Question

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)=75 - 12 ln(t + 1), 0≤t≤80 a) what was the average score when they initially took the test? the average score was 75 %. (round to one decimal place as needed.) b) what was the average score after 4 months? the average score after 4 months was 55.7 %. (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.)

Explanation:

Step1: Substitute $t = 24$ into the formula

We have the formula $S(t)=75 - 12\ln(t + 1)$. Substitute $t = 24$ into it, getting $S(24)=75-12\ln(24 + 1)$.

Step2: Calculate the natural - logarithm value

First, calculate $\ln(25)$. Using a calculator, $\ln(25)\approx3.218876$.

Step3: Calculate the product

Then, calculate $12\times\ln(25)$. So, $12\times3.218876 = 38.626512$.

Step4: Calculate the final score

Finally, calculate $S(24)=75-38.626512\approx36.4$.

Answer:

$36.4$