To calculate the monthly payment for a 30-year mortgage of \[tex]$300,000 at an annual interest rate of 7%, we use the provided mortgage payment formula:
\[ R = \frac{P \cdot \left( \frac{r}{m} \cdot (1 + \frac{r}{m})^n \right)}{\left( (1 + \frac{r}{m})^n - 1 \right)} \]
First, let's list all the necessary variables and their given values:
- \( P \) (Principal amount) = \$[/tex]300,000
- [tex]\( r \)[/tex] (Annual interest rate) = 7% = 0.07 in decimal form
- [tex]\( m \)[/tex] (Number of payments per year) = 12 (since payments are monthly)
- [tex]\( n \)[/tex] (Total number of payments) = [tex]\( m \times t \)[/tex], where [tex]\( t \)[/tex] is the term in years. For a 30-year mortgage:
[tex]\[ n = 12 \times 30 = 360 \][/tex]
Next, we need to calculate the monthly interest rate:
[tex]\[ \frac{r}{m} = \frac{0.07}{12} = 0.005833333333333334 \][/tex]
Substitute the values back into the formula to solve for [tex]\( R \)[/tex]:
[tex]\[ R = \frac{300000 \cdot \left(0.005833333333333334 \cdot (1 + 0.005833333333333334)^{360} \right)}{ \left( (1 + 0.005833333333333334)^{360} - 1 \right)} \][/tex]
We recognize that the formula's components include the exponential term:
[tex]\[ (1 + \frac{r}{m})^n = (1 + 0.005833333333333334)^{360} \approx 10.294637732843006 \][/tex]
Now calculating the numerator:
[tex]\[ \left( 0.005833333333333334 \cdot 10.294637732843006 \right) = 0.06005614787011975 \][/tex]
For the denominator part:
[tex]\[ (1 + 0.005833333333333334)^{360} - 1 \approx 10.294637732843006 - 1 = 9.294637732843006 \][/tex]
Putting this all together in the formula, we get:
[tex]\[ R = \frac{300000 \cdot 0.06005614787011975}{9.294637732843006} \approx \frac{18016.84436103592}{9.294637732843006} = 1995.9074855375472 \][/tex]
Therefore, the monthly payment is:
[tex]\[ \boxed{1995.91} \][/tex]
So, the monthly payment for a \[tex]$300,000 mortgage at a 7% annual interest rate over 30 years is approximately \$[/tex]1995.91.
