Question 1

Choose the one alternative that best completes the statement or answers the question.
-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.
$\begin{array}{l|ccccc}
\text { Time, } \text { (in months) } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Score, } \mathrm{y} \text { (in percentage) } & 86.2 & 85.7 & 85.4 & 85.2 & 85.0
\end{array}$
 Using a graphing utility, fit a logarithmic function $y = a + b \ln x$ to the data. Using the function you found, estin 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 Choose the one alternative that best completes...

Question 2

Choose the one alternative that best completes the statement or answers the question.
-The logistic growth function $\mathrm { f } ( \mathrm { t } ) = \frac { 50,000 } { 1 + 2,499.0 \mathrm { e } ^ { - 1.9 t } }$ 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 logistic growth function and solve for $f(0)$. Substituting $t = 0$ into the function gives $f(0) = \frac{50,000}{1 + 2,499.0e^{-1.9(0)}} = \frac{50,000}{1 + 2,499} = \frac{50,000}{2,500} = 20$.

Question 3

Choose the one alternative that best completes the statement or answers the question.
-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) $y = \frac { 306.53 } { 1 + 7.92 e ^ { - 0.254 x } } , 307 ^ { \circ } \mathrm { F }$
B) $\mathrm { y } = \frac { 311.63 } { 1 + 8.1 \mathrm { e } ^ { - 0.253 \mathrm { x } } } , 312 ^ { \circ } \mathrm { F }$
C)$ \mathrm { y } = \frac { 314.79 } { 1 + 7.86 \mathrm { e } ^ { - 1.22 \mathrm { x } } } , 315 ^ { \circ } \mathrm { F }$
D) $\mathrm { y } = \frac { 314.79 } { 1 + 7.86 \mathrm { e } ^ { - 0.246 \mathrm { x } } } , 315 ^ { \circ } \mathrm { F }$

Accepted Answer

The carrying capacity in a logistic function is the maximum value the function approaches, which is the numerator in the logistic equation. In option D, the carrying capacity is 314.79, which is closest to the observed data trend.

Question 4

Choose the one alternative that best completes the statement or answers the question.
-The logistic growth function $\mathrm { f } ( \mathrm { t } ) = \frac { 24,000 } { 1 + 599.0 \mathrm { e } ^ { - 1.8 \mathrm { t } } }$ 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 determined by the carrying capacity of the logistic growth function, which is the coefficient of the highest power of t in the denominator when t approaches infinity. In this case, it is 24,000 people.

Question 5

Choose the one alternative that best completes the statement or answers the question.
-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) $y = 2.353 \mathrm { e } ^ { 0.610 x , 91 \text { days } }$
B) $\mathrm { y } = 1.669 \mathrm { x } ^ { 1.706,46 \text { days } }$
C) $\mathrm { y } = 1.278 \mathrm { e } ^ { 0.611 \mathrm { x } , 92 \mathrm { days } }$
D) $\mathrm { y } = 1.849 \mathrm { e } ^ { 0.531 \mathrm { x } , 76 \mathrm { days } }$

Accepted Answer

The exponential function that best fits the data is $y = 2.353 \mathrm { e } ^ { 0.610 x }$. To predict the percentage of trading days in 2001 having such swings in value, we substitute $x = 6$ into the equation:$$y = 2.353 \mathrm { e } ^ { 0.610 ( 6 ) } \approx 91$$Therefore, the predicted percentage is 91 days.

Question 6

Choose the one alternative that best completes the statement or answers the question.
-A nuclear scientist has a sample of $100 \mathrm { mg }$ 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) $y = 92 \mathrm { e } ^ { - 0.0686 x } , 5.9 \mathrm { mg }$
B) $y = 86 e ^ { - 0.071 x , 5.0 \mathrm { mg } }$
C) $y = 100 \mathrm { e } ^ { - 0.077 x } , 4.6 \mathrm { mg }$
D) $y = 100 \mathrm { e } ^ { - 0.077 x } , 6.7 \mathrm { mg }$

Accepted Answer

The exponential decay formula for radioactive material is $y = y_0 e^{-kt}$, where $y_0$ is the initial amount, $k$ is the decay constant, and $t$ is time. The option that correctly represents the decay of the material with the initial amount of 100 mg and fits the data provided is $y = 100 e^{-0.077x}$. To predict the amount of material remaining at 40 hours, substituting $x = 40$ into the equation gives approximately 4.6 mg, which matches option C.

