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Find the amount \(A\) in the account as a function of the term of the investment \(t\) in years.
To the nearest cent, determine how much is in the account after \(5\), \(10\), \(30\) and \(35\) years.
To the nearest year, determine how long will it take for the initial investment to double.
Find and interpret the average rate of change of the amount in the account from the end of the fourth year to
the end of the fifth year, and from the end of the thirty-fourth year to the end of the thirty-fifth year. Round your
answer to two decimal places.
\(\$500\) is invested in an account which offers \(0.75 \%\), compounded monthly.
It will take approximately \(92\) years for the investment to double.
The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(3.88\). This means that the
investment is growing at an average rate of \(\$3.88\) per year at this point. The average rate of change from the end of the thirty-fourth
year to the end of the thirty-fifth year is approximately \(4.85\). This means that the investment is growing at an average rate of \(\$4.85\) per
year at this point.
\(\$500\) is invested in an account which offers \(0.75 \%\), compounded continuously.
It will take approximately \(92\) years for the investment to double.
The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(3.88\). This means that the
investment is growing at an average rate of \(\$3.88\) per year at this point. The average rate of change from the end of the thirty-fourth
year to the end of the thirty-fifth year is approximately \(4.86\). This means that the investment is growing at an average rate of \(\$4.86\) per
year at this point.
\(\$1000\) is invested in an account which offers \(1.25 \%\), compounded monthly.
It will take approximately \(55\) years for the investment to double.
The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(13.22\). This means that the
investment is growing at an average rate of \(\$13.22\) per year at this point. The average rate of change from the end of the thirty-fourth
year to the end of the thirty-fifth year is approximately \(19.23\). This means that the investment is growing at an average rate of \(\$19.23\) per
year at this point.
\(\$1000\) is invested in an account which offers \(1.25 \%\), compounded continuously.
It will take approximately \(55\) years for the investment to double.
The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(13.22\). This means that the
investment is growing at an average rate of \(\$13.22\) per year at this point. The average rate of change from the end of the thirty-fourth
year to the end of the thirty-fifth year is approximately \(19.24\). This means that the investment is growing at an average rate of \(\$19.24\) per
year at this point.
\(\$5000\) is invested in an account which offers \(2.125 \%\), compounded monthly.
It will take approximately \(33\) years for the investment to double.
The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(116.80\). This means that the
investment is growing at an average rate of \(\$116.80\) per year at this point. The average rate of change from the end of the thirty-fourth
year to the end of the thirty-fifth year is approximately \(220.83\). This means that the investment is growing at an average rate of \(\$220.83\) per
year at this point.
\(\$5000\) is invested in an account which offers \(2.125 \%\), compounded continuously.
It will take approximately \(33\) years for the investment to double.
The average rate of change from the end of the fourth year to the end of the fifth year is approximately \(116.91\). This means that the
investment is growing at an average rate of \(\$116.91\) per year at this point. The average rate of change from the end of the thirty-fourth
year to the end of the thirty-fifth year is approximately \(221.17\). This means that the investment is growing at an average rate of \(\$221.17\) per
year at this point.
Look back at your answers to Exercises basicinterestexfirst - basicinterestexlast. What can be said about the difference between monthly compounding and
continuously compounding the interest in those situations? With the help of your classmates, discuss scenarios
where the difference between monthly and continuously compounded interest would be more dramatic. Try
varying the interest rate, the term of the investment and the principal. Use computations to support your answer.
How much money needs to be invested now to obtain \(\$2000\) in 3 years if the interest rate in a savings account is \(0.25 \%\), compounded
continuously? Round your answer to the nearest cent.
How much money needs to be invested now to obtain \(\$5000\) in 10 years if the interest rate in a CD is \(2.25 \%\), compounded monthly?
Round your answer to the nearest cent.
On May, 31, 2009, the Annual Percentage Rate listed at Jeff’s bank for regular savings accounts was \(0.25\%\) compounded monthly.
Use Equation compoundinterest to answer the following.
The Annual Percentage Yield is the simple interest rate that returns the same amount of interest after
one year as the compound interest does. With the help of your classmates, compute the APY for this
investment.
