\( 1 \leqslant \begin{array}{l}\text { Tamara invested } \$ 15,000 \text { in a money market account with an interest rate of } 2.25 \% \text { compounded semiannually. Five } \\ \text { years later, Tamara withdrew the full amount to put toward the down payment on a new house. How much did Tamara } \\ \text { withdraw from the account? } \\ \text { Tamara withdrew } \$ \square \text {. } \\ \text { (Round to the nearest cent as needed.) }\end{array} \). \( \quad \).

To find out how much Tamara withdrew from her money market account after 5 years, 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 investment), - \( r \) 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. In this case: - \( P = 15,000 \) - \( r = 0.0225 \) (2.25% as a decimal), - \( n = 2 \) (since the interest is compounded semiannually), - \( t = 5 \). Plugging in the values: \[ A = 15000 \left(1 + \frac{0.0225}{2}\right)^{2 \cdot 5} \] This simplifies to: \[ A = 15000 \left(1 + 0.01125\right)^{10} \] \[ A = 15000 \left(1.01125\right)^{10} \] Calculating \( (1.01125)^{10} \): \[ (1.01125)^{10} \approx 1.115686 \] Now substituting back to find \( A \): \[ A \approx 15000 \times 1.115686 \approx 16735.29 \] Thus, Tamara withdrew approximately **$16,735.29**. So, the final answer is: Tamara withdrew \$16,735.29.