malika just graduated from college and owes $49,000 on her student loans. the bank charges an annual…

malika just graduated from college and owes $49,000 on her student loans. the bank charges an annual interest rate of 2.1%, compounded monthly. if malika wants to pay off her student loans using equal monthly payments over the next 9 years, what would the monthly payment be, to the nearest dollar?\n\n$m = \\frac{pr}{1 - (1 + r)^{-n}}$\n\n$m =$ the monthly payment\n$p =$ the amount owed\n$r =$ the interest rate per month\n$n =$ the number of payments

malika just graduated from college and owes $49,000 on her student loans. the bank charges an annual interest rate of 2.1%, compounded monthly. if malika wants to pay off her student loans using equal monthly payments over the next 9 years, what would the monthly payment be, to the nearest dollar?\n\n$m = \\frac{pr}{1 - (1 + r)^{-n}}$\n\n$m =$ the monthly payment\n$p =$ the amount owed\n$r =$ the interest rate per month\n$n =$ the number of payments

Answer

Explanation:

Step1: Calculate the monthly interest rate (r)

The annual interest rate is (2.1%=0.021). Since the interest is compounded monthly, (r=\frac{0.021}{12}=0.00175).

Step2: Calculate the number of payments (n)

The loan is to be paid off over (9) years. Since there are (12) months in a year, (n = 9\times12=108).

Step3: Substitute the values into the formula

We know that (P = 49000), (r=0.00175), and (n = 108). First, calculate ((1 + r)^{-n}=(1 + 0.00175)^{-108}). Let (x=(1 + 0.00175)^{-108}), then (\ln(x)=- 108\times\ln(1.00175)). (\ln(1.00175)\approx0.001748), so (\ln(x)=-108\times0.001748=-0.188784), and (x = e^{-0.188784}\approx0.8287). Then (1-(1 + r)^{-n}=1 - 0.8287 = 0.1713). And (Pr=49000\times0.00175 = 85.75). Now, (M=\frac{Pr}{1-(1 + r)^{-n}}=\frac{85.75}{0.1713}\approx501.75).

Answer:

(502)