\(\left (1 + \frac {0.0225}{12}\right )^{12} \approx 1.0227\) so the APY is 2.27%
A finance company offers a promotion on \(\$5000\) loans. The borrower does not have to make any payments for the first three years,
however interest will continue to be charged to the loan at \(29.9 \%\) compounded continuously. What amount will be due at the end of
the three year period, assuming no payments are made? If the promotion is extended an additional three years, and no
payments are made, what amount would be due?
Use Equation compoundinterest to show that the time it takes for an investment to double in value does not depend on the principal \(P\), but
rather, depends only on the APR and the number of compoundings per year. Let \(n = 12\) and with the help of your classmates
compute the doubling time for a variety of rates \(r\). Then look up the Rule of 72 and compare your answers to what that rule
says. If you’re really interested (Awesome pun!) in Financial Mathematics, you could also compare and contrast the Rule
of 72 with the Rule of 70 and the Rule of 69.
In Exercises radioactivefirst - radioactivelast, we list some radioactive isotopes and their associated half-lives. Assume that each decays
according to the formula \(A(t) = A_{0}e^{kt}\) where \(A_{0}\) is the initial amount of the material and \(k\) is the decay constant. For each
isotope:
Find the decay constant \(k\). Round your answer to four decimal places.
Find a function which gives the amount of isotope \(A\) which remains after time \(t\). (Keep the units of \(A\) and \(t\) the same
as the given data.)
Determine how long it takes for \(90 \%\) of the material to decay. Round your answer to two decimal places. (HINT: If \(90 \%\)
of the material decays, how much is left?)
Cobalt 60, used in food irradiation, initial amount 50 grams, half-life of \(5.27\) years.
\(k = \frac {\ln (1/2)}{5.27} \approx -0.1315\)
\(A(t) = 50e^{-0.1315t}\)
\(t = \frac {\ln (0.1)}{-0.1315} \approx 17.51\) years.
Phosphorus 32, used in agriculture, initial amount 2 milligrams, half-life \(14\) days.
Americium 241, used in smoke detectors, initial amount 0.29 micrograms, half-life \(432.7\) years.
\(k = \frac {\ln (1/2)}{432.7} \approx -0.0016\)
\(A(t) = 0.29e^{-0.0016t}\)
\(t = \frac {\ln (0.1)}{-0.0016} \approx 1439.11\) years.
Uranium 235, used for nuclear power, initial amount \(1\) kg, half-life \(704\) million years.
\(k = \frac {\ln (1/2)}{704} \approx -0.0009846\)
\(A(t) = e^{-0.0009846t}\)
\(t = \frac {\ln (0.1)}{-0.0009846} \approx 2338.60\) million years, or \(2.339\) billion years.
With the help of your classmates, show that the time it takes for \(90 \%\) of each isotope listed in Exercises radioactivefirst - radioactivelast to decay does not
depend on the initial amount of the substance, but rather, on only the decay constant \(k\). Find a formula, in terms of \(k\) only, to
determine how long it takes for \(90 \%\) of a radioactive isotope to decay.
In Example cardepreciationex in Section ExponentialFunctions, the exponential function \(V(x) = 25 \left (\frac {4}{5}\right )^{x}\) was used to model the value of a car over time. Use a change of base
formula to rewrite the model in the form \(V(t) = 25e^{kt}\).
The Gross Domestic Product (GDP) of the US (in billions of dollars) \(t\) years after the year 2000 can be modeled by:
\[ G(t) = 9743.77 e^{0.0514t}\]
Find and interpret \(G(0)\).
\(G(0) = 9743.77\) This means that the GDP of the US in 2000 was \(\$9743.77\) billion dollars.
According to the model, what should have been the GDP in 2007? In 2010? (According to the US Department of
Commerce, the 2007 GDP was \(\$14,369.1\) billion and the 2010 GDP was \(\$14,657.8\) billion.)
\(G(7) = 13963.24\) and \(G(10) = 16291.25\), so the model predicted a GDP of \(\$ 13,963.24\) billion in 2007 and \(\$ 16,291.25\) billion in 2010.
The diameter \(D\) of a tumor, in millimeters, \(t\) days after it is detected is given by:
\[D(t) = 15e^{0.0277t} \]
What was the diameter of the tumor when it was originally detected?
\(D(0) = 15\), so the tumor was 15 millimeters in diameter when it was first detected.
How long until the diameter of the tumor doubles?
\(t = \frac {\ln (2)}{0.0277} \approx 25\) days.
