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2 years ago

5

Sky needs $6700 for school and has two options. Option A: 5 years at 5% interest with monthly payments of $126.44 Option B: 6 ye

ars at 3.5% interest with monthly payments of $103.30 Calculate the total payback on each loan and calculate how much Sky will save by choosing option B.

1 answer:

zimovet [89]2 years ago

4 0

Answer:

148.80

Step-by-step explanation:

Take 126.44 multiply by 60 months equals 7586.40

Take 103.30 multiply by 72 months equals 7437.60

Subtract 7437.60 from 7586.40

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5x=25 could be one.
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There are 28.35 grams in an ounce and 2.21 pounds in a kilogram. Miriam converted 7 kilograms to ounces, but her answer is not c

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Answer: OPTION C.

Step-by-step explanation:

For this exercise it is important to remember the following:

$1\ kilogram=1,000\ grams$

Knowing this and based on the information provided in the exercise, you can make the conversion from kilograms to ounces. This is:

$(7\ Kg)(\frac{1,000\ gr}{1\ Kg})(\frac{1\ oz}{28.35\ gr})=246.913\ oz$

According to the exercise, Miriam's procedure was:

$(7\ Kg)(\frac{1,000\ gr}{1\ Kg})(\frac{28.35\ gr}{1\ oz})= 198,450\ oz$

So you can identify that she made a mistake.

Notice that:

1. Miriaim multiplied correctly.

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2 years ago

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Rodolpho uses a prepaid gas card to spend $35 each week for gas. After the first week, he has $140 left on the card. After every

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I got this T_T y=175-35t

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Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

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Answer:

Therefore $k= \frac{ln2 }{18}$ , A=184

Step-by-step explanation:

Given function is

$T(t)=230 -e^{-kt}$

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

$138=230 -Ae^{-k\times 18}$

$\Rightarrow -Ae^{-18k}=138-230$

$\Rightarrow Ae^{-18k}=92$ .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

$184=230-Ae^{-k\times 36}$

$\Rightarrow Ae^{-36k}=230-184$

$\Rightarrow Ae^{-36k}=46$ .......(2)

Dividing (2) by (1)

$\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}$

$\Rightarrow e^{-18k}=\frac{46}{92}$

Taking ln both sides

$ln e^{-18k}=ln\frac{46}{92}$

$\Rightarrow -18k =ln (\frac12)$

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