To determine the balance of an account after a certain number of years with interest compounded monthly, we will use the compound interest formula:
[tex]\[
F = P \left(1 + \frac{r}{n}\right)^{nt}
\][/tex]
where:
- [tex]\( F \)[/tex] is the future value of the investment/loan, including interest.
- [tex]\( P \)[/tex] is the principal investment amount (the initial deposit or loan amount).
- [tex]\( r \)[/tex] is the annual interest rate (decimal).
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year.
- [tex]\( t \)[/tex] is the number of years the money is invested or borrowed for.
Given:
- [tex]\( P = \$1200 \)[/tex]
- [tex]\( r = 11\% = 0.11 \)[/tex]
- [tex]\( n = 12 \)[/tex] (since the interest is compounded monthly)
- [tex]\( t = 11 \)[/tex] years
Plugging these values into the formula:
[tex]\[
F = 1200 \left(1 + \frac{0.11}{12}\right)^{12 \times 11}
\][/tex]
First, calculate the monthly interest rate:
[tex]\[
\frac{0.11}{12} \approx 0.0091667
\][/tex]
Then, add 1 to the monthly interest rate:
[tex]\[
1 + 0.0091667 \approx 1.0091667
\][/tex]
Next, raise this sum to the power of the total number of compounding periods:
[tex]\[
1.0091667^{12 \times 11} = 1.0091667^{132}
\][/tex]
Calculating this power gives:
[tex]\[
1.0091667^{132} \approx 3.33505
\][/tex]
Now, multiply this result by the principal amount [tex]\( P \)[/tex]:
[tex]\[
F = 1200 \times 3.33505 = 4002.06
\][/tex]
Thus, the balance after 11 years, rounded to the nearest cent, is:
[tex]\[
F = \$4002.06
\][/tex]
