Juan invested \( \$ 5300 \) in an account that pays an annual interest rate of \( 3.7 \% \), compounded daily. Assume there are 365 days in each year. Answer each part. If necessary, refer to the list of financial formulas. (a) Find the amount in the account after one year, assuming no withdrawals are made. Do not round any intermediate computations, and round your answer to the nearest cent. s! (b) Find the effective annual interest rate, expressed as a percentage. Do not round any intermediate computations, and round your answer to the nearest hundredth of a percent. \( \square \) \( \% \)

To find the amount in the account after one year, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount in the account after \( t \) years, - \( P \) is the principal amount (initial investment), - \( r \) is the annual interest rate, - \( n \) is the number of times the interest is compounded per year, - \( t \) is the time in years. Given: - \( P = \$5300 \), - \( r = 3.7\% = 0.037 \), - \( n = 365 \) (compounded daily), - \( t = 1 \) year. Substitute the values into the formula: \[ A = 5300 \left(1 + \frac{0.037}{365}\right)^{365 \cdot 1} \] Now, we can calculate the amount in the account after one year. Calculate the value by following steps: - step0: Calculate: \(5300\left(1+\frac{0.037}{365}\right)^{365}\) - step1: Divide the terms: \(5300\left(1+\frac{37}{365000}\right)^{365}\) - step2: Add the numbers: \(5300\left(\frac{365037}{365000}\right)^{365}\) - step3: Simplify: \(5300\times \frac{365037^{365}}{365000^{365}}\) - step4: Rewrite the expression: \(100\times 53\times \frac{365037^{365}}{365000^{365}}\) - step5: Rewrite the expression: \(100\times 53\times \frac{365037^{365}}{1000^{365}\times 365^{365}}\) - step6: Rewrite the expression: \(10^{2}\times 53\times \frac{365037^{365}}{10^{1095}\times 365^{365}}\) - step7: Reduce the numbers: \(53\times \frac{365037^{365}}{10^{1093}\times 365^{365}}\) - step8: Multiply the fractions: \(\frac{53\times 365037^{365}}{10^{1093}\times 365^{365}}\) The amount in the account after one year is approximately $5499.76. Now, to find the effective annual interest rate, we can use the formula: \[ r_{\text{effective}} = \left(1 + \frac{r}{n}\right)^n - 1 \] Substitute the values into the formula: \[ r_{\text{effective}} = \left(1 + \frac{0.037}{365}\right)^{365} - 1 \] Now, we can calculate the effective annual interest rate. Calculate the value by following steps: - step0: Calculate: \(\left(1+\frac{0.037}{365}\right)^{365}-1\) - step1: Divide the terms: \(\left(1+\frac{37}{365000}\right)^{365}-1\) - step2: Add the numbers: \(\left(\frac{365037}{365000}\right)^{365}-1\) - step3: Rewrite the expression: \(\frac{365037^{365}}{365000^{365}}-1\) - step4: Reduce fractions to a common denominator: \(\frac{365037^{365}}{365000^{365}}-\frac{365000^{365}}{365000^{365}}\) - step5: Transform the expression: \(\frac{365037^{365}-365000^{365}}{365000^{365}}\) The effective annual interest rate is approximately 3.7691%. Therefore, the amount in the account after one year is $5499.76, and the effective annual interest rate is approximately 3.7691%.