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3 years ago

Select the correct option from each drop-down menu.

Mathematics

1 answer:

lys-0071 [83]3 years ago

4 0

9514 1404 393

Answer:

The function r(x)=(2+.25x)(240-20x) reveals that the revenue earned by the cafe from coffee sales will be $500 after 2 $0.25 price increases.

Step-by-step explanation:

The revenue from sales is the product of the price per cup and the number of cups sold.

For x increases of $0.25 each, the price per cup is (2 +0.25x).

For x increases of $0.25 each, the number of cups sold is (240 -20x).

Then the revenue as a function of x is ...

r(x) = (2 +0.25x)(240 -20x)

The revenue function can be simplified to better show its zeros.

r(x) = 5(8 +x)(12 -x)

The maximum is halfway between the zeros, so is at x = (-8 +12)/2 = 2.

When x=2, the revenue is ...

r(2) = 5(8+2)(12-2) = 5(10)(10) = 500

The function r(x)=(2+.25x)(240-20x) reveals that the revenue earned by the cafe from coffee sales will be $500 after 2 $0.25 price increases.

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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

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Step-by-step explanation:

From the given information:

The present value of the house = 128000

interest rate compounded monthly r = 7.8% = 0.078

number of months in a year n= 12

duration of time t = 30 years

To find their regular monthly payment, we have:

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

$128000 = P \begin {bmatrix} \dfrac{1 - (1 + \dfrac{0.078}{12})^{- 12*30}}{\dfrac{0.078}{12}} \end {bmatrix}$

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P = 128000/138.914

P = $921.433

∴ Their regular monthly payment P = $921.433

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

when they begin the payment ,

t = 30 year - 5years

t= 25 years

$PV= 921.433 \begin {bmatrix} \dfrac{1 - (1 - \dfrac{0.078}{12})^{25*30}}{\dfrac{0.078}{12}} \end {bmatrix}$

PV = $121718.2714

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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