Rememberme 19. How much money should be deposited today in an account that earns \( 3.5 \% \) compounded monthly so that it will accumulate to \( \$ 10,000 \) in three years?

**Step 1. Write the formula for compound interest** The future value \( FV \) with a present deposit \( P \) compounded periodically is given by: \[ FV = P \left(1 + \frac{r}{m}\right)^{mt} \] where \( r \) is the annual interest rate, \( m \) is the number of compounding periods per year, and \( t \) is the time in years. **Step 2. Substitute the given values** We are given: \( FV = 10,\!000 \) dollars, \( r = 0.035 \) (which is \( 3.5\% \)), \( m = 12 \) (monthly compounding), and \( t = 3 \) years. Substitute these into the formula: \[ 10,\!000 = P \left(1 + \frac{0.035}{12}\right)^{12 \times 3} \] **Step 3. Simplify the equation** First, calculate the periodic rate: \[ \frac{0.035}{12} \approx 0.00291667 \] and the total number of compounding periods is: \[ 12 \times 3 = 36 \] Thus, the equation becomes: \[ 10,\!000 = P \left(1.00291667\right)^{36} \] **Step 4. Solve for \( P \)** Rearrange the equation to solve for \( P \): \[ P = \frac{10,\!000}{\left(1.00291667\right)^{36}} \] **Step 5. Calculate \(\left(1.00291667\right)^{36}\)** Taking natural logarithms (or using a calculator), we have: \[ \left(1.00291667\right)^{36} \approx 1.11047 \] **Step 6. Compute \( P \)** Now substitute the computed value: \[ P = \frac{10,\!000}{1.11047} \approx 9000 \] **Answer** The amount that should be deposited today is approximately \(\$9000\).