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trapecia [35]
3 years ago
13

Pastries made out of filo dough and brushed with either olive oil or butter (but not both). Pastries made out of shortcrust doug

h are not brushed with anything. Rashid and Mikhail submitted a total of x pastries to a baking competition. Mikhail used filo dough for all of his pastries, Rashid used shortcrust dough for all of his pastries, and each pastry was made using only one kind of dough. If Rashid made 2323 as many pastries as Mikhail, and 5858 of the filo dough pastries were brushed with olive oil, then how many of the pastries submitted by Rashid and Mikhail, in terms of x, were brushed with butter?
A. 3X203X20
B. 9X409X40
C. 1X41X4
D. 3X83X8
E. 5X12
Mathematics
1 answer:
Genrish500 [490]3 years ago
6 0

Answer:

\frac{9x}{40} is the answer.

Step-by-step explanation:

The question has some typo errors.

Rashid and Mikhail submitted a total of x pastries to a baking competition.

Suppose say :

Mikhail made x pastries .

Hence Rashid made \frac{2x}{3} pastries.

And in total they made \frac{5x}{3} pastries.

Now we have that out of x, \frac{5}{8} of x were brushed with olive oil.

So, \frac{3}{8} of x that is \frac{3x}{8} are brushed by butter.

So, it becomes (\frac{3}{8}) / ( \frac{5x}{3} )

= \frac{3}{8} \times\frac{3x}{5} = \frac{9x}{40}

Hence, answer is option B.

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Find the difference between (x+7) & (3x+4)
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3 years ago
What is the 93rd term of the arithmetic sequence-11,-7,-3
Bogdan [553]

Answer: 361

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3 0
3 years ago
Miss Croft made snack bags for the picnic that contain three types of snacks: packages of crackers, packages of cookies, and can
jek_recluse [69]

Answer: $1.25

Step-by-step explanation:

Let the cost of cracker be a.

Let the cost of cookies be b.

Let the cost of candy bars be c.

From the question,

6a + 6b + 6c = $21

8a + 5b + 10c = $26

5a + 4b + 7c = $18.50

To solve this, we first pick any two pairs of equation. This will be:

6a + 6b + 6c = $21 ....... i

8a + 5b + 10c = $26 ....... ii

Multiply equation i by 8

Multiply equation ii by 6

48a + 48b + 48c = $168 ....... iii

48a + 30b + 60c = $156 ....... iv

Subtract equation iv from iii

18b - 12c = 12

We then pick another two pairs

8a + 5b + 10c = $26

5a + 4b + 7c = $18.50

Multiply equation i by 5

Multiply equation ii by 8

40a + 25b + 50c = $130

40a + 32b + 56c = $148

Subtract the equations

-7b - 6c = -18

Then, solve the new equations formed

18b - 12c = 12 ....... v

-7b - 6c = -18 ....... vi

Multiply equation i by -7

Multiply equation ii by 18

-126b + 84c = -84

-126b - 108c = -324

Subtract the equations

192c = 240

c = 240/129

c = $1.25

From equation v, put the value of c

18b - 12c = 12

18b - 12($1.25) = 12

18b - $15 = $12

18b = $27

b = $27/18

b = $1.5

Since,

6a + 6b + 6c = $21

6a + 6($1.5) + 6($1.25) = $21

6a + $9 + $7.5 = $21

6a + $16.5 = $21

6a = $21 - $16.5

6a = $4.5

a = $4.5/6

a = $0.75

One candy bar cost $1.25

6 0
3 years ago
Calculate the percent change in estimated life expectancy for the African American female from 2001 to 2002, correct to nearest
weeeeeb [17]

Answer:

0.132%

Step-by-step explanation:

In 2001, the life expectancy for African American female is 75.5. In 2002, the life expectancy for African American female is 75.6.

Therefore, the change in the life expectancy for African American female from 2001 to 2002 is (75.6 - 75.5) = 0.1

Therefore, the percent change in estimated life expectancy for the African American female between the given period will be \frac{0.1}{75.5} \times 100\% = 0.132\% (Answer)

3 0
3 years ago
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