A bank features a savings account that has an annual percentage rate of \( 5.8 \% \) with interest compounded semiannually. Laila deposits \( \$ 10,000 \) into the account. How much money will Laila have in the account in 1 year? Answer = \( \$ \square \) What is the annual percentage yield (APY) for the savings account? APY \( =\square \). Round to the nearest hundredth of a percent.

To calculate how much money Laila will have in her account after 1 year with semiannual compounding, you can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) = the amount of money accumulated after n years, including interest. - \( P \) = the principal amount (the initial deposit). - \( r \) = the annual interest rate (decimal). - \( n \) = the number of times that interest is compounded per year. - \( t \) = the number of years the money is invested or borrowed. Substituting the known values: - \( P = 10,000 \) - \( r = 0.058 \) (5.8% as a decimal) - \( n = 2 \) (since it is compounded semiannually) - \( t = 1 \) The formula becomes: \[ A = 10,000 \left(1 + \frac{0.058}{2}\right)^{2 \cdot 1} \] \[ A = 10,000 \left(1 + 0.029\right)^{2} \] \[ A = 10,000 \left(1.029\right)^{2} \] \[ A = 10,000 \times 1.059841 \] \[ A \approx 10,598.41 \] So, Laila will have approximately \( \$ 10,598.41 \) in the account after 1 year. To find the annual percentage yield (APY), use the formula: \[ APY = \left(1 + \frac{r}{n}\right)^{n} - 1 \] Substituting the values: \[ APY = \left(1 + \frac{0.058}{2}\right)^{2} - 1 \] \[ APY = \left(1.029\right)^{2} - 1 \] \[ APY \approx 1.059841 - 1 \] \[ APY \approx 0.059841 \] Converting this to a percentage: \[ APY \approx 5.98\% \] Thus, the annual percentage yield (APY) for the savings account is approximately \( 5.98\% \). So, the final answers are: Answer = \( \$ 10,598.41 \) APY \( = 5.98 \% \)