Question

months, the average score​ S(t), as a​ percentage, was found to be given by the following​ equation, whereStudents 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.
Required:
a. What was the average score when they initially took the test, t=0?
b. What was the average score after 4 months?
c. What was the average score after 24 months?
d. What percentage of their original answers did the students retain after 2 years?
e. The maximum of the function is_____%.

Answer 1

Answer:

a. 73; b. 48.9; c. 2; d. 33.8; e. 73

Step-by-step explanation:

Assume the function was

S(t)= 73 - 15 ln(t + 1), t  ≥ 0

a. Average score at t = 0

S(0) = 73 - 15 ln(0 + 1) = 73 - 15 ln(1) = 73 - 15(0) =73 - 0 = 73

b. Average score at t = 4

S(4) = 73 - 15 ln(4 + 1) = 73 - 15 ln(5) = 73 - 15(1.61) =73 - 24.14 = 48.9

c. Average score at t =24

S(24) = 73 - 15 ln(24 + 1) = 73 - 15 ln(25) = 73 - 15(3.22) =73 - 48.28 = 24.7

d. Percent of answers retained

At t = 0. the students retained 73 % of the answers.

At t = 24, they retained 24.7 % of the answers.

$\text{Percent retention} = \dfrac{\text{24.7}}{\text{73}} \times 100 \, \% = \text{33.8 \%}\\\\\text{The students retained \large \boxed{\mathbf{33.8 \, \%}} of their original knowledge after two years.}$

e. Maximum of the function

The maximum of the function is at t= 0.

Max = 73 %

The graph below shows your knowledge decay curve. Knowledge decays rapidly at first but slows as time goes on.