Answer
a. $88$
b. $71.5$,
$ 63.9$,
$ 58.8$,
$ 55.0$,
$ 52.0$,
$ 49.5$,
c. See graph, decrease with time.
Work Step by Step
a. Given the average score function $f(t) = 88 - 15 ln(t + 1)$, the original exam happened at $t=0$; we have $f(0) = 88 - 15 ln(0 + 1)=88-0=88$
b. For $t=2,4,6,8,10,12$, we have
$f(2) = 88 - 15 ln(2 + 1)\approx71.5$,
$f(4) = 88 - 15 ln(4 + 1)\approx63.9$,
$f(6) = 88 - 15 ln(6 + 1)\approx58.8$,
$f(8) = 88 - 15 ln(8 + 1)\approx55.0$,
$f(10) = 88 - 15 ln(10 + 1)\approx52.0$,
$f(12) = 88 - 15 ln(12 + 1)\approx49.5$,
c. See graph. We can see that the material retained by the students, represented by the average scores, decrease with time.
