm kg}.$$
\bigskip
\noindent
B.  In 2002, 10731 species were listed on the World
Conservation Union ``red list'' of endangered species.
In 2004, the corresponding number was 15042.
\medskip
\item{1.}  What was the ratio of the size of the 
2004 list to the size of the 2002 list?
\bigskip
\item{} {\it Solution:}  The ratio is ${15042\over 10731} \doteq 1.402$.
\bigskip
\item{2.}  What was the percentage increase
in the size of the list from 2002 to 2004?
\bigskip
\item{} {\it Solution:}  The percentage increase is
$${15042 - 10731\over 10731} \times 100\% = 40.2 \%$$
(Note that you can ``see this'' in the answer to part 1 if 
you subtract the leading 1.)
\item{3.}  It is not known exactly how many
species of plants and animals there are on the 
Earth today.  But say that number is estimated as
12 million. What percentage of that total number
appeared on the red list in each year?
\bigskip
\item{} {\it Solution:}  In 2002, the percent was
$${10731\over 12000000} \times 100\% \doteq .089 \%$$
Similarly, in 2004, the percent was
$${15042\over 12000000} \times 100\% \doteq .125 \%$$
(around 1 tenth of one percent in each case).  These are not 
huge numbers of course, but any loss of biodiversity
means that potentially valuable or useful biochemical
products of these plants and animals would be lost.   
\bigskip
\noindent
C.  The following table gives several African countries,
their total forest areas, and their areas of legally protected 
forests (in square kilometers)
\bigskip
$$\matrix{{\rm Country} & {\rm Total} & {\rm Protected}\cr
{\rm Cameroon} & 289965 & 17845\cr
{\rm Congo} & 278979 & 12935\cr
{\rm Gabon} & 239369 & 8975\cr
{\rm Zaire} & 1439178 & 93160 \cr}$$
\item{1.}  Which country has the largest protected forest area?
\bigskip 
\item{} {\it Solution:}  The numerically largest protected forest area is in Zaire.
\bigskip
\item{2.}  Which country has the largest percentage of protected forest?
\bigskip
\item{}  {\it Solution:}  The percentages of protected forest are
$$\eqalign{
{\rm Cameroon:\ \ }  {17845\over 289965} \times 100\% &= 6.15\%\cr
{\rm Congo:\ \ }  {12935\over 278979} \times 100\% &= 4.64\%\cr
{\rm Gabon:\ \ }  {8975\over 239369} \times 100\% &= 3.75\%\cr
{\rm Zaire:\ \ }  {93160\over 1439178} \times 100\% &= 6.47\%\cr
}$$
So Zaire also has the largest percentage of protected forest.  
\bigskip
\noindent
D.  A cubic meter of air has a mass of about 1.25 kg.  About 2 mg of
that mass is methane.  What is the mass concentration of methane in 
air, expressed in parts per million?
\bigskip
\noindent
{\it Solution:}  The mass concentration can be computed
like this, converting the numerator and denominator
to units of kilograms:
$${2 \times 10^{-6}  \over 1.25} = 1.6 \times 10^{-6} = 
{1.6 \over 10^6},$$
so: 1.6 parts per million methane.  (Note:  Methane is a
much more ``potent'' greenhouse gas than carbon dioxide.  It has 25 times
the impact of carbon dioxide in trapping heat from solar radiation in 
the atmosphere.)
\bigskip
\noindent
E.  The US EPA Toxic Release Inventory for 2000 includes the 
following information:
$$\matrix{{\rm State} & {\rm Toxics\ (in\ 10^6 lb)}\cr
{\rm Nevada} & 1000\cr
{\rm Utah} & 956 \cr
{\rm Arizona} & 744\cr
{\rm Alaska} & 535 \cr
{\rm Texas} & 302 \cr}$$
\item{1.}  Construct a bar chart showing this information.  Label your vertical axis.
\bigskip
\item{} {\it Solution:}  Mark the vertical axis in units of $100 \times 10^6 = 10^8$
lb of toxics released.  Then the bar chart will have 1 bar of 
height 10 for Nevada, one of height 9.56 for Utah, etc.  
\bigskip 
\item{2.}  Construct a pie chart showing the portions of the 
total toxics from these five states that each state contributes.
\bigskip
\item{} {\it Solution:}  The total is $3537 \times 10^6$ lb of toxics
released.  So the portions are:
$$\eqalign{
{\rm Nevada:\ \ }  {1000\over 3537} \times 100\% &\doteq 28\%\cr
{\rm Utah:\ \ }  {956\over 3537} \times 100\% &\doteq 27\%\cr
{\rm Arizona:\ \ }  {744\over 3537} \times 100\% &\doteq 21\%\cr
{\rm Alaska:\ \  }  {535\over 3537} \times 100\% &\doteq 15\%\cr
{\rm Texas:\ \ }  {302\over 3537} \times 100\% &\doteq 8.5\%\cr
}$$
(Note that the total is not quite $100\%$ because of rounding.)
\bigskip
\item{3.}  Which state or states is approximately ``twice as
toxic'' as Alaska.
\bigskip
\item{} {\it Solution:} Both Nevada and Utah are close to ``twice
as toxic'' as Alaska.   
\bigskip
\noindent
F.  Look at the plot in Figure 3-28 on page 71 of the text, which 
shows the gray wolf population in Wisconsin from 1980 to 2002.
\medskip
\noindent
\item{1.}  Note that the portion of the graph representing
the years 1980 to 1992 looks quite linear.  Estimate
a linear model for this period by using the line
through the two points $(1980,50)$ and $(1990,60)$.  (Determine
the equation of the line through those points.)
\bigskip
\item{} {\it Solution:}  If we make the actual year be $x$ and the 
wolf population $y$, then the equation of the line would be
$$y - 50 = \left({60 - 50\over 1990 - 1980}\right) (x - 1980), {\rm\ or\ } y = x - 1930$$
(It would also be possible to use $x = $ years since $1980$, 
so $x = 0 \Leftrightarrow 1980$ and $x = 10 \Leftrightarrow 1990$.
Doing it that way, the equation would be $y = x + 50$.)
\bigskip
\item{2.}  Do the same for the period 1993-2002
using the two points $(1993,70)$ and $(2002,325)$.
\bigskip
\item{} {\it Solution:}  With $x = $ actual year and $y =$  wolf population,
we have
$$y - 70 = \left({325 - 70 \over 2002 - 1993}\right)(x - 1993),
{\rm\ or\ } y = 28.3 x - 56331.9$$
\item{3.}  What are the slopes of these two lines?  What are the 
units of the slope and what do they represent in real world
terms?
\bigskip
\item{} {\it Solution:}  The slope of the first line is 1 and the slope
of the second is 28.3.  These represent the growth rate of the wolf
population per year in the two different time periods.
\bigskip
\item{4.}  The graph in the spreadsheet (see course homepage) shows the 
least squares regression line for the whole data set, 
together with the computed slope and intercept values
for the regression line.  
Which method seems superior in this case -- using two
different lines for different time periods, or one 
regression line for the whole time?  Explain how 
you are making this decision.
\bigskip
\item{} {\it Solution:}  In this case it seems pretty clear
that the two separate lines do a better job of approximating
the actual wolf populations over the years in the two 
time periods.  Using a single line for the whole period
gives a consistent pattern in the residuals and is not
a close fit.
\bye
