Alecia wants \( \$ 7,000 \) saved in 4 years to make a down payment on a house. How much money should she invest now at \( 6.1 \% \) compounded annually in order to meet her goal? Round value to the nearest cent.

To solve this problem, we can use the formula for compound interest: \[ A = P \left( 1 + \frac{r}{n} \right)^{nt} \] where: - \( A \) is the amount of money Alecia wants to save (in this case, $7,000), - \( P \) is the principal amount (the amount of money Alecia invests now), - \( r \) is the annual interest rate (in decimal form, so 6.1% becomes 0.061), - \( n \) is the number of times the interest is compounded per year (in this case, 1 since it's compounded annually), - \( t \) is the time in years (in this case, 4 years). We need to solve for \( P \) to find out how much money Alecia should invest now to meet her goal. Let's substitute the given values into the formula and solve for \( P \). Solve the equation by following steps: - step0: Solve for \(P\): \(7000=P\left(1+0.061\right)^{4}\) - step1: Simplify: \(7000=\frac{1061^{4}}{1000^{4}}\times P\) - step2: Swap the sides: \(\frac{1061^{4}}{1000^{4}}\times P=7000\) - step3: Rewrite the expression: \(\frac{1061^{4}P}{1000^{4}}=7000\) - step4: Cross multiply: \(1061^{4}P=1000^{4}\times 7000\) - step5: Simplify the equation: \(1061^{4}P=7000\times 1000^{4}\) - step6: Divide both sides: \(\frac{1061^{4}P}{1061^{4}}=\frac{7000\times 1000^{4}}{1061^{4}}\) - step7: Divide the numbers: \(P=\frac{7000\times 1000^{4}}{1061^{4}}\) The solution to the equation is \( P = \frac{7000 \times 1000^4}{1061^4} \). Now, we can calculate the value of \( P \) to find out how much money Alecia should invest now to meet her goal. Calculate the value by following steps: - step0: Calculate: \(\frac{7000\times 1000^{4}}{1061^{4}}\) Alecia should invest approximately $5,523.78 now at 6.1% compounded annually to meet her goal of saving $7,000 in 4 years.