Question 10 of 12 (1 point) I Question Attempt 1 of Unlimited Suppose that \( \$ 2000 \) is invested at a rate of \( 3.2 \% \), compounded quarterly. Assuming that no withdrawals are made, find the total amount after 10 yea Do not round any intermediate computations, and round your answer to the nearest cent.

To find the total amount after 10 years with a principal of $2000, an annual interest rate of 3.2%, and compounded quarterly, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the total amount after \( t \) years, - \( P \) is the principal amount ($2000 in this case), - \( r \) is the annual interest rate (3.2% or 0.032 in decimal form), - \( n \) is the number of times the interest is compounded per year (4 times for quarterly compounding), - \( t \) is the time in years (10 years in this case). Substitute the given values into the formula: \[ A = 2000 \left(1 + \frac{0.032}{4}\right)^{4 \times 10} \] Now, we can calculate the total amount after 10 years. Calculate the value by following steps: - step0: Calculate: \(2000\left(1+\frac{0.032}{4}\right)^{4\times 10}\) - step1: Divide the terms: \(2000\left(1+\frac{1}{125}\right)^{4\times 10}\) - step2: Add the numbers: \(2000\left(\frac{126}{125}\right)^{4\times 10}\) - step3: Multiply the numbers: \(2000\left(\frac{126}{125}\right)^{40}\) - step4: Simplify: \(2000\times \frac{126^{40}}{125^{40}}\) - step5: Rewrite the expression: \(125\times 16\times \frac{126^{40}}{125^{40}}\) - step6: Reduce the numbers: \(16\times \frac{126^{40}}{125^{39}}\) - step7: Multiply: \(\frac{16\times 126^{40}}{125^{39}}\) The total amount after 10 years with a principal of $2000, an annual interest rate of 3.2%, and compounded quarterly is approximately $2750.75.