To find the balance in Neil's savings account after 9 years with an initial deposit of \[tex]$400.00 and an annual interest rate of 15% compounded monthly, we can use the compound interest formula:
\[ A = P \left( 1 + \frac{r}{n} \right)^{nt} \]
where:
- \( P \) is the principal amount (\$[/tex]400.00)
- [tex]\( r \)[/tex] is the annual interest rate (15%, or 0.15 as a decimal)
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year (12 times per year, since it's compounded monthly)
- [tex]\( t \)[/tex] is the time the money is invested for in years (9 years)
Let's break down the steps to find the final amount [tex]\( A \)[/tex].
1. Identify the values:
- [tex]\( P = 400.00 \)[/tex]
- [tex]\( r = 0.15 \)[/tex]
- [tex]\( n = 12 \)[/tex]
- [tex]\( t = 9 \)[/tex]
2. Substitute these values into the formula:
[tex]\[ A = 400.00 \left( 1 + \frac{0.15}{12} \right)^{12 \cdot 9} \][/tex]
3. Calculate the interest rate per compounding period:
[tex]\[ 1 + \frac{0.15}{12} \][/tex]
[tex]\[ 1 + 0.0125 = 1.0125 \][/tex]
4. Calculate the exponent [tex]\( nt \)[/tex]:
[tex]\[ 12 \cdot 9 = 108 \][/tex]
5. Raise the base to the power of the exponent:
[tex]\[ 1.0125^{108} \approx 3.82528 \][/tex]
6. Multiply this result by the principal:
[tex]\[ 400.00 \times 3.82528 \approx 1530.11 \][/tex]
7. Round to the nearest cent:
[tex]\[ 1530.11 \][/tex]
Therefore, the balance in Neil's savings account after 9 years is:
[tex]\[
\boxed{1530.11}
\][/tex]
