Questions: Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After t months, the average score S(t), as a percentage, was found to be given by the following equation, where t ≥ 0. Complete parts (a) through (e) below. S(t)=79-13 ln (t+1), 0 ≤ t ≤ 80 a) What was the average score when they initially took the test? The average score was 79%. (Round to one decimal place as needed.) b) What was the average score after 4 months? The average score after 4 months was 58.1%. (Round to one decimal place as needed.) c) What was the average score after 24 months? The average score after 24 months was %. (Round to one decimal place as needed.)

Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After t months, the average score S(t), as a percentage, was found to be given by the following equation, where t ≥ 0. Complete parts (a) through (e) below.

S(t)=79-13 ln (t+1), 0 ≤ t ≤ 80

a) What was the average score when they initially took the test?

The average score was 79%. (Round to one decimal place as needed.)

b) What was the average score after 4 months?

The average score after 4 months was 58.1%. (Round to one decimal place as needed.)

c) What was the average score after 24 months?

The average score after 24 months was %. (Round to one decimal place as needed.)

Transcript text: Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After t months, the average score $\mathrm{S}(\mathrm{t})$, as a percentage, was found to be given by the following equation, where $\mathrm{t} \geq 0$. Complete parts (a) through (e) below. \[ S(t)=79-13 \ln (t+1), 0 \leq t \leq 80 \] a) What was the average score when they initially took the test? The average score was $79 \%$. (Round to one decimal place as needed.) b) What was the average score after 4 months? The average score after 4 months was $58.1 \%$. (Round to one decimal place as needed.) c) What was the average score after 24 months? The average score after 24 months was \%. (Round to one decimal place as needed.)

Solution

Solution Steps

Step 1: Initial Score Calculation

The initial average score, $S(0)$, is calculated by substituting $t = 0$ into the equation, yielding $S(0) = A - B \ln(1) = A$, since $\ln(1) = 0$. Therefore, the initial average score is $S(0) = 79$.

Step 2: Score at Any Time t

To find the average score at any given time $t$, we substitute the value of $t$ into the equation $S(t) = A - B \ln(t + 1)$. For $t = 24$, the average score is $S(24) = 79 - 13 \ln(24 + 1) = 37.2$.

Step 3: Rate of Change of Score

The rate of change of the average score with respect to time, $S'(t)$, is found by differentiating $S(t)$ with respect to $t$, yielding $S'(t) = -\frac{B}{t + 1} = -0.5$. This represents how quickly the average score is changing at any given time $t$.

Step 4: Interpreting S'(t)

At $t = 24$, the rate of change of the average score, $S'(24) = -0.5$, indicates that the average score is decreasing. A more negative value would indicate a faster decrease in average score.