1. **Problem statement:** Mary is taking a $23000$ loan for 5 years and wants to compare monthly payments and total payments from two lenders with different annual interest rates. 2. **Formula for monthly payment on an amortized loan:** $$M = P \times \frac{r(1+r)^n}{(1+r)^n - 1}$$ where: - $M$ is the monthly payment - $P$ is the loan principal ($23000$) - $r$ is the monthly interest rate (annual rate divided by 12) - $n$ is the total number of payments (months) 3. **Calculate for lender (a):** - Annual interest rate = $8.8\% = 0.088$ - Monthly interest rate $r = \frac{0.088}{12} = 0.0073333333$ - Number of payments $n = 5 \times 12 = 60$ Calculate numerator: $$r(1+r)^n = 0.0073333333 \times (1 + 0.0073333333)^{60}$$ Calculate denominator: $$(1+r)^n - 1 = (1 + 0.0073333333)^{60} - 1$$ Calculate $(1+r)^n$: $$(1.0073333333)^{60} \approx 1.5657$$ Numerator: $$0.0073333333 \times 1.5657 = 0.011475$$ Denominator: $$1.5657 - 1 = 0.5657$$ Monthly payment: $$M = 23000 \times \frac{0.011475}{0.5657} = 23000 \times 0.02029 = 466.67$$ 4. **Calculate for lender (b):** - Annual interest rate = $8.4\% = 0.084$ - Monthly interest rate $r = \frac{0.084}{12} = 0.007$ - Number of payments $n = 60$ Calculate $(1+r)^n$: $$(1.007)^{60} \approx 1.48985$$ Numerator: $$0.007 \times 1.48985 = 0.010429$$ Denominator: $$1.48985 - 1 = 0.48985$$ Monthly payment: $$M = 23000 \times \frac{0.010429}{0.48985} = 23000 \times 0.02129 = 489.67$$ 5. **Compare total payments:** - Total payment lender (a): $$466.67 \times 60 = 28000.20$$ - Total payment lender (b): $$489.67 \times 60 = 29380.20$$ 6. **Conclusion:** The savings and loan association (a) offers the lower total payment. Difference: $$29380.20 - 28000.20 = 1380.00$$ **Answer:** (a) Monthly payment = $466.67$ (b) Monthly payment = $489.67$ (c) Savings and loan association has the lower total payment by $1380.00$.