Jamal puts \( \$ 4,000.00 \) into an account to use for school expenses. The account earns \( 10 \% \) interest, compounded monthly. How much will be in the account after 9 years? Use the formula \( A=P\left(1+\frac{r}{n}\right)^{n t} \), where \( A \) is the balance (final amount), \( P \) is the principal (starting amount), \( r \) is the interest rate expressed as a decimal, \( n \) is the number of times per year that the interest is compounded, and \( t \) is the time in years. Round your answer to the nearest cent.

To find the amount in the account after 9 years, we will use the formula for compound interest: \[ A = P\left(1+\frac{r}{n}\right)^{nt} \] Where: - \( A \) is the balance (final amount) - \( P \) is the principal (starting amount) - \( r \) is the interest rate expressed as a decimal - \( n \) is the number of times per year that the interest is compounded - \( t \) is the time in years ### Step 1: Identify the known values - \( P = 4000 \) - \( r = 10\% = 0.10 \) - \( n = 12 \) (since the interest is compounded monthly) - \( t = 9 \) ### Step 2: Substitute the known values into the formula Now we can substitute these values into the formula: \[ A = 4000\left(1+\frac{0.10}{12}\right)^{12 \times 9} \] ### Step 3: Calculate the expression inside the parentheses First, we calculate \( \frac{0.10}{12} \): \[ \frac{0.10}{12} = 0.0083333 \] Now, we can calculate \( 1 + 0.0083333 \): \[ 1 + 0.0083333 = 1.0083333 \] ### Step 4: Calculate the exponent Next, we calculate \( 12 \times 9 \): \[ 12 \times 9 = 108 \] ### Step 5: Calculate the final amount Now we can calculate \( A \): \[ A = 4000 \times (1.0083333)^{108} \] Let's perform this calculation. Calculate the value by following steps: - step0: Calculate: \(4000\times 1.0083333^{108}\) - step1: Convert the expressions: \(4000\left(\frac{10083333}{10000000}\right)^{108}\) - step2: Simplify: \(4000\times \frac{10083333^{108}}{10000000^{108}}\) - step3: Rewrite the expression: \(4000\times \frac{10083333^{108}}{4000^{108}\times 2500^{108}}\) - step4: Reduce the numbers: \(1\times \frac{10083333^{108}}{4000^{107}\times 2500^{108}}\) - step5: Multiply the fractions: \(\frac{10083333^{108}}{4000^{107}\times 2500^{108}}\) The calculation gives us: \[ A \approx 9801.76 \] ### Step 6: Round to the nearest cent Thus, the amount in the account after 9 years will be approximately: \[ \boxed{9801.76} \] So, Jamal will have \( \$ 9,801.76 \) in the account after 9 years.