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 %.(round to one decimal place as needed)

Explanation:

Step1: Substitute t = 4 into the formula

We are given $S(t)=75 - 12\ln(t + 1)$. Substitute $t = 4$ into it, so we get $S(4)=75-12\ln(4 + 1)$.

Step2: Calculate the natural - logarithm part

First, calculate $\ln(5)$. We know that $\ln(5)\approx1.6094$.

Step3: Calculate the product part

Then, calculate $12\times\ln(5)$. So $12\times1.6094 = 19.3128$.

Step4: Calculate the final result

Finally, calculate $S(4)=75-19.3128=55.6872\approx55.7$.

Answer:

$55.7$