4. Mr. Garza deposited \( \$ 1,350 \) into an account that earns \( 5.25 \% \) interest compounded annually. He made no additional deposits or withdrawals. What will be the balance in Mr. Garza's account in dollars and cents af the end of 6 years?

To find the balance in Mr. Garza's account after 6 years with compound interest, we can use the formula for compound interest: \[ A = P \times \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) is the amount of money accumulated after \( t \) years, including interest. - \( P \) is the principal amount (initial deposit). - \( r \) is the annual interest rate (in decimal form). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for, in years. Given: - \( P = \$1,350 \) - \( r = 5.25\% = 0.0525 \) (in decimal form) - \( n = 1 \) (compounded annually) - \( t = 6 \) years Substitute these values into the formula and calculate the balance in Mr. Garza's account after 6 years. Calculate the value by following steps: - step0: Calculate: \(1350\left(1+\frac{0.0525}{1}\right)^{1\times 6}\) - step1: Divide the terms: \(1350\left(1+\frac{21}{400}\right)^{1\times 6}\) - step2: Add the terms: \(1350\left(\frac{421}{400}\right)^{1\times 6}\) - step3: Calculate: \(1350\left(\frac{421}{400}\right)^{6}\) - step4: Simplify: \(1350\times \frac{421^{6}}{400^{6}}\) - step5: Rewrite the expression: \(50\times 27\times \frac{421^{6}}{400^{6}}\) - step6: Rewrite the expression: \(50\times 27\times \frac{421^{6}}{50^{6}\times 8^{6}}\) - step7: Reduce the numbers: \(27\times \frac{421^{6}}{50^{5}\times 8^{6}}\) - step8: Multiply the fractions: \(\frac{27\times 421^{6}}{50^{5}\times 8^{6}}\) The balance in Mr. Garza's account after 6 years will be approximately $1,835.13.