Question 7

Choose the one alternative that best completes the statement or answers the question.
-The logistic growth function $f ( t ) = \frac { 440 } { 1 + 6.3 e ^ { - 0.2 t } }$ 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 determined by the carrying capacity of the habitat, which is the coefficient of the numerator in the logistic growth function, $440$ in this case.

Question 8

Write the word or phrase that best completes each statement or answers the question.
-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 is $\mathrm { A } = 150 \cdot 1.79 ^ { \mathrm { t } }$. Express the function of best fit in the form
$$A = A _ { 0 } e ^ { k t } \text {. }$$

Accepted Answer

The answer of Write the word or phrase that best...

Question 9

Write the word or phrase that best completes each statement or answers the question.
-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 $p = 25 \cdot 0.89 q$. 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 Write the word or phrase that best...

Question 10

Choose the one alternative that best completes the statement or answers the question.
-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) $y = 8.94 \mathrm { e } ^ { 0.068 x , } \$ 1044$
C) $y = 6.367 \mathrm { e } ^ { 0.068 \mathrm { x } } , \$ 743$
D) $y = - 9306.4 + 2516.3 \ln ( x ) , \$ 1723$

Accepted Answer

The exponential function $y = 8.94 \mathrm{e}^{0.068x}$ fits the data well, and using it to predict the premium for age 70 gives approximately \$1044.

Question 11

Write the word or phrase that best completes each statement or answers the question.
-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 $p = 398 - 73 \ln ( q )$. 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 Write the word or phrase that best...

Question 12

Write the word or phrase that best completes each statement or answers the question.
-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 is $y = 822.99 - 167.55 \ln x$ 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 Write the word or phrase that best...

Question 13

Choose the one alternative that best completes the statement or answers the question.
-The logistic growth function $f ( t ) = \frac { 12,000 } { 1 + 199 e ^ { - 1.7 t } }$ 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 logistic growth function and solve for $f(9)$:$$f(9) = \frac{12,000}{1 + 199e^{-1.7 \times 9}}$$$$f(9) = \frac{12,000}{1 + 199e^{-15.3}}$$$$f(9) \approx \frac{12,000}{1 + 199 \times 2.2 \times 10^{-7}}$$$$f(9) \approx \frac{12,000}{1 + 0.0438}$$$$f(9) \approx \frac{12,000}{1.0438}$$$$f(9) \approx 11,999$$Therefore, approximately 11,999 people were ill after 9 weeks.

Question 14

Write the word or phrase that best completes each statement or answers the question.
-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 is $y = 57.76 + 48.56 \ln x$ where $x$ represents the number of years sinc 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 Write the word or phrase that best...

Question 15

Choose the one alternative that best completes the statement or answers the question.
-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) $y = 98.75 - 32.66 \ln ( \mathrm { x } ) , 0.91$
B) $y = 107.55 - 41 \ln ( x ) , - 15.27$
C) $\mathrm { y } = 112.97 - 45.97 \ln ( \mathrm { x } ) , - 24.74$
D) $y = 100.5 - 32.7 \ln ( x ) , 2.54$

Accepted Answer

The correct choice is determined by fitting a logarithmic function to the given data points. Among the options, choice A provides a logarithmic function that could reasonably fit the trend of decreasing luminosity over time, as indicated by the data. The other options either have coefficients that don't align as well with the general trend of the data or include values not relevant to the function's form (e.g., an extra number outside the function). Logarithmic functions are often used to model phenomena that decrease rapidly at first and then level off, which seems to match the description of the luminosity decrease in the bacteria culture.

Question 16

Write the word or phrase that best completes each statement or answers the question.
-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 Write the word or phrase that best...

Question 17

Choose the one alternative that best completes the statement or answers the question.
-The logistic growth function $f ( t ) = \frac { 800 } { 1 + 9.0 \mathrm { e } ^ { - 0.16 t } }$ 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 logistic growth function and solve for $f(15)$:$$f(15) = \frac{800}{1 + 9.0e^{-0.16 \times 15}}$$$$= \frac{800}{1 + 9.0e^{-2.4}}$$$$= \frac{800}{1 + 9.0 \times 0.09071795328941247}$$ (using a calculator for $e^{-2.4}$)$$= \frac{800}{1 + 0.8162614796055123}$$$$= \frac{800}{1.8162614796055123}$$$$= 440.53$$Rounding to the nearest whole number, the expected number of butterflies after 15 months is approximately 440.

Quiz 4: Exponential and Logarithmic Functions