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Mkey [24]
3 years ago
5

A 1 quart container of oil cost $2.89 a case of 12 1 quart containers cost $28.56 how much is saved per container by buying the

oil by the case
Mathematics
1 answer:
FromTheMoon [43]3 years ago
3 0
1= 2.89
12= 28.56


so, 28.56/2.89= 9.88 so per each container saved by buying oil is 9.88 dollars
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What fraction of a liter is 270 mL?
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2 years ago
Mr. Zelman wants to give 50% of his comic book collection to his nephew Alex. If Mr. Zelman has 800 comic books, then how many w
BARSIC [14]

Mr. Zelman gave 400 books to his nephew

<em><u>Solution:</u></em>

Given that, Mr. Zelman wants to give 50% of his comic book collection to his nephew Alex

Mr. Zelman has 800 comic books

Number of books with Zelman = 800

Mr. Zelman wants to give 50% of books to Alex

Number of books given to nephew = 50 % of books of Zelman

Therefore, we have to find 50 % of 800

\rightarrow 50 \% \text{ of } 800

\rightarrow 50 \% \times 800\\\\\rightarrow \frac{50}{100} \times 800 = 400

Thus he gave 400 books to his nephew

4 0
3 years ago
Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on
boyakko [2]

<u>Answer:</u>

<em>Mathew invested</em><em> $600 and $2400</em><em> in each account.</em>

<u>Solution:</u>

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest (I_1) earned in first account for one year,

\text {simple interest}=\frac{\text {pnr}}{100}

Where  

p = amount invested in first account

n = number of years  

r = rate of interest

hence, by using above equation we get (I_1) as,  

I_{1}=\frac{P \times 1 \times 3}{100} ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest (I_2) earned in second account,

I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

I = I_1+ I_2 ---- eqn 4

By substituting eqn 1 , 2, 3 in eqn 4

\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}

\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400  

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600  

Hence, Mathew invested $600 and $2400 in each account.

3 0
3 years ago
Read 2 more answers
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