Under optimal conditions, the growth of a certain strain of E. Coli is modeled by the Law of Uninhibited Growth \(N(t) = N_{0} e^{kt}\) where \(N_{0}\) is the
initial number of bacteria and \(t\) is the elapsed time, measured in minutes. From numerous experiments, it has
been determined that the doubling time of this organism is 20 minutes. Suppose 1000 bacteria are present
initially.
Find the growth constant \(k\). Round your answer to four decimal places.
\(k = \frac {\ln (2)}{20} \approx 0.0346\)
Find a function which gives the number of bacteria \(N(t)\) after \(t\) minutes.
\(N(t) = 1000e^{0.0346 t}\)
How long until there are 9000 bacteria? Round your answer to the nearest minute.
Yeast is often used in biological experiments. A research technician estimates that a sample of yeast suspension contains
2.5 million organisms per cubic centimeter (cc). Two hours later, she estimates the population density to be 6 million
organisms per cc. Let \(t\) be the time elapsed since the first observation, measured in hours. Assume that the yeast growth
follows the Law of Uninhibited Growth \(N(t) = N_{0} e^{kt}\).
Find the growth constant \(k\). Round your answer to four decimal places.
The Law of Uninhibited Growth also applies to situations where an animal is re-introduced into a suitable environment. Such a
case is the reintroduction of wolves to Yellowstone National Park. According to the National Park Service, the wolf population
in Yellowstone National Park was 52 in 1996 and 118 in 1999. Using these data, find a function of the form \(N(t) = N_{0} e^{kt}\) which models the
number of wolves \(t\) years after 1996. (Use \(t = 0\) to represent the year 1996. Also, round your value of \(k\) to four decimal
places.) According to the model, how many wolves were in Yellowstone in 2002? (The recorded number is
272.)
During the early years of a community, it is not uncommon for the population to grow according to the Law of Uninhibited
Growth. According to the Painesville Wikipedia entry, in 1860, the Village of Painesville had a population of 2649. In
1920, the population was 7272. Use these two data points to fit a model of the form \(N(t) = N_{0} e^{kt}\) were \(N(t)\) is the number of
Painesville Residents \(t\) years after 1860. (Use \(t = 0\) to represent the year 1860. Also, round the value of \(k\) to four decimal
places.) According to this model, what was the population of Painesville in 2010? (The 2010 census gave the
population as 19,563) What could be some causes for such a vast discrepancy? For more on this, see Exercise
PainesvillePopulationManyPoints.
\(N_{0} = 2649\), \(k = \frac {1}{60} \ln \left ( \frac {7272}{2649}\right ) \approx 0.0168\), \(N(t) = 2649e^{0.0168t}\). \(N(150) \approx 32923\), so the population of Painesville in 2010 based on this model would have been 32,923.
The population of Sasquatch in Bigfoot county is modeled by
\[P(t) = \dfrac {120}{1 + 3.167e^{-0.05t}}\]
where \(P(t)\) is the population of Sasquatch \(t\) years after
\(2010\).
Find and interpret \(P(0)\).
\(P(0) = \frac {120}{4.167} \approx 29\). There are 29 Sasquatch in Bigfoot County in 2010.
Find the population of Sasquatch in Bigfoot county in 2013 rounded to the nearest Sasquatch.
To the nearest year, when will the population of Sasquatch in Bigfoot county reach 60?
\(t = 20 \ln (3.167) \approx 23\) years.
Find and interpret \(\ds {\lim _{t \rightarrow \infty } P(t)}\) analytically. Check your answer using a graphing utility.
We find \(\ds {\lim _{t \rightarrow \infty } P(t) = 120}\). As time goes by, the Sasquatch Population in Bigfoot County will approach 120. Graphically, \(y = P(x)\) has a horizontal
asymptote \(y=120\).
Let \(f(x) = \dfrac {10}{1+e^{-x+1}}\).
From Calculus, we know the inflection point of the graph of \(y=f(x)\) is \((1,5)\). This means the function is increasing the
fastest at \(x=1\), or, equivalently, the slope at \((1,5)\) is the largest anywhere on the graph. Graph \(y=f(x)\) using a graphing utility
and convince yourself of the reasonableness of this claim.
Find average rate of change of \(f\) over each of the intervals below. What do you guess the slope of the curve is
at \((1,5)\)? Zoom in on the graph near \((1,5)\) to check your guess.
