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mars1129 [50]
4 years ago
12

Students in a psychology class took a final examination. As part of an experiment to see how much of the course content they rem

embered over​ time, they took equivalent forms of the exam in monthly intervals thereafter. The average score for the group f (t )​, after t months was modeled by the function f (t )equals 88 minus 14 ln (t plus 1 )​, 0 less than or equals t less than or equals 12. A. What was the average score on the original​ exam? B. What was the average score after 2 ​months? C. Sketch the graph of f​ (either by hand or with a graphing​ utility.) Describe what the graph indicates in terms of the material retained by the students.
Mathematics
1 answer:
VikaD [51]4 years ago
8 0

The question is a bit messy. So I tidy it up as follows. Please tell if it's different from the original

Students in a psychology class took a final examination. As part of an experiment to see how much of the course content they remembered over time, they took equivalent forms of the exam in monthly intervals thereafter. The average score for the group, f(t), after t months was modeled by the function

F (t) = 88-15 In(t + 1), 0 ≤ t ≤ 12

Answer the following questions and show all work to obtaining that answer.

1. What was the average score on the original exam?

2. What was the average score, to the nearest tenth, after 2 months?

3. Sketch the graph of f (either by hand or with a graphing utility). Describe what the graph indicates in terms of the material retained by the students. You do not need to include a picture of the graph in your assignment.

Note: ln represents the natural logarithm in the function.

Answer:

Step-by-step explanation:

1) The original score is when t = 0. substitue t into the original function:

F(0) = 88-15 In(0 + 1)

       = 88-15 In(1)

       = 88 - 15(0)

       = 88

2)

After 2 month, t = 2,

F(2) = 88-15 In(2+ 1)

       = 88-15 In(3)

       = 88 - 15(1.099)

       = 88 - 16.5

       = 71.5

3) Graphing can be done by graphic calculator. The graph will be seen as a natural logarithm function with assymptote of x = 0.

From the graph it can be seen that as t increase i.e. as the number of month increase, the average marks drop approaching a certain value

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