Luestion 2 (1 point) Find " \( m \) " for the investment of \( \$ 1000.00 \) for 2 years at \( 1.8 \% \) compounde semi-annually.

To find the value of \( m \) for the investment of $1000.00 for 2 years at 1.8% compounded semi-annually, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (initial investment). - \( r \) 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: - \( P = \$1000.00 \) - \( r = 1.8\% = 0.018 \) - \( n = 2 \) (compounded semi-annually) - \( t = 2 \) years Substitute the given values into the formula: \[ A = 1000 \left(1 + \frac{0.018}{2}\right)^{2 \cdot 2} \] Now, we can calculate the value of \( A \) and then find the value of \( m \) by dividing \( A \) by \( P \). Calculate the value by following steps: - step0: Calculate: \(1000\left(1+\frac{0.018}{2}\right)^{2\times 2}\) - step1: Divide the terms: \(1000\left(1+\frac{9}{1000}\right)^{2\times 2}\) - step2: Add the numbers: \(1000\left(\frac{1009}{1000}\right)^{2\times 2}\) - step3: Multiply the numbers: \(1000\left(\frac{1009}{1000}\right)^{4}\) - step4: Simplify: \(1000\times \frac{1009^{4}}{1000^{4}}\) - step5: Reduce the numbers: \(1\times \frac{1009^{4}}{1000^{3}}\) - step6: Multiply: \(\frac{1009^{4}}{1000^{3}}\) The value of \( A \) is approximately $1036.49. Now, we can find the value of \( m \) by dividing \( A \) by \( P \): \[ m = \frac{A}{P} = \frac{1036.49}{1000} \] Therefore, the value of \( m \) is approximately 1.03649.