Finding the final amount in a word problem on compound interest Suppose that \( \$ 2000 \) is loaned at a rate of \( 11 \% \), compounded semiannually. Assuming that no payments are made, find the amount owed after 6 years. Do not round any intermediate computations, and round your answer to the nearest cent.

To find the final amount owed after 6 years with compound interest, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial amount of money). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the number of years the money is invested or borrowed. ### Given values: - \( P = 2000 \) - \( r = 0.11 \) (11% as a decimal) - \( n = 2 \) (since the interest is compounded semiannually) - \( t = 6 \) ### Step 1: Substitute the values into the formula We will substitute the known values into the compound interest formula: \[ A = 2000 \left(1 + \frac{0.11}{2}\right)^{2 \times 6} \] ### Step 2: Calculate the expression inside the parentheses First, calculate \( \frac{0.11}{2} \): \[ \frac{0.11}{2} = 0.055 \] Now, substitute this back into the formula: \[ A = 2000 \left(1 + 0.055\right)^{12} \] ### Step 3: Calculate \( 1 + 0.055 \) \[ 1 + 0.055 = 1.055 \] ### Step 4: Raise \( 1.055 \) to the power of \( 12 \) Now we need to calculate \( (1.055)^{12} \): \[ (1.055)^{12} \] ### Step 5: Multiply by the principal amount Finally, multiply the result by \( 2000 \): \[ A = 2000 \times (1.055)^{12} \] Now, let's perform the calculations to find the final amount \( A \). Calculate the value by following steps: - step0: Calculate: \(2000\times 1.055^{12}\) - step1: Convert the expressions: \(2000\left(\frac{211}{200}\right)^{12}\) - step2: Simplify: \(2000\times \frac{211^{12}}{200^{12}}\) - step3: Rewrite the expression: \(200\times 10\times \frac{211^{12}}{200^{12}}\) - step4: Reduce the numbers: \(10\times \frac{211^{12}}{200^{11}}\) - step5: Rewrite the expression: \(10\times \frac{211^{12}}{100^{11}\times 2^{11}}\) - step6: Rewrite the expression: \(10\times \frac{211^{12}}{10^{22}\times 2^{11}}\) - step7: Reduce the numbers: \(1\times \frac{211^{12}}{10^{21}\times 2^{11}}\) - step8: Multiply the fractions: \(\frac{211^{12}}{2048\times 10^{21}}\) The calculation gives us: \[ A \approx 3802.414972 \] ### Step 6: Round to the nearest cent Rounding \( 3802.414972 \) to the nearest cent, we get: \[ A \approx 3802.41 \] ### Final Answer The amount owed after 6 years is approximately **\$3802.41**.