How much would you need to deposit in an account now in order to have \( \$ 5000 \) in the account in 10 years? Assume the account earns \( 3 \% \) interest compounded monthly. Round value to the nearest cent.

To find out how much you would need to deposit now to have \( \$5000 \) in 10 years with an interest rate of \( 3\% \) compounded monthly, we can use the formula for compound interest: \[ P = \frac{A}{(1 + \frac{r}{n})^{nt}} \] Where: - \( P \) is the present value (the deposit now), - \( A \) is the amount of money accumulated after n years, including interest, - \( r \) is the annual interest rate (decimal), - \( n \) is the number of times that interest is compounded per year, and - \( t \) is the number of years the money is invested or borrowed. In your case: - \( A = 5000 \) - \( r = 0.03 \) - \( n = 12 \) (monthly), - \( t = 10 \). Plugging in the numbers: \[ P = \frac{5000}{(1 + \frac{0.03}{12})^{12 \times 10}} \] Calculating: \[ P = \frac{5000}{(1 + 0.0025)^{120}} = \frac{5000}{(1.0025)^{120}} \approx \frac{5000}{1.349353} \approx 3706.44 \] So, you would need to deposit approximately **$3706.44** now to have **$5000** in your account in **10 years** at a **3%** interest rate compounded monthly.