\([0.75, 1]\)
\(\approx 2.487\)
\([0.9, 1]\)
\(\approx 2.498\)
\([0.99,1]\)
\(\approx 2.500\)
\([1, 1.01]\)
\(\approx 2.500\)
\([1, 1.1]\)
\(\approx 2.498\)
\([1, 1.25]\)
\(\approx 2.487\)
The average rates of change are listed in order above. They suggest slope at \((1,5)\) is \(\answer {2.5}\).
The half-life of the radioactive isotope Carbon-14 is about 5730 years.
Use Equation radioactivedecay to express the amount of Carbon-14 left from an initial \(N\) milligrams as a function of time \(t\) in years.
What percentage of the original amount of Carbon-14 is left after 20,000 years?
\(A(20000) \approx 0.088978 \cdot N\) so about 8.9% remains
If an old wooden tool is found in a cave and the amount of Carbon-14 present in it is estimated to be only 42% of the
original amount, approximately how old is the tool?
\(t \approx \dfrac {\ln (.42)}{-0.00012097} \approx 7171\) years old
Radiocarbon dating is not as easy as these exercises might lead you to believe. With the help of your classmates,
research radiocarbon dating and discuss why our model is somewhat over-simplified.
Carbon-14 cannot be used to date inorganic material such as rocks, but there are many other methods of radiometric dating
which estimate the age of rocks. One of them, Rubidium-Strontium dating, uses Rubidium-87 which decays to Strontium-87
with a half-life of 50 billion years. Use Equation radioactivedecay to express the amount of Rubidium-87 left from an initial 2.3 micrograms as a
function of time \(t\) in billions of years. Research this and other radiometric techniques and discuss the margins of error for
various methods with your classmates.
\(A(t) = 2.3e^{-0.0138629t}\)
Find and interpret the relative rate of change of \(A(t)\) in Equation compoundinterest over the interval \(\left [t, t+\frac {1}{n} \right ]\).
The relative rate of change of \(A(t)\) over \(\left [t, t+\frac {1}{n} \right ]\) is \(\frac {r}{n}\) which is the annual percentage rate divided by the number of compoundings per year
– that is, the percentage growth rate over one compounding.
Use Equation radioactivedecay to show that \(k = -\dfrac {\ln (2)}{h}\) where \(h\) is the half-life of the radioactive isotope.
A pork roast (This roast was enjoyed by Jeff and his family on June 10, 2009. This is real data, folks!) was taken out of
a hardwood smoker when its internal temperature had reached \(180^{\circ }\)F and it was allowed to rest in a \(75^{\circ }\)F house for 20 minutes after
which its internal temperature had dropped to \(170^{\circ }\)F. Assuming that the temperature of the roast follows Newton’s Law of Cooling
(Equation newtonslawofcooling),
Express the temperature \(T\) (in \(^{\circ }\)F) as a function of time \(t\) (in minutes).
\(T(t) = 75 + 105e^{-0.005005t}\)
Find the time at which the roast would have dropped to \(140^{\circ }\)F had it not been eaten.
The roast would have cooled to \(140^{\circ }\)F in about 95 minutes.
In reference to Exercise pursuitfurther in Section PowerFunctions, if Fritzy the Fox’s speed is the same as Chewbacca the Bunny’s speed, Fritzy’s pursuit
curve is given by
Graph this path for \(x > 0\) using a graphing utility. Investigate \(\ds {\lim _{x \rightarrow 0^{+}} y(x)}\) and interpret.
From the graph, it appears that \(\ds {\lim _{x \rightarrow 0^{+}} y(x) = \infty }\). This is due to the presence of the \(\ln (x)\) term in the function. This means that Fritzy will never
catch Chewbacca, which makes sense since Chewbacca has a head start and Fritzy only runs as fast as he does.
The current \(i\) measured in amps in a certain electronic circuit with a constant impressed voltage of 120 volts is given by \(i(t) = 2 - 2e^{-10t}\)
where \(t \geq 0\) is the number of seconds after the circuit is switched on. Determine \(\ds {\lim _{t \rightarrow \infty } i(t)}\). (This is called the steady state
current.)
The steady state current is 2 amps.
