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3 years ago

Please answer this thanks

Mathematics

2 answers:

olganol [36]3 years ago

8 0

Answer:

It's C! 75%

Step-by-step explanation:

matrenka [14]3 years ago

3 0

Answer: 75%

Step-by-step explanation:

Just form the simple equation of 14/8=x/100

cross mulitply: 8x=1400

=> x=175

175-100(super important step)=75

75%

:) Tada your answer. Have a nice day

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Answer:

The answer is B, C, D, and E

Step-by-step explanation:

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Clark and Lana take a 30-year home mortgage of $128,000 at 7.8%, compounded monthly. They make their regular monthly payments fo

svp [43]

Answer:

Step-by-step explanation:

From the given information:

The present value of the house = 128000

interest rate compounded monthly r = 7.8% = 0.078

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To find their regular monthly payment, we have:

$PV = P \begin {bmatrix} \dfrac{1 - (1 + \dfrac{r}{n})^{-nt}}{\dfrac{r}{n}} \end {bmatrix}$

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To find the unpaid balance when they begin paying the $1400.

when they begin the payment ,

t = 30 year - 5years

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C) In order to estimate how many payments of $1400 it will take to pay off the loan, we have:

$121718.2714 = \begin {bmatrix} \dfrac{1300 (1 - \dfrac{12.078}{12}))^{-nt}}{\dfrac{0.078}{12}} \end {bmatrix}$

$121718.2714 = 200000 \begin {bmatrix} (1 - \dfrac{12.078}{12}))^{-nt} \end {bmatrix}$

$\dfrac{121718.2714}{200000 } = \begin {bmatrix} (1 - \dfrac{12.078}{12}))^{-nt} \end {bmatrix}$

$0.60859 = \begin {bmatrix} (1 - \dfrac{12}{12.078}))^{nt} \end {bmatrix}$

$0.60859 = (0.006458)^{nt}$

$nt = \dfrac{0.60859}{0.006458}$

nt = 94.238 payments is required to pay off the loan.

How much interest will they save by paying the loan using the number of payments from part (c)?

The total amount of interest payed on $921.433 = 921.433 × 30(12) years

= 331715.88

The total amount paid using 921.433 and 1300 = (921.433 × 60 )+( 1300 + 94.238)

= 177795.38

The amount of interest saved = 331715.88 - 177795.38

The amount of interest saved = $153920.5

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Umnica [9.8K]

Answer:

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The sum of the first n terms of a geometric series is:

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$a_1$ is the first term of the series,
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