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Nata [24]
3 years ago
10

Alice purchased 4 1⁄2 kilograms of olive oil for $27. What is the price per kilogram?

Mathematics
1 answer:
Ede4ka [16]3 years ago
8 0
<h3>Answer: 6 dollars per kilogram</h3>

=====================================================

Explanation:

The mixed number 4 & 1/2 converts to the decimal form 4.5

The unit price is 27/(4.5) = 6 dollars per kilogram

------

If you wanted, you can think of it like this ratio

4.5 kg : 27 dollars

Dividing both parts of that ratio by 4.5 will lead to

1 kg : 6 dollars

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Please use the picture and help. The topic is scale factor
Elden [556K]

Answer:

6.5

Step-by-step explanation:

divide 32.5 and 5 i believe this is the awnser

8 0
3 years ago
Read 2 more answers
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

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}

128000 = 138.914 P

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

6 0
4 years ago
Please help me!
Alexandra [31]
In constructing the equation, you need to know the following:

1. What don't we know? How many minutes you must talk to have the same cost for both calling plans. So, let x be the number of minutes.
2. What do we know? Plan 1 charges $17.50 per month plus $0.17 per minute used and Plan 2 charges $32 per month plus $0.07 per minute used.

So the equation must look like this: 17.50 + .17x = 32 + 0.07x

Solving the equation:

1. Multiply both sides by 100
(100) 17.5 + .17x = 32 + 0.07x (100)
1750 + 17x = 3200 + 7x

2. Subtract 1750 from both sides
1750 + 17x - 1750 = 3200 + 7x - 1750
17x = 7x +1450

3. Subtract 7x from both sudes
17x - 7x = 7x + 1450 - 7x
10x = 1450

4. Divide both sides by 100
10x / 10 = 1450/10

x= 145 minutes

145 minutes is the number of minutes you must talk to have the same cost for both calling plans.
7 0
3 years ago
Combine the like terms to make a simplified expression:<br><br> - 2w - 5w=
timofeeve [1]

Answer:

-7w

Step-by-step explanation:

-2w-5w

-2w + (-5w)

-7w

5 0
3 years ago
Read 2 more answers
Petra and Robyn work part-time on weekends. The ratio of Petra's wage to
Marianna [84]

Step-by-step explanation:

   4    +       3 = 7

petra   robyn    total

  210     /   7          =30

wage     total

              ratio

30× 4  = $120

     <em>Petra's part</em>

30× 3  =  $90

      <em>Robyn’s part</em>

6 0
3 years ago
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