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Maricela was recently approved for an [tex]$\$[/tex] 18,000[tex]$ loan for 5 years at an interest rate of $[/tex]6.2\%[tex]$. Use the monthly payment formula to complete the statement.

\[
\begin{array}{l}
M=\frac{P \left(\frac{i}{n}\right) \left(1+\frac{i}{n}\right)^{nt}}{\left(1+\frac{i}{n}\right)^{nt}-1} \\
M=\text{monthly payment} \\
P=\text{principal} \\
i=\text{annual interest rate} \\
n=\text{number of payments per year} \\
t=\text{number of years}
\end{array}
\]

Maricela's monthly payment for the loan is $[/tex]\[tex]$ \square$[/tex].


Sagot :

To solve for Maricela's monthly payment for her [tex]$18,000 loan over 5 years with a 6.2% annual interest rate, we need to use the monthly payment formula for an installment loan, which is given by: \[ M = \frac{P \cdot i \cdot (1 + i)^{n \cdot t}}{(1 + i)^{n \cdot t} - 1} \] where: - \( M \) is the monthly payment - \( P \) is the principal amount ($[/tex]18,000)
- [tex]\( r \)[/tex] is the annual interest rate (6.2%)
- [tex]\( t \)[/tex] is the number of years (5)
- [tex]\( n \)[/tex] is the number of payments per year (12 monthly payments)
- [tex]\( i \)[/tex] is the monthly interest rate ([tex]\(\frac{r}{n}\)[/tex])

Let's break down the steps:

1. Determine the monthly interest rate [tex]\( i \)[/tex]:
[tex]\[\text{Annual interest rate } r = 6.2\%\][/tex]
[tex]\[\text{Monthly interest rate } i = \frac{r}{n} = \frac{6.2\%}{12} = \frac{6.2}{100 \cdot 12} = 0.0051667\][/tex]

2. Plug the values into the monthly payment formula:
- [tex]\( P = 18,000 \)[/tex]
- [tex]\( i = 0.0051667 \)[/tex]
- [tex]\( n = 12 \)[/tex]
- [tex]\( t = 5 \)[/tex]

[tex]\[ M = \frac{P \cdot i \cdot (1 + i)^{n \cdot t}}{(1 + i)^{n \cdot t} - 1} \][/tex]
[tex]\[ M = \frac{18000 \cdot 0.0051667 \cdot (1 + 0.0051667)^{12 \cdot 5}}{(1 + 0.0051667)^{12 \cdot 5} - 1} \][/tex]

After calculating this step by step, the result yields:

[tex]\[ M \approx 349.67 \][/tex]

Therefore, Maricela's monthly payment for the loan is \[tex]$349.67. Let's complete the statement: Maricela's monthly payment for the loan is $[/tex]349.67.