Question 1

Solve the problem.
-The logistic growth functi $f ( t ) = \frac { 12,000 } { 1 + 199 e ^ { - 1.7 t } }$ models the number of people who have become ill with a particular infection t weeks after its initial outbreak in a particular community. How many people were ill after
9 weeks?
A) 12,000 people
B) 11,999 people
C) 12,200 people
D) 540 people

Accepted Answer

To find the number of people ill after 9 weeks, substitute $t = 9$ into the given logistic growth function: $f(9) = \frac{12,000}{1 + 199e^{-1.7 \times 9}}$. Calculating this gives approximately 11,999 people.

Question 2

Solve the problem.
-Money magazine reports that the percentage of trading days in which the Dasdaq loses or gains 2% or more has been increasing since 1995 indicating more volatility in the Dasdaq. Use a graphing utility to fit an exponential
Function to the data. Predict the percentage of trading days in 2001 having such swings in value. $\begin{array}{l|l}
\text { Year } & \% \text { of trading days } \
\hline 1995,0 & 2 \
1996,1 & 5 \
1997,2 & 8 \
1998,3 & 18 \
1999,4 & 23 \
2000,5 & 49
\end{array}$

A) $y=2.353 e^{0.610 x}, 91 \text { days }$
B) $\mathrm { y } = 1.669 \mathrm { x } 1.706,46$ days
C) $\mathrm { y } = 1.278 \mathrm { e } ^ { 0.611 \mathrm { x } } , 92 \text { days }$
D) $\mathrm { y } = 1.849 \mathrm { e } ^ { 0.531 \mathrm { x } } , 76$ days

Accepted Answer

Given the exponential nature of the increase in the percentage of trading days with significant swings, an exponential function is the most appropriate model. Choice A provides an exponential function and predicts 91 days, aligning with the trend of increasing volatility in the dataset.

Question 3

Solve the problem.
-A mechanic is testing the cooling system of a boat engine. He measures the engine's temperature over time. Use a graphing utility to fit a logistic function to the data. What is the carrying capacity of the cooling system? $\begin{array}{l|lllll}
\text { time, min } & 5 & 10 & 15 & 20 & 25 \
\hline \text { temperature, }^{\circ} \mathrm{F} & 100 & 180 & 270 & 300 & 305
\end{array}$

A) $y = \frac { 306.53 } { 1 + 7.92 \mathrm { e } ^ { - 0.254 x } } , 307 ^ { \circ } \mathrm { F }$

B) $y = \frac { 311.63 } { 1 + 8.1 \mathrm { e } ^ { - 0.253 \mathrm { x } } } , 312 ^ { \circ } \mathrm { F }$

C) $y = \frac { 314.79 } { 1 + 7.86 \mathrm { e } ^ { - 1.22 \mathrm { x } } } , 315 ^ { \circ } \mathrm { F }$

D) $y = \frac { 314.79 } { 1 + 7.86 \mathrm { e } ^ { - 0.246 \mathrm { x } } } , 315 ^ { \circ } \mathrm { F }$

Accepted Answer

The answer of Solve the problem.
-A mechanic is testing the...

Question 4

Solve the problem.
-A music store manager collected data regarding price and quantity demanded of cassette tapes every week for
10 weeks, and found that the exponential function of best fit to the data was p = 25 ·0.89q. Express the function
of best fit in the form $\mathrm { p } = \mathrm { p } _ { 0 } \mathrm { e } ^ { \mathrm { kq } }$ , and use this expression to predict the quantity demanded if the price is $8.50.

Accepted Answer

The answer of Solve the problem.
-A music store manager collected...

Question 5

Solve the problem.
-The logistic growth functi $f ( t ) = \frac { 440 } { 1 + 6.3 e ^ { - 0.2 t } }$ describes the population of a species of butterflies t months after they are introduced to a non-threatening habitat. What is the limiting size of the butterfly population that the
Habitat will sustain?
A) 60 butterflies
B) 440 butterflies
C) 6 butterflies
D) 880 butterflies

Accepted Answer

The limiting size of the butterfly population is found by examining the logistic growth function as $t$ approaches infinity. The term $6.3e^{-0.2t}$ approaches 0 as $t$ becomes very large, leaving the function to approach $\frac{440}{1+0} = 440$. Therefore, the limiting size of the butterfly population the habitat will sustain is 440 butterflies.

Question 6

Solve the problem.
-The population (in hundred thousands) for the Colonial United States in ten-year increments for the years
1700-1780 is given in the table. (Source: 1998 Information Please Almanac)  State whether the data can be more accurately modeled using an exponential function or a logarithmic function. Using a
graphing utility, find a model for population (in hundred thousands) as a function of decades since 1700.

Accepted Answer

The answer of Solve the problem.
-The population (in hundred thousands)...

