Suppose \( \$ 5,000 \) is deposited into an account paying \( 7.5 \% \) interest, compounded annually. How much money is in the account after four years if no withdrawals or additional deposits are made?

To find the amount of money in the account after four years with no withdrawals or additional deposits, we can use the formula for compound interest: \[ A = P \left( 1 + \frac{r}{100} \right)^n \] where: - \( A \) is the amount of money in the account after \( n \) years, - \( P \) is the principal amount (initial deposit), - \( r \) is the annual interest rate, - \( n \) is the number of years. Given: - Principal amount (\( P \)) = $5,000 - Annual interest rate (\( r \)) = 7.5% - Number of years (\( n \)) = 4 Substitute the given values into the formula: \[ A = 5000 \left( 1 + \frac{7.5}{100} \right)^4 \] Now, we can calculate the amount of money in the account after four years. Calculate the value by following steps: - step0: Calculate: \(5000\left(1+\frac{7.5}{100}\right)^{4}\) - step1: Divide the terms: \(5000\left(1+\frac{3}{40}\right)^{4}\) - step2: Add the numbers: \(5000\left(\frac{43}{40}\right)^{4}\) - step3: Simplify: \(5000\times \frac{43^{4}}{40^{4}}\) - step4: Rewrite the expression: \(40\times 125\times \frac{43^{4}}{40^{4}}\) - step5: Reduce the numbers: \(125\times \frac{43^{4}}{40^{3}}\) - step6: Rewrite the expression: \(125\times \frac{43^{4}}{5^{3}\times 8^{3}}\) - step7: Rewrite the expression: \(5^{3}\times \frac{43^{4}}{5^{3}\times 8^{3}}\) - step8: Reduce the numbers: \(1\times \frac{43^{4}}{8^{3}}\) - step9: Multiply: \(\frac{43^{4}}{8^{3}}\) - step10: Simplify: \(\frac{43^{4}}{512}\) The amount of money in the account after four years is $6677.345703125.