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Suppose that ​19,665$ is invested at an interest rate of ​6.8% per​ year, compounded continuously.

​a) Find the exponential function that describes the amount in the account after time​ t, in years.
​b) What is the balance after 1​ year? 2​ years? 5​ years? 10​ years?
​c) What is the doubling​ time?

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Answer: a) The exponential function that describes the amount in the account after time t, in years, is given by:

A(t) = Pe^(rt)

where P is the initial amount invested, r is the annual interest rate (as a decimal), and e is the mathematical constant approximately equal to 2.71828.

Substituting the given values, we get:

A(t) = 19665e^(0.068t)

b) To find the balance after 1 year, we substitute t = 1 in the above formula:

A(1) = 19665e^(0.068*1) = $20,983.88

To find the balance after 2 years, we substitute t = 2:

A(2) = 19665e^(0.068*2) = $22,429.45

To find the balance after 5 years, we substitute t = 5:

A(5) = 19665e^(0.068*5) = $29,137.27

To find the balance after 10 years, we substitute t = 10:

A(10) = 19665e^(0.068*10) = $43,127.22

c) The doubling time can be found using the formula:

t = ln(2)/r

where ln is the natural logarithm function. Substituting the given values, we get:

t = ln(2)/0.068 ≈ 10.20 years

Therefore, the doubling time is approximately 10.20 years.

Explanation:

User Daniel Nelson
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