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LenaWriter [7]
3 years ago
15

It took Amir 2 hours to hike 5 miles. On the first part of the hike, Amir averaged 3 miles per hour. For the second part of the

hike, the terrain was more difficult so his average speed decreased to 1.5 mile per hour. Which equation can be used to find t, the amount of time Amir spent hiking during the second, more difficult part of the hike?
Mathematics
1 answer:
fiasKO [112]3 years ago
8 0

The formula to find time spent hiking during the second part of hike was derived from the basic formula of average speed which is the total number of speed during both parts of hike equals to the total distance traveled divided by total time taken. 

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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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