Question 19 Find the final amount of money in an account if \( \$ 8,300 \) is deposited at \( 5.5 \% \) interest compounded annually and the money is left for 9 years. The final amount is \( \$ \square \) Next Question Round answer to 2 decimal places P

To find the final amount of money in an account after depositing $8,300 at 5.5% 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) - \( r \) is the annual interest rate - \( n \) is the number of years the money is left in the account Given: - \( P = \$8,300 \) - \( r = 5.5\% \) - \( n = 9 \) years Substitute the given values into the formula: \[ A = 8300 \left( 1 + \frac{5.5}{100} \right)^9 \] Now, we can calculate the final amount of money in the account. Calculate the value by following steps: - step0: Calculate: \(8300\left(1+\frac{5.5}{100}\right)^{9}\) - step1: Divide the terms: \(8300\left(1+\frac{11}{200}\right)^{9}\) - step2: Add the numbers: \(8300\left(\frac{211}{200}\right)^{9}\) - step3: Simplify: \(8300\times \frac{211^{9}}{200^{9}}\) - step4: Rewrite the expression: \(100\times 83\times \frac{211^{9}}{200^{9}}\) - step5: Rewrite the expression: \(100\times 83\times \frac{211^{9}}{100^{9}\times 2^{9}}\) - step6: Reduce the numbers: \(83\times \frac{211^{9}}{100^{8}\times 2^{9}}\) - step7: Multiply the fractions: \(\frac{83\times 211^{9}}{512\times 100^{8}}\) The final amount of money in the account after depositing $8,300 at 5.5% interest compounded annually for 9 years is approximately $13,438.48.