Suppose Melissa places \( \$ 9500 \) in an account that pays \( 11 \% \) interest compounded each year. Assume that no withdrawals are made from the account. Follow the instructions below. Do not do any rounding. (a) Find the amount in the account at the end of 1 year. \( \$ \square \) (b) Find the amount in the account at the end of 2 years. \( \$ \square \)

To solve this problem, we will use the formula for compound interest, which is given by: \[ A = P(1 + r)^t \] where: - \( A \) is the amount in the account after \( t \) years, - \( P \) is the principal amount (initial investment), - \( r \) is the annual interest rate (as a decimal), - \( t \) is the number of years the money is invested or borrowed. Given: - \( P = 9500 \) - \( r = 0.11 \) (which is \( 11\% \)) - \( t \) will be \( 1 \) for part (a) and \( 2 \) for part (b). ### (a) Find the amount in the account at the end of 1 year. Substituting the values into the formula for \( t = 1 \): \[ A = 9500(1 + 0.11)^1 \] Now, let's calculate this step by step. 1. Calculate \( 1 + 0.11 \): \[ 1 + 0.11 = 1.11 \] 2. Now, substitute back into the formula: \[ A = 9500 \times 1.11 \] 3. Calculate \( 9500 \times 1.11 \): \[ A = 10545 \] So, the amount in the account at the end of 1 year is \( \$ 10545 \). ### (b) Find the amount in the account at the end of 2 years. Now, we will calculate for \( t = 2 \): \[ A = 9500(1 + 0.11)^2 \] 1. We already know \( 1 + 0.11 = 1.11 \), so we need to calculate \( (1.11)^2 \): \[ (1.11)^2 = 1.2321 \] 2. Substitute back into the formula: \[ A = 9500 \times 1.2321 \] 3. Calculate \( 9500 \times 1.2321 \): \[ A = 11656.45 \] So, the amount in the account at the end of 2 years is \( \$ 11656.45 \). ### Final Answers: (a) \( \$ 10545 \) (b) \( \$ 11656.45 \)