An account is opened with an initial deposit of \( \$ 5,500 \) and earns \( 2.5 \% \) interest compounded annually. What will the account be worth in 10 years? Round your answer to the nearest cent. Do NOT round until you have calculated the final answer.

To find the worth of the account after 10 years with an initial deposit of $5,500 and an annual interest rate of 2.5% compounded annually, we can use the formula for compound interest: \[ A = P \left( 1 + \frac{r}{100} \right)^n \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (initial deposit). - \( r \) is the annual interest rate (in %). - \( n \) is the number of years. Given: - Initial deposit \( P = \$5,500 \) - Annual interest rate \( r = 2.5\% \) - Number of years \( n = 10 \) Substitute the given values into the formula: \[ A = 5500 \left( 1 + \frac{2.5}{100} \right)^{10} \] Now, we can calculate the amount of money accumulated after 10 years using the formula. Calculate the value by following steps: - step0: Calculate: \(5500\left(1+\frac{2.5}{100}\right)^{10}\) - step1: Divide the terms: \(5500\left(1+\frac{1}{40}\right)^{10}\) - step2: Add the numbers: \(5500\left(\frac{41}{40}\right)^{10}\) - step3: Simplify: \(5500\times \frac{41^{10}}{40^{10}}\) - step4: Rewrite the expression: \(20\times 275\times \frac{41^{10}}{40^{10}}\) - step5: Rewrite the expression: \(20\times 275\times \frac{41^{10}}{20^{10}\times 2^{10}}\) - step6: Reduce the numbers: \(275\times \frac{41^{10}}{20^{9}\times 2^{10}}\) - step7: Rewrite the expression: \(25\times 11\times \frac{41^{10}}{20^{9}\times 2^{10}}\) - step8: Rewrite the expression: \(25\times 11\times \frac{41^{10}}{5^{9}\times 4^{9}\times 2^{10}}\) - step9: Rewrite the expression: \(5^{2}\times 11\times \frac{41^{10}}{5^{9}\times 4^{9}\times 2^{10}}\) - step10: Reduce the numbers: \(11\times \frac{41^{10}}{5^{7}\times 4^{9}\times 2^{10}}\) - step11: Multiply the fractions: \(\frac{11\times 41^{10}}{5^{7}\times 4^{14}}\) - step12: Calculate: \(\frac{11\times 41^{10}}{80^{7}}\) The account will be worth approximately $7,040.46 after 10 years, rounded to the nearest cent.