10 Multiple Choice 1 point hboard An investor deposited \( \$ 6,100 \) in an investment account. - The account pays \( 3.1 \% \) interest compounded annually. - The investor leaves the money in the account for 2 years. - The investor make no additional deposits or withdrawals. What is the balance in the account at the end of 2 years? \$6,686.21 \( \$ 6,478.20 \) \$6,484.06 History 5 \( \$ 6,289.10 \) 88 6 Mastery 7 Clear my selection Help Previous

To find the balance in the investment account at the end of 2 years with an initial deposit of \( P = 6100 \) and an annual interest rate of \( r = 3.1\% \) compounded annually, we can use the formula for compound interest: \[ A = P(1 + r)^t \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial deposit). - \( r \) is the annual interest rate (decimal). - \( t \) is the number of years the money is invested or borrowed. ### Step 1: Convert the interest rate from percentage to decimal \[ r = \frac{3.1}{100} = 0.031 \] ### Step 2: Substitute the known values into the formula \[ A = 6100(1 + 0.031)^2 \] ### Step 3: Calculate \( (1 + r) \) \[ 1 + r = 1 + 0.031 = 1.031 \] ### Step 4: Calculate \( (1.031)^2 \) \[ (1.031)^2 = 1.062061 \] ### Step 5: Calculate the final amount \( A \) \[ A = 6100 \times 1.062061 \] Now, let's perform the calculation for \( A \). Calculate the value by following steps: - step0: Calculate: \(6100\times 1.062061\) - step1: Multiply the numbers: \(6478.5721\) The calculation gives us: \[ A \approx 6478.5721 \] ### Final Step: Round to two decimal places The balance in the account at the end of 2 years is approximately \( \$ 6,478.57 \). ### Conclusion Among the provided options, the closest match is: \(\$ 6,478.20\) Thus, the correct answer is \( \$ 6,478.20 \).