Question 10 Find the final amount of money in an account if \( \$ 4,000 \) is deposited at \( 3 \% \) interest compounded annually and the money is left for 9 years. The final amount is \( \$ \square \) Round answer to 2 decimal places Submit Question

To find the final amount of money in an account after depositing $4,000 at 3% interest compounded annually for 9 years, we can use the formula for compound interest: \[ A = P \left( 1 + \frac{r}{100} \right)^n \] where: - \( A \) is the final amount of money in the account - \( P \) is the principal amount (initial deposit) = $4,000 - \( r \) is the annual interest rate = 3% - \( n \) is the number of years = 9 Substitute the given values into the formula: \[ A = 4000 \left( 1 + \frac{3}{100} \right)^9 \] Now, we can calculate the final amount of money in the account. Calculate the value by following steps: - step0: Calculate: \(4000\left(1+\frac{3}{100}\right)^{9}\) - step1: Add the numbers: \(4000\left(\frac{103}{100}\right)^{9}\) - step2: Simplify: \(4000\times \frac{103^{9}}{100^{9}}\) - step3: Rewrite the expression: \(1000\times 4\times \frac{103^{9}}{100^{9}}\) - step4: Rewrite the expression: \(10^{3}\times 4\times \frac{103^{9}}{10^{18}}\) - step5: Reduce the numbers: \(4\times \frac{103^{9}}{10^{15}}\) - step6: Rewrite the expression: \(4\times \frac{103^{9}}{2^{15}\times 5^{15}}\) - step7: Rewrite the expression: \(2^{2}\times \frac{103^{9}}{2^{15}\times 5^{15}}\) - step8: Reduce the numbers: \(1\times \frac{103^{9}}{2^{13}\times 5^{15}}\) - step9: Multiply the fractions: \(\frac{103^{9}}{8192\times 5^{15}}\) The final amount of money in the account after depositing $4,000 at 3% interest compounded annually for 9 years is approximately $5,219.09. Therefore, the final amount is $5,219.09.