Question 7

Solve the problem.
-The logistic growth functi $f ( t ) = \frac { 440 } { 1 + 7.8 e ^ { - 0.17 t } }$ describes the population of a species of butterflies t months after they are introduced to a non-threatening habitat. How many butterflies were initially introduced to the
Habitat?
A) 2 butterflies
B) 50 butterflies
C) 8 butterflies
D) 440 butterflies

Accepted Answer

The answer of Solve the problem.
-The logistic growth functi \(f...

Question 8

Solve the problem.
-The rates of death (in number of deaths per 100,000 population) for 20-24 year olds in the United States
between 1985-1993 are given below. (Source: NCHS Data Warehouse) $$\begin{array} { c | c } 
\text { Year } & \text { Rate of Death } \
\hline 1985 & 134.9 \
\hline 1987 & 154.7 \
\hline 1989 & 162.9 \
\hline 1991 & 174.5 \
\hline 1993 & 182.2
\end{array}$$ A logarithmic equation that models this data i $y = 57.76 + 48.56 \ln x$ 56 ln x where x represents the number of years since
1980 and y represents the rate of death in that year. Use this equation to predict the year in which the rate of
death for 20-24 year olds first exceeds 200.

Accepted Answer

The answer of Solve the problem.
-The rates of death (in...

Question 9

Solve the problem.
-The logistic growth functi $f ( t ) = \frac { 50,000 } { 1 + 2,499.0 e ^ { - 1.9 t } }$ models the number of people who have become ill with a particular infection t weeks after its initial outbreak in a particular community. How many people became ill
With this infection when the epidemic began?
A) 20 people
B) 2,500 people
C) 50,000 people
D) 2,499 people

Accepted Answer

To find the number of people who became ill when the epidemic began, we set $t = 0$ in the given logistic growth function: $f(0) = \frac{50,000}{1 + 2,499.0e^{-1.9(0)}} = \frac{50,000}{1 + 2,499} = \frac{50,000}{2,500} = 20$.

Question 10

Solve the problem.
-A nuclear scientist has a sample of 100 mg of a radioactive material which has a half-life in hours. She monitors the amount of radioactive material over a period of a day and obtains the following data. Use a graphing utility
To fit an exponential function to the data. Predict the amount of material remaining at 40 hours. $\begin{array}{l|lcccccc}
\text { Hours } & 0 & 5 & 10 & 15 & 20 & 25 & 30 \
\hline \mathrm{mg} & 100 & 68.3 & 45.2 & 31.3 & 21.5 & 14.6 & 9.8
\end{array}$
A) $\mathrm { y } = 92 \mathrm { e } ^ { - 0.0686 \mathrm { x } } , 5.9 \mathrm { mg }$
B) $\mathrm { y } = 100 \mathrm { e } ^ { - 0.077 \mathrm { x } } , 4.6 \mathrm { mg }$
C) $\mathrm { y } = 100 \mathrm { e } ^ { - 0.077 \mathrm { x } } , 6.7 \mathrm { mg }$

Accepted Answer

Given the data, the correct exponential function that models the decay of the radioactive material is $y = 100e^{-0.077x}$. Using this model to predict the amount of material remaining at 40 hours, we substitute $x = 40$ into the equation, resulting in approximately 4.6 mg of material remaining.

Question 11

Solve the problem.
-The logistic growth functi $f ( t ) = \frac { 800 } { 1 + 9.0 e ^ { - 0.16 t } }$ describes the population of a species of butterflies t months after they are introduced to a non-threatening habitat. How many butterflies are expected in the habitat after 15
Months?
A) 801 butterflies
B) 1,200 butterflies
C) 440 butterflies
D) 12,000 butterflies

Accepted Answer

To find the expected number of butterflies after 15 months, substitute $t = 15$ into the given logistic growth function: $f(15) = \frac{800}{1 + 9.0e^{-0.16 \times 15}}$. Calculating this gives approximately 440 butterflies.

Question 12

Solve the problem.
-After introducing an inhibitor into a culture of luminescent bacteria, a scientist monitors the luminosity produced by the culture. Use a graphing utility to fit a logarithmic function to the data. Predict the luminosity
After 20 hours. $\begin{array}{l|ccccccc}
\text { Time, hrs } & 2 & 3 & 4 & 5 & 8 & 10 & 15 \
\hline \text { Luminosity } & 77.4 & 60.8 & 54.5 & 45.8 & 30.0 & 24.3 & 10.5
\end{array}$

A) $y = 98.75 - 32.66 \ln ( x ) , 0.91$
B) $y = 107.55 - 41 \ln ( x ) , - 15.27$
C) $y = 112.97 - 45.97 \ln ( x ) , - 24.74$
D) $y = 100.5 - 32.7 \ln ( x ) , 2.54$

Accepted Answer

Given the data, a logarithmic function can be fitted using a graphing utility or statistical software. The correct choice, A, represents a logarithmic function that could be derived from such a process. The function $y = 98.75 - 32.66 \ln(x)$ likely fits the given data points well, and the number following the function, 0.91, could represent the correlation coefficient or another statistic related to the fit quality, but it's not part of the function itself for predicting luminosity.

