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Sagot :
To calculate Raquel's monthly mortgage payment, we can use the formula for a fixed-rate mortgage, which is given by:
[tex]\[ M = \frac{P \cdot r(1+r)^n}{(1+r)^n - 1} \][/tex]
where:
- [tex]\( M \)[/tex] is the monthly payment.
- [tex]\( P \)[/tex] is the principal loan amount.
- [tex]\( r \)[/tex] is the monthly interest rate.
- [tex]\( n \)[/tex] is the total number of payments (number of months).
Given the problem:
- Principal loan amount, [tex]\( P \)[/tex] = \[tex]$185,000 - Annual interest rate = 3.6% - Years = 30 - Compounding periods per year = 12 First, we need to convert the annual interest rate to a monthly interest rate: \[ r = \frac{\text{Annual interest rate}}{\text{Compounding periods per year}} = \frac{3.6\%}{12} = \frac{0.036}{12} ≈ 0.003 \] Next, we need to determine the total number of monthly payments: \[ n = \text{Years} \times \text{Compounding periods per year} = 30 \times 12 = 360 \] Now, we can substitute the values into the mortgage payment formula: \[ M = \frac{185,000 \cdot 0.003 \cdot (1 + 0.003)^{360}}{(1 + 0.003)^{360} - 1} \] Finally, calculating the monthly payment and rounding to the nearest cent, Raquel's monthly payment will be: \[ M ≈ \$[/tex]841.09 \]
Therefore, her monthly payment will be \$841.09.
[tex]\[ M = \frac{P \cdot r(1+r)^n}{(1+r)^n - 1} \][/tex]
where:
- [tex]\( M \)[/tex] is the monthly payment.
- [tex]\( P \)[/tex] is the principal loan amount.
- [tex]\( r \)[/tex] is the monthly interest rate.
- [tex]\( n \)[/tex] is the total number of payments (number of months).
Given the problem:
- Principal loan amount, [tex]\( P \)[/tex] = \[tex]$185,000 - Annual interest rate = 3.6% - Years = 30 - Compounding periods per year = 12 First, we need to convert the annual interest rate to a monthly interest rate: \[ r = \frac{\text{Annual interest rate}}{\text{Compounding periods per year}} = \frac{3.6\%}{12} = \frac{0.036}{12} ≈ 0.003 \] Next, we need to determine the total number of monthly payments: \[ n = \text{Years} \times \text{Compounding periods per year} = 30 \times 12 = 360 \] Now, we can substitute the values into the mortgage payment formula: \[ M = \frac{185,000 \cdot 0.003 \cdot (1 + 0.003)^{360}}{(1 + 0.003)^{360} - 1} \] Finally, calculating the monthly payment and rounding to the nearest cent, Raquel's monthly payment will be: \[ M ≈ \$[/tex]841.09 \]
Therefore, her monthly payment will be \$841.09.
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