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?

Respuesta :

Answer:

4%

Step-by-step explanation:

[tex]A = P(1+r/n)^{nt}[/tex]

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

[tex]A = P(1+r/n)^{nt}\\3932.82 = 3101(1+r/2)^{12}[/tex]

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

[tex]3932.82/3101 = (1+r/2)^{12}[/tex]

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

[tex](3932.82/3101)^{1/12}= (1+r/2)[/tex]

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

[tex](3932.82/3101)^{1/12}-1= (r/2)[/tex]

multiply both sides by 2 to get r

[tex]((3932.82/3101)^{1/12}-1) * 2= r[/tex]

r ≈ 0.04 = 4%