If the voltage in the circuit in Exercise explogsappcircuitone above is switched off after 30 seconds, the current is given by the piecewise-defined
function
With the help of a graphing utility, graph \(y = i(t)\) and discuss with your classmates the physical significance of the two parts of the
graph \(0 \leq t < 30\) and \(t \geq 30\).
In Exercise parabolicbridgecable in Section QuadraticFunctions, we stated that the cable of a suspension bridge formed a parabola but that a free hanging cable did
not. A free hanging cable forms a catenary and its basic shape is given by \(y = \frac {1}{2}\left (e^{x} + e^{-x}\right )\).
where \(x\) and \(y\) are measured in feet and \(-315 \leq x \leq 315\). Find the highest point on the arch.
630 feet
In Exercise APLcats in Section QuadraticFunctions, we examined the data set given below which showed how two cats and their surviving offspring can
produce over 80 million cats in just ten years.
Plot \(x\) versus \(\ln (x)\) as was done on page ?? using a graphing utility.
Find a linear model for this new data and comment on its goodness of fit and find an exponential model for the
original data and comment on its goodness of fit.
Year \(x\)
1
2
3
4
5
6
7
8
9
10
Number of
Cats \(N(x)\)
12
66
382
2201
12680
73041
420715
2423316
13968290
80399780
The linear regression on the data below is \(y = 1.74899x + 0.70739\) with \(r^{2} \approx 0.999995\).
This is an excellent fit.
\(x\)
1
2
3
4
5
6
7
8
9
10
\(\ln (N(x))\)
2.4849
4.1897
5.9454
7.6967
9.4478
11.1988
12.9497
14.7006
16.4523
18.2025
\(N(x) = 2.02869(5.74879)^{x} = 2.02869e^{1.74899x}\) with \(r^{2} \approx 0.999995\). This is also an excellent fit and corresponds to our linearized model because \(\ln (2.02869) \approx 0.70739\).
In Example LorenzEx in Section PowerFunctions, we fit a power function of the form \(L(x) = a x^{p}\) to a set of data, \((x, L(x))\). In this exercise, we use logs to linearize this
data using the same methods presented on page ??, but with a slight difference in interpretation.
Starting with \(L(x) = a x^{p}\), take natural logs of both sides of the equation and use log properties to rewrite the resulting
equation as: \(\ln (L(x)) = p \ln (x) + \ln (a)\).
Use a graphing utility to find a least squares regression line using the data \((\ln (x), \ln (L(x)))\).
NOTE: In this situation, we are plotting \(\ln (x)\) versus \(\ln (L(x))\) instead of \(x\) versus \(\ln (L(x))\).
The linearized model is: \(\ln (L(x)) \approx 2.106 \ln (x) - 5.268\) with an \(r^2 \approx 0.9914\).
Find the slope \(p\) of the regression line and the intercept \(\ln (a)\). Use these to construct a model of the form \(L(x) = a x^{p}\). Find and interpret
\(L(90)\).
\(L(x) = 0.005154 x^{2.106}\). \(L(90) \approx 67.3\) meaning the bottom \(90 \%\) of wage earners take home \(67.3 \%\) of the total national income. Said differently, according to this
model, the top \(10 \%\) of wage earners take home \(32.7 \%\) of the total national income.
Graph both the model obtained in Example LorenzEx and the model obtained in part newlorenzepart along with the original data. What do you
notice?
This exercise is a follow-up to Exercise PainesvillePopulationTwoPoint which more thoroughly explores the population growth of Painesville, Ohio.
According to Wikipedia, the population of Painesville, Ohio is given by
Year \(t\)
1860
1870
1880
1890
1900
1910
1920
1930
1940
1950
Population
2649
3728
3841
4755
5024
5501
7272
10944
12235
14432
Year \(t\)
1960
1970
1980
1990
2000
Population
16116
16536
16351
15699
17503
Use a graphing utility to perform an exponential regression on the data from 1860 through 1920 only, letting \(t = 0\)
represent the year 1860 as before. How does this model compare with the model you found in Exercise PainesvillePopulationTwoPoint? Use
the graphing utility’s exponential model to predict the population in 2010. (The 2010 census gave the population
as 19,563)
We get: \(y = 2895.06 (1.0147)^{x}\). Graphing this along with our answer from Exercise PainesvillePopulationTwoPoint over the interval \([0,60]\) shows that they are pretty close. From
this model, \(y(150) \approx 25840\) which once again overshoots the actual data value.
