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? 20. How much

**Step 1. Write down the compound interest formula** The compound interest formula is \[ A = P\left(1+\frac{r}{n}\right)^{n t} \] where \(A\) is the accumulated amount, \(P\) is the principal (the amount deposited today), \(r\) is the annual interest rate (in decimal form), \(n\) is the number of compounding periods per year, and \(t\) is the time in years. **Step 2. Identify the given values** For this problem, we have: \[ \begin{aligned} A &= 10\,000,\\[1mm] r &= 0.035,\\[1mm] n &= 12,\\[1mm] t &= 3. \end{aligned} \] **Step 3. Solve for \(P\)** We need to rearrange the formula to solve for \(P\): \[ P = \frac{A}{\left(1+\frac{r}{n}\right)^{n t}} \] Substitute the given values: \[ P = \frac{10\,000}{\left(1+\frac{0.035}{12}\right)^{12 \cdot 3}} \] Notice that \(12 \cdot 3 = 36\), so the formula becomes \[ P = \frac{10\,000}{\left(1+\frac{0.035}{12}\right)^{36}} \] **Step 4. Simplify the expression inside the parentheses** Calculate the monthly interest rate: \[ \frac{0.035}{12} \approx 0.00291667 \] Thus, \[ 1+\frac{0.035}{12} \approx 1.00291667 \] **Step 5. Compute the exponent** Raise the value to the 36th power: \[ \left(1.00291667\right)^{36} \] Using logarithms or a calculator: \[ \left(1.00291667\right)^{36} \approx 1.1106 \] **Step 6. Find the principal \(P\)** Now, substitute back to find \(P\): \[ P \approx \frac{10\,000}{1.1106} \approx 9\,000.54 \] Thus, the amount that should be deposited today is approximately \[ \boxed{\$9\,000.54} \]