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When a deposit of $1000 is made into an account paying 2% interest, compounded annually, the balance, $B, in the account after t years is given by B = 1000(1.02)t. Find the average rate of change in the balance over the interval t = 0 to t = 5. Give units and interpret your answer in terms of the balance in the account.

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Answer:

The average rate of change in the balance over the interval t = 0 to t = 5 is of $20.82 a year. This means that the balance increased by $20.82 a year over the interval t = 0 to t = 5.

Explanation:

Given a function y, the average rate of change S of y=f(x) in an interval
(x_(s), x_(f)) will be given by the following equation:


S = (f(x_(f)) - f(x_(s)))/(x_(f) - x_(s))

In this problem, we have that:


B(t) = 1000(1.02)^(t)

Find the average rate of change in the balance over the interval t = 0 to t = 5.


B(0) = 1000(1.02)^(0) = 1000


B(5) = 1000(1.02)^(5) = 1104.08

Then


S = (1104.08 - 1000)/(5-0) = 20.82

The average rate of change in the balance over the interval t = 0 to t = 5 is of $20.82 a year. This means that the balance increased by $20.82 a year over the interval t = 0 to t = 5.

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