Question 13

Solve the problem.
-Data representing the price and quantity demanded for hand-held electronic organizers were analyzed every
day for 15 days. The logarithmic function of best fit to the data was found to be $$\mathrm { p } = 398 - 73 \ln ( \mathrm { q } )$$ Use this to
predict the number of hand-held electronic organizers that would be demanded if the price were $275.

Accepted Answer

The answer of Solve the problem.
-Data representing the price and...

Question 14

Solve the problem.
-The logistic growth functi $f ( t ) = \frac { 24,000 } { 1 + 599.0 \mathrm { e } ^ { - 1.8 t } }$ models the number of people who have become ill with a particular infection t weeks after its initial outbreak in a particular community. What is the limiting size of the
Population that becomes ill?
A) 599 people
B) 24,000 people
C) 600 people
D) 48,000 people

Accepted Answer

The limiting size of the population that becomes ill is the horizontal asymptote of the logistic growth function, which is the value of the function as $t$ approaches infinity. In the given function $f(t) = \frac{24,000}{1 + 599.0e^{-1.8t}}$, as $t$ approaches infinity, $e^{-1.8t}$ approaches 0, making the function approach $\frac{24,000}{1 + 0} = 24,000$. Therefore, the limiting size of the population that becomes ill is 24,000 people.

Question 15

Solve the problem.
-The rates of death (in number of deaths per 100,000 population) for 1-4 year olds in the United States between
1980-1995 are given below. (Source: NCHS Data Warehouse) $$\begin{array} { c | c } 
\text { Year } & \text { Rate of Death } \
\hline 1980 & 91.4 \
\hline 1985 & 74.5 \
\hline 1990 & 69.3 \
\hline 1995 & 61.3
\end{array}$$ A logarithmic equation that models this data i $$y = 822.99 - 167.55 \ln x$$ 167.55 ln x where x represents the number of years
since 1900. Use this equation to predict the rate of death for 1-4 year olds in 2005.

Accepted Answer

The answer of Solve the problem.
-The rates of death (in...

Question 16

Solve the problem.
-A biologist has a bacteria sample. She records the amount of bacteria every week for 8 weeks and finds that the
exponential function of best fit to the data i $\mathrm { A } = 150 \cdot 1.79 \mathrm { t }$ . Express the function of best fit in the form $A = A_{ 0}e ^ { k t }$

Accepted Answer

The answer of Solve the problem.
-A biologist has a bacteria...

Question 17

Solve the problem.
-In a Psychology class, the students were tested at the end of the course on a final exam. Then they were retested with an equivalent test at subsequent time intervals. Their average scores after t months are given in the table. $$\begin{array} { l | c c c c c } 
\text { Time, t ( in months) } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Score, y ( in percentage) } & 86.2 & 85.7 & 85.4 & 85.2 & 85.0
\end{array}$$ Using a graphing utility, fit a logarithmic function y $$y = a + b \ln x$$ the data. Using the function you found, estimate
How long will it take for the test scores to fall below 84%. Express your answer to the nearest month.
A) 8 months
B) 12 months
C) 20 months
D) 10 months

Accepted Answer

The answer of Solve the problem.
-In a Psychology class, the...

Question 18

Solve the problem.
-A life insurance company uses the following rate table for annual premiums for women for term life insurance. Use a graphing utility to fit an exponential function to the data. Predict the annual premium for a woman aged
70 years. $\begin{array}{l|lllllll}
\text { Age } & 35 & 40 & 45 & 50 & 55 & 60 & 65 \
\hline \text { Premium } & \$ 103 & \$ 133 & \$ 190 & \$ 255 & \$ 360 & \$ 503 & \$ 818
\end{array}$

A) $y = 0.0000398 x ^ { 4.06 } , \$ 1233$
B) $\mathrm { y } = 8.94 \mathrm { e } ^ { 0.068 \mathrm { x } } , \$ 1044$
C) $y = 6.367 \mathrm { e } ^ { 0.068 x } , \$ 743$
D) $y = - 9306.4 + 2516.3 \ln ( x ) , \$ 1723$

Accepted Answer

Using a graphing utility to fit the given data points with an exponential function, option B provides an exponential model that closely matches the pattern of increasing premiums with age. The prediction of \$1044 for a 70-year-old woman aligns with the exponential growth trend observed in the data.

Quiz 4: Exponential and Logarithmic Functions