An account pays \( 7.2 \% \) per year Suppose 18 years from now you would like to have \( \$ 4000 \) in the account, with no additional money added to the account other than interest. (a) If it pays \( 7.2 \% \) compounded annually, what minimum amount must be deposited into this account today in order to have \( \$ 4000 \) in the account 18 years from now? (Round to the nearest 0.01 dollars) Number (b) If it pays \( 7.2 \% \) compounded daily ( 365 times per year), what minimum amount must be deposited into this account today in order to have \( \$ 4000 \) in the account 18 years from now? \( \$ \) Number (Round to the nearest 0.01 dollars) (c) If it pays \( 7.2 \% \) compounded continuously, what minimum amount must be deposited into this account today in order to have \( \$ 4000 \) in the account 18 years from now? \( \$ \) Number (Round to the nearest 0.01 dollars)

We start by recalling that if a principal \( P \) is invested at an annual interest rate \( r \) for \( t \) years, then: 1. For annual compounding the future value is \[ FV = P (1 + r)^t. \] 2. For compounding \( n \) times per year the future value is \[ FV = P \left(1 + \frac{r}{n}\right)^{nt}. \] 3. For continuous compounding the future value is \[ FV = P\, e^{rt}. \] In each case we solve for \( P \) given that we want \( FV = \$4000 \) in \( t = 18 \) years with an annual interest rate \( r = 7.2\% = 0.072 \). --- **(a) Compounded Annually:** We have \[ 4000 = P (1 + 0.072)^{18}. \] Solving for \( P \): \[ P = \frac{4000}{(1.072)^{18}}. \] Using a calculator to evaluate \( (1.072)^{18} \) we get approximately 3.483. Therefore, \[ P \approx \frac{4000}{3.483} \approx 1148.01. \] So, the minimum amount to deposit is approximately **\$1148.01**. --- **(b) Compounded Daily:** For daily compounding, \( n = 365 \), so the equation becomes \[ 4000 = P \left(1 + \frac{0.072}{365}\right)^{365 \times 18}. \] That is, \[ P = \frac{4000}{\left(1 + \frac{0.072}{365}\right)^{6570}}. \] Evaluating the expression, we note that the effective growth factor is slightly higher than the annual compound due to more frequent compounding. A calculator gives a value for the denominator of approximately 3.651. Thus, \[ P \approx \frac{4000}{3.651} \approx 1095.86. \] So, the minimum amount to deposit is approximately **\$1095.86**. --- **(c) Compounded Continuously:** For continuous compounding, the equation is \[ 4000 = P\, e^{0.072 \times 18}. \] Since \( 0.072 \times 18 = 1.296 \), solving for \( P \) we obtain \[ P = \frac{4000}{e^{1.296}}. \] Numerically, \( e^{1.296} \) is approximately 3.655, hence \[ P \approx \frac{4000}{3.655} \approx 1093.50. \] So, the minimum amount to deposit is approximately **\$1093.50**. --- Thus, the answers are: (a) \$1148.01 (b) \$1095.86 (c) \$1093.50