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 % (round to one decimal place as needed.)

Explanation:

Step1: Identify the value of t for initial test

At the initial test, $t = 0$.

Step2: Substitute t = 0 into the function

Substitute $t=0$ into $S(t)=75 - 12\ln(t + 1)$. We get $S(0)=75-12\ln(0 + 1)$.
Since $\ln(1)=0$, then $S(0)=75-12\times0$.

Step3: Calculate the result

$S(0)=75$.

Answer:

75.0