Joah puts \( \$ 3,084.00 \) into an account to use for school expenses. The account earns \( 3.65 \% \) nterest, compounded annually. How much will be in the account after 4 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 4 years, we can use the formula for compound interest: \[ A = P\left(1 + \frac{r}{n}\right)^{nt} \] Given: - Principal (P) = $3,084.00 - Interest rate (r) = 3.65% = 0.0365 (as a decimal) - Compounded annually, so n = 1 - Time (t) = 4 years Substitute the given values into the formula: \[ A = 3084\left(1 + \frac{0.0365}{1}\right)^{1 \times 4} \] Now, we can calculate the amount in the account after 4 years. Calculate the value by following steps: - step0: Calculate: \(3084\left(1+0.0365\right)^{4}\) - step1: Add the numbers: \(3084\times 1.0365^{4}\) - step2: Convert the expressions: \(3084\left(\frac{2073}{2000}\right)^{4}\) - step3: Simplify: \(3084\times \frac{2073^{4}}{2000^{4}}\) - step4: Rewrite the expression: \(4\times 771\times \frac{2073^{4}}{2000^{4}}\) - step5: Rewrite the expression: \(4\times 771\times \frac{2073^{4}}{16^{4}\times 125^{4}}\) - step6: Rewrite the expression: \(4\times 771\times \frac{2073^{4}}{4^{8}\times 125^{4}}\) - step7: Reduce the numbers: \(771\times \frac{2073^{4}}{4^{7}\times 125^{4}}\) - step8: Multiply the fractions: \(\frac{771\times 2073^{4}}{16384\times 125^{4}}\) The amount in the account after 4 years will be approximately $3,559.52. Therefore, after 4 years, Joah will have $3,559.52 in the account.