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15 January, 01:58

1.) The function f (x) = 12,500 (0.87) ^x models the value of a car x years after it is purchased.

How does the average rate of change between years 11 and 15 compare to the average rate of change between years 1 and 5?

A. The average rate of change between years 11 and 15 is about 2 times the rate between years 1 and 5.

B. The average rate of change between years 11 and 15 is about 3 times the rate between years 1 and 5.

C. The average rate of change between years 11 and 15 is about 13 the rate between years 1 and 5.

D. The average rate of change between years 11 and 15 is about 14 the rate between years 1 and 5.

2. A population of beavers decreases exponentially at a rate of 7.5% per year.

What is the equivalent monthly rate to the nearest hundredth of a percent?

A. 0.65%

B. 0.81%

C. 0.91%

D. 2.37%

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Answers (1)
  1. 15 January, 05:16
    0
    1] Given that the value of x has been modeled by f (x) = 12500 (0.87) ^x, then:

    the rate of change between years 1 and 5 will be:

    rate of change is given by:

    [f (b) - f (a) ] / (b-a)

    thus:

    f (1) = 12500 (0.87) ^1=10875

    f (5) = 12500 (0.87) ^5=6230.3

    rate of change will be:

    (6230.3-10875) / (5-1)

    =-1161.2

    rate of change in years 11 to 15 will be:

    f (11) = 12500 (0.87) ^11=2701.6

    f (15) = 12500 (0.87) ^15=1,547.74

    thus the rate of change will be:

    (1547.74-2701.6) / (15-11)

    =-288

    dividing the two rates of change we get:

    -288/-1161.2

    -=1/4

    comparing the two rate of change we conclude that:

    The average rate of change between years 11 and 15 is about 1/4 the rate between years 1 and 5.

    The answer is D]

    2] Given that the population of beavers decreases exponentially at the rate of 7.5% per year, the monthly rate will be:

    monthly rate = (n/12)

    where n is the number of months

    =7.5/12

    =0.625

    This is approximately equal to 0.65%. The correct answer is A. 0.65%
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