k: MLM #8 Question 13, 2.3.29-SS Part 1 of 5 points Points: 0 of 5 st View an example | All parts showing X ing 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 2 0. Complete parts (a) through (e) below. S(t) = 72 - 12 In (t+ 1), Osts 80 a) What was the average score when they initially took the test? To calculate the initial average score, let t = 0 and evaluate the function. S(t) = 72 - 12 In (t+1) S(0) = 72 - 12 In (0+ 1) = 72 Therefore, the average score when they initially took the test was 72%. b) What was the average score after 4 months? To calculate the average score, let t = 4 and evaluate the function, rounding to one decimal place. S(t) = 72 - 12 In (t+1) S(4) = 72 - 12 In (4+1) = 52.7 Therefore, the average score after 4 months was 52.7%. c) What was the average score after 24 months? To calculate the average score, let t = 24 and evaluate the function, rounding to one decimal place. S(t) = 72 - 12 In (t + 1) S(24) = 72 - 12 In (24+1) = 33.4 Therefore, the average score after 24 months was 33.4%. d) Find S'(t). To find S'(t), take the derivative of each term and then find the difference of the derivatives. S(t) = 72 - 12 In (t+1) S' (1) = - 17 at [72 - 12 In (t + 1)] = at 72 - 12 of In (t +1) Next, recall that the derivative of the natural logarithm of a function is the derivative of the function divided by the function. S' (t) = - 72 - 12- In (t+ 1) = 0 -12 .- t + 1 12 t + 1 12 Thus, S'(t) = - - t + 1 e) Find S'(4) and S'(24), and interpret the meaning of these numbers. Recall that if a derivative is negative at a point, the function is decreasing at that point. If a derivative is positive at a point, the function is increasing at that point. Substitute the value t = 4 into the derivative function calculated in part (d) to obtain the value of S'(4). 12 S'(4) = - 7 4+1 = - 2.4 Therefore, after 4 months, the average score is decreasing by 2.4 points a month. Substitute the value t = 24 into the derivative function calculated in part (d) to obtain the value of S'(24). 12 S'(24) = - 24 + 1 - 0.48 Therefore, after 24 months, the average score is decreasing by 0.48 points a month.Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. Question list K After t months, the average score S(t), as a percentage, was found to be given by the following equation, where t 2 0. Complete parts (a) through (e) below. S(t) = 80 - 12 In (t + 1), Osts 80 Question 1 a) What was the average score when they initially took the test? Question 2 The average score was %. (Round to one decimal place as needed.) Question 3