To solve the problem of finding the future value of an investment with compound interest, we can use the compound interest formula:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( P \)[/tex] is the principal amount (initial investment),
- [tex]\( r \)[/tex] is the annual interest rate (in decimal form),
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year,
- [tex]\( t \)[/tex] is the number of years the money is invested,
- [tex]\( A \)[/tex] is the amount of money accumulated after [tex]\( t \)[/tex] years, including interest.
Let's identify the given values:
- Principal amount ([tex]\( P \)[/tex]) = [tex]$1,179
- Annual interest rate (\( r \)) = 8% = 0.08 (in decimal form)
- Number of times interest is compounded per year (\( n \)) = 12
- Number of years (\( t \)) = 15
Substituting these values into the formula, we have:
\[ A = 1179 \left(1 + \frac{0.08}{12}\right)^{12 \cdot 15} \]
Calculate the value inside the parentheses first:
\[ \left(1 + \frac{0.08}{12}\right) \]
\[ 1 + \frac{0.08}{12} = 1 + 0.00666667 \approx 1.00666667 \]
Now, raise this value to the power of \( 12 \cdot 15 \):
\[ \left(1.00666667\right)^{180} \]
Use a calculator to find the value of \( 1.00666667^{180} \):
\[ 1.00666667^{180} \approx 3.308286 \]
Finally, multiply this value by the principal amount:
\[ A = 1179 \times 3.308286 \approx 3,900.26 \]
So, the account balance after 15 years will be approximately $[/tex]3,900.26.
