Find the monthly payment for a mortgage of \( \$ 85,500 \) at \( 6^{\frac{3}{4}} \% \) for 20 years. Select one: a. \( \$ 6,500.00 \) b. \( \$ 650.66 \) c. \( \$ 876.54 \) d. \( \$ 393.60 \)

We use the monthly payment formula for a fixed-rate mortgage: \[ M = P \cdot \frac{i (1+i)^n}{(1+i)^n-1} \] where - \(P = 85500\) is the principal, - \(i\) is the monthly interest rate, - \(n\) is the total number of payments. **Step 1. Calculate the monthly interest rate** The annual interest rate is given as \(6^{\frac{3}{4}}\%\), which is equivalent to \(6.75\%\) in decimal form \(0.0675\). The monthly rate \(i\) is: \[ i = \frac{0.0675}{12} \approx 0.005625 \] **Step 2. Calculate total number of payments** For a term of 20 years, with 12 payments per year: \[ n = 20 \times 12 = 240 \] **Step 3. Evaluate \((1+i)^n\)** \[ (1+i)^n = (1.005625)^{240} \] Using logarithms or a calculator, we find: \[ (1.005625)^{240} \approx 3.845 \] **Step 4. Substitute values into the payment formula** Substitute \(P = 85500\), \(i = 0.005625\), and \((1+i)^n \approx 3.845\) into the formula: \[ M = 85500 \cdot \frac{0.005625 \times 3.845}{3.845 - 1} \] Calculate the numerator: \[ 0.005625 \times 3.845 \approx 0.02164 \] Multiply with the principal: \[ 85500 \times 0.02164 \approx 1849.62 \] Calculate the denominator: \[ 3.845 - 1 = 2.845 \] Thus, the monthly payment \(M\) is: \[ M \approx \frac{1849.62}{2.845} \approx 650.66 \] The correct monthly payment is: \[ \boxed{\$650.66} \] So, the answer is option b: \( \$650.66 \).