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Neil opened a savings account and deposited \$400.00 as principal. The account earns 15% interest, compounded monthly. What is the balance after 9 years?

Use the formula [tex]A = P\left(1 + \frac{r}{n}\right)^{nt}[/tex], where:
- [tex]A[/tex] is the balance (final amount),
- [tex]P[/tex] is the principal (starting amount),
- [tex]r[/tex] is the interest rate expressed as a decimal,
- [tex]n[/tex] is the number of times per year that the interest is compounded,
- [tex]t[/tex] is the time in years.

Round your answer to the nearest cent.

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Answer :

GinnyAnswer
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]