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CaHeK987 [17]
3 years ago
12

Leila is considering buying her first home. The house she is interested in buying is priced at $125,000. Leila can put down a $2

0,000 payment, and she qualifies for a 30-year mortgage at 6%. What will her monthly mortgage payment be?

Mathematics
2 answers:
serg [7]3 years ago
6 0

Answer:

$629.53

Step-by-step explanation:

A P E X

PIT_PIT [208]3 years ago
3 0

Answer:

* The monthly mortgage payment is $629.53 ⇒ answer C

Step-by-step explanation:

* Lets explain how to solve the problem

- Leila is considering buying her first home

- The house she is interested in buying is priced at $125,000

∴ She can put $20000 down  payment

* Lets find the balance to be paid off on mortgage

∴ The balance = 125000 - 20000 = 105000

- She qualifies for a 30-year mortgage at 6%

* Lets find the rule of the monthly payment

∵ pmt=\frac{\frac{r}{n}[P(1+\frac{r}{n})^{tn}]}{(1+\frac{r}{n})^{tn}-1} , where  

- pmt is the monthly mortgage payment

- P = the initial amount  

- r = the annual interest rate (decimal)

- n = the number of times that interest is compounded per unit t

- t = the time the money is invested or borrowed for

∵ P = 105000

∵ r = 6/100 = 0.06

∵ n = 12

∵ t = 30

∴ pmt=\frac{\frac{0.06}{12}[105000(1+\frac{0.06}{12})^{30(12)}}{(1+\frac{0.06}{12})^{30(12)}-1}

∴  pmt=\frac{0.005[105000(1.005)^{360}]}{(1.005)^{360}-1} =629.528

* The monthly mortgage payment is $629.53

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Answer:

1. The Venn diagrams are attached

2. When the statistics students number = 10·x + 3, we have;

The number of students that study

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The number of students that study

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Step-by-step explanation:

The parameters given are;

Total number of students = 30

Number of students that study algebra n(A) = x + 10

Number of students that study statistics n(B) = 10·x + 3

Number of student that study both algebra and statistics n(A∩B) = 4

Number of student that study only algebra n(A\B) = 2·x

Number of students that study neither algebra or statistics n(A∪B)' = 3

Therefore;

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n(A∪B) = n(A) + n(B) - n(A∩B)

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n(A) = 18/11 + 10 = 128/11

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n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

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