Question

7) Mofor invests $3,101 in a savings account

with a fixed annual interest rate

compounded 2 times per year. After 6

years, the balance reaches $3. 932. 82.

What is the interest rate of the account?

Answer 1

4%

Explanation:

`A = P(1+r/n)^(nt)`

A = final amount

P = initial amount

r = interest rate per time period

n = number of times compounded per time period

t = number of time period

A = 3932.82

r = what we want to find

n = 2

t = 6

`A = P(1+r/n)^(nt)\\3932.82 = 3101(1+r/2)^(12)`

divide both sides by 3101 to isolate the exponent and its parts

`3932.82/3101 = (1+r/2)^(12)`

put both sides to the power of (1/12) to help isolate the variable

`(3932.82/3101)^(1/12)= (1+r/2)`

subtract 1 from both sides to isolate the variable and its coefficient

`(3932.82/3101)^(1/12)-1= (r/2)`

multiply both sides by 2 to get r

`((3932.82/3101)^(1/12)-1) * 2= r`

r ≈ 0.04 = 4%