Suson has R1 000 that she invests in an occount that pays \( 7.5 \% \) interest compounded yearly. How much money does Susan have at the end of 5 ye

To find out how much money Susan has at the end of 5 years with an investment of R1,000 and an annual interest rate of 7.5% compounded yearly, we can use the formula for compound interest: \[ A = P \left( 1 + \frac{r}{100} \right)^n \] Where: - \( A \) is the amount of money at the end of the investment period - \( P \) is the principal amount (initial investment) - \( r \) is the annual interest rate - \( n \) is the number of years Given: - \( P = R1,000 \) - \( r = 7.5\% \) - \( n = 5 \) years Substitute the values into the formula: \[ A = 1000 \left( 1 + \frac{7.5}{100} \right)^5 \] Now, we can calculate the amount of money Susan has at the end of 5 years. Calculate the value by following steps: - step0: Calculate: \(1000\left(1+\frac{7.5}{100}\right)^{5}\) - step1: Divide the terms: \(1000\left(1+\frac{3}{40}\right)^{5}\) - step2: Add the numbers: \(1000\left(\frac{43}{40}\right)^{5}\) - step3: Simplify: \(1000\times \frac{43^{5}}{40^{5}}\) - step4: Rewrite the expression: \(40\times 25\times \frac{43^{5}}{40^{5}}\) - step5: Reduce the numbers: \(25\times \frac{43^{5}}{40^{4}}\) - step6: Rewrite the expression: \(25\times \frac{43^{5}}{5^{4}\times 8^{4}}\) - step7: Rewrite the expression: \(5^{2}\times \frac{43^{5}}{5^{4}\times 8^{4}}\) - step8: Reduce the numbers: \(1\times \frac{43^{5}}{5^{2}\times 8^{4}}\) - step9: Multiply: \(\frac{43^{5}}{320^{2}}\) Susan will have approximately R1,435.63 at the end of 5 years.