se PMT \( =\frac{\mathrm{P}\left(\frac{r}{n}\right)}{\left[1-\left(1+\frac{r}{n}\right)^{-n t}\right]} \) to determine the regular payment amount, rounded to the nearest dollar. The price of a ome is \( \$ 183,000 \). The bank requires a \( 20 \% \) down payment and three points at the time of closing. The cost of the ome is financed with a 30 -year fixed-rate mortgage at \( 10 \% \). Complete parts (a) through (e) below. a. Find the required down payment. \( \$ 36,600 \) b. Find the amount of the mortgage. \( \$ 146,400 \) c. How much must be paid for the three points at closing? \( \$ 4,392 \) (Round to the nearest dollar as needed.) d. Find the monthly payment (excluding escrowed taxes and insurance). \( \$ 1,285 \) (Round to the nearest dollar as needed.) e. Find the total cost of interest over 30 years. \( \$ \square \) (Round to the nearest dollar as needed.)

**Step 1. Down Payment** The down payment is \(20\%\) of the home price. So, \[ \text{Down Payment} = 0.20 \times \$183,\!000 = \$36,\!600. \] **Step 2. Mortgage Amount** The mortgage amount is the remaining \(80\%\) of the home price: \[ \text{Mortgage Amount} = \$183,\!000 - \$36,\!600 = \$146,\!400. \] **Step 3. Points at Closing** Three points cost \(3\%\) of the mortgage amount. Hence, \[ \text{Cost of Points} = 0.03 \times \$146,\!400 = \$4,\!392. \] **Step 4. Monthly Payment Calculation** We use the formula: \[ \text{PMT} = \frac{P\left(\frac{r}{n}\right)}{1-\left(1+\frac{r}{n}\right)^{-n t}}, \] where \(P = \$146,\!400\) (mortgage amount), \(r = 0.10\) (annual interest rate), \(n = 12\) (monthly compounding), \(t = 30\) (years). Substitute the values: \[ \text{PMT} = \frac{146,\!400\left(\frac{0.10}{12}\right)}{1-\left(1+\frac{0.10}{12}\right)^{-12 \times 30}}. \] Compute the monthly rate: \[ \frac{r}{n} = \frac{0.10}{12} \approx 0.0083333. \] The numerator becomes: \[ 146,\!400 \times 0.0083333 \approx 1,\!220. \] For the denominator compute: \[ 1-\left(1+0.0083333\right)^{-360}. \] First, note: \[ 1+0.0083333 \approx 1.0083333. \] Then, \[ 1.0083333^{360} \approx 19.82, \] so \[ \left(1.0083333\right)^{-360} \approx \frac{1}{19.82} \approx 0.05045. \] Thus, the denominator is: \[ 1-0.05045 = 0.94955. \] Now, \[ \text{PMT} \approx \frac{1,\!220}{0.94955} \approx 1,\!285. \] So, the monthly payment is approximately: \[ \$1,\!285. \] **Step 5. Total Cost of Interest Over 30 Years** The total amount paid over 30 years is: \[ \text{Total Paid} = \text{Monthly Payment} \times \text{Number of Months} = 1,\!285 \times 360 = \$462,\!600. \] Subtract the mortgage amount to find the total interest cost: \[ \text{Total Interest} = \$462,\!600 - \$146,\!400 = \$316,\!200. \] **Final Answers:** a. Required down payment: \(\$36,\!600\) b. Mortgage amount: \(\$146,\!400\) c. Points at closing: \(\$4,\!392\) d. Monthly payment: \(\$1,\!285\) e. Total cost of interest over 30 years: \(\$316,\!200\)