Suppose Miguel places \( \$ 7000 \) in an account that pays \( 2 \% \) 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 find the amount in the account at the end of each year with compounded interest, we can use the formula: \[ A = P(1 + r)^n \] where: - \( A \) is the amount in the account after n years, - \( P \) is the principal amount (initial investment), - \( r \) is the annual interest rate (in decimal), - \( n \) is the number of years the money is invested for. Given \( P = 7000 \) and \( r = 0.02 \): (a) For the end of 1 year (n = 1): \[ A = 7000(1 + 0.02)^1 = 7000(1.02) = 7140 \] So, the amount in the account at the end of 1 year is \( \$ 7140 \). (b) For the end of 2 years (n = 2): \[ A = 7000(1 + 0.02)^2 = 7000(1.02)^2 = 7000(1.0404) = 7282.80 \] Therefore, the amount in the account at the end of 2 years is \( \$ 7282.80 \).