calculating your monthly payment\npractice the following problems in your notebook and then upload the…

Explanation:

Step1: Recall the loan - payment formula

The formula for the monthly payment of a loan is $M = P\frac{r(1 + r)^n}{(1 + r)^n-1}$, where $M$ is the monthly payment, $P$ is the principal loan amount, $r$ is the monthly interest rate, and $n$ is the total number of payments.

Step2: Solve for problem 1

First, find the monthly interest rate $r_1$. The annual interest rate is $6.04%=0.0604$, so $r_1=\frac{0.0604}{12}$. The principal $P_1 = 14287$ and the number of payments $n_1 = 36$. $r_1=\frac{0.0604}{12}\approx0.005033$ $M_1=14287\times\frac{0.005033(1 + 0.005033)^{36}}{(1 + 0.005033)^{36}-1}$ $(1 + 0.005033)^{36}\approx1.1977$ $0.005033\times(1 + 0.005033)^{36}\approx0.005033\times1.1977 = 0.00603$ $(1 + 0.005033)^{36}-1\approx0.1977$ $M_1=14287\times\frac{0.00603}{0.1977}\approx14287\times0.0305 = 435.75$

Step3: Solve for problem 2

First, find the principal loan amount $P_2$. The cost of the car is $22687$ and the down - payment is $1500$, so $P_2=22687 - 1500=21187$. The annual interest rate is $5.58% = 0.0558$, so the monthly interest rate $r_2=\frac{0.0558}{12}=0.00465$. The number of payments $n_2 = 60$. $(1 + 0.00465)^{60}\approx1.3137$ $0.00465\times(1 + 0.00465)^{60}\approx0.00465\times1.3137=0.00611$ $(1 + 0.00465)^{60}-1\approx0.3137$ $M_2=21187\times\frac{0.00611}{0.3137}\approx21187\times0.0195 = 413.15$

Answer:

1. The monthly payment is approximately $$435.75$.
2. The monthly payment is approximately $$413.15$.