The logistic model fit to all of the given data points for the population of Painesville \(t\) years after 1860 (again, using \(t = 0\) as
1860) is
According to this model, what should the population of Painesville have been in 2010? (The 2010 census gave
the population as 19,563.) What is the population limit of Painesville?
\(P(150) \approx 17914\), so this model predicts 17,914 people in Painesville in 2010, a more conservative number than was recorded in the
2010 census. We have \(\ds {\lim _{t \rightarrow \infty } P(t) = 18691}\), so the limiting population of Painesville based on this model is 18,691 people.
According to OhioBiz, the census data for Lake County, Ohio is as follows:
Year \(t\)
1860
1870
1880
1890
1900
1910
1920
1930
1940
1950
Population
15576
15935
16326
18235
21680
22927
28667
41674
50020
75979
Year \(t\)
1960
1970
1980
1990
2000
Population
148700
197200
212801
215499
227511
Use a graphing utility to fit a logistic model to these data with \(x = 0\) representing the year 1860.
\(y = \frac {242526}{1+874.63e^{-0.07113x}}\), where \(x\) is the number of years since 1860.
Graph the data and your model using a graphing utility to judge the reasonableness of the fit.
Use this model to estimate the population of Lake County in 2010. (The 2010 census gave the population to be
230,041.)
\(y(140) \approx 232884\), so this model predicts 232,884 people in Lake County in 2010.
According to your model, what is the population limit of Lake County, Ohio?
We get \(\ds {\lim _{x \rightarrow \infty } y = 242526}\), so the limiting population of Lake County based on this model is 242,526 people.
According to facebook, the number of active users of facebook has grown significantly since its initial launch from a Harvard
dorm room in February 2004. The chart below has the approximate number \(U(x)\) of active users, in millions, \(x\) months after
February 2004. For example, the first entry \((10, 1)\) means that there were \(1\) million active users in December 2004 and the last entry \((77, 500)\)
means that there were \(500\) million active users in July 2010.
Month \(x\)
10
22
34
38
44
54
59
60
62
65
67
70
72
77
Active Users in
Millions \(U(x)\)
1
5.5
12
20
50
100
150
175
200
250
300
350
400
500
With the help of your classmates, find a model for this data.
Each Monday during the registration period before the Fall Semester at LCCC, the Enrollment Planning Council gets a report
prepared by the data analysts in Institutional Effectiveness and Planning. (Thanks to Dr. Wendy Marley and her staff for
this data and Dr. Marcia Ballinger for the permission to use it in this problem.) While the ongoing enrollment
data is analyzed in many different ways, we shall focus only on the overall headcount. Below is a chart of the
enrollment data for Fall Semester 2008. It starts 21 weeks before “Opening Day” and ends on “Day 15” of the
semester, but we have relabeled the top row to be \(x = 1\) through \(x = 24\) so that the math is easier. (Thus, \(x = 22\) is Opening
Day.)
Week \(x\)
1
2
3
4
5
6
7
8
Total
Headcount
1194
1564
2001
2475
2802
3141
3527
3790
Week \(x\)
9
10
11
12
13
14
15
16
Total
Headcount
4065
4371
4611
4945
5300
5657
6056
6478
Week \(x\)
17
18
19
20
21
22
23
24
Total
Headcount
7161
7772
8505
9256
10201
10743
11102
11181
With the help of your classmates, find a model for this data. Unlike most of the phenomena we have studied in
this section, there is no single differential equation which governs the enrollment growth. Thus there is no
scientific reason to rely on a logistic function even though the data plot may lead us to that model. What are
some factors which influence enrollment at a community college and how can you take those into account
mathematically?
When we wrote this exercise, the Enrollment Planning Report for Fall Semester 2009 had only 10 data points for the first 10
weeks of the registration period. Those numbers are given below.
Week \(x\)
1
2
3
4
5
6
7
8
9
10
Total
Headcount
1380
2000
2639
3153
3499
3831
4283
4742
5123
5398
With the help of your classmates, find a model for this data and make a prediction for the Opening Day enrollment as well as
the Day 15 enrollment. (WARNING: The registration period for 2009 was one week shorter than it was in 2008 so Opening
Day would be \(x = 21\) and Day 15 is \(x = 23\).)
