A depositor opens a new savings account with \( \$ 10,000 \) at \( 7 \% \) compounded semiannually. At the beginning of year 3, an additional \( \$ 5000 \) is deposited. At the end of four years, what is the balance in the accou The balance in the account at the end of 4 years is \( \$ \square \) (Round to the nearest cent as needed.)

To calculate the balance in the account at the end of 4 years, we'll break it down into two parts: the first two years prior to the additional deposit and the subsequent two years after the deposit. ### Step 1: Calculate the balance at the end of Year 2 The formula for compound interest is: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial 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. For the first two years: - \( P = 10,000 \) - \( r = 0.07 \) - \( n = 2 \) (since it's compounded semiannually) - \( t = 2 \) Plugging in these values: \[ A = 10,000 \left(1 + \frac{0.07}{2}\right)^{2 \times 2} = 10,000 \left(1 + 0.035\right)^4 \] \[ A = 10,000 \left(1.035\right)^4 \approx 10,000 \left(1.148882\right) \approx 11,488.82 \] ### Step 2: Add the additional deposit at the beginning of Year 3 Now at the beginning of Year 3, an additional \( \$5000 \) is deposited, making the new principal: \[ P = 11,488.82 + 5,000 = 16,488.82 \] ### Step 3: Calculate the balance at the end of Year 4 Now we calculate for the next two years with the new principal: \[ A = 16,488.82 \left(1 + \frac{0.07}{2}\right)^{2 \times 2} = 16,488.82 \left(1.035\right)^4 \] \[ A = 16,488.82 \times 1.148882 \approx 18,906.89 \] Rounding to the nearest cent, the balance in the account at the end of 4 years is: \(\$18,906.89\)