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Thepotemich [5.8K]
4 years ago
12

Chocolates costing $8 per pound are to be mixed with chocolates costing $3 per pound to make a 20 pound mixture. If the mixture

is to sell for $5 per pound, how many pounds of each chocolate should be used?
Which of the following equations could be used to solve the problem?



A. 8x + 3x = 5(20)
B. 8x + 3(20) = 5(x + 20)
C. 8x + 3(20 - x) = 5(20)
Mathematics
1 answer:
rewona [7]4 years ago
3 0
  put the amount of $8 a pound chocolate as say it as x
 As the total number of pounds is 20
so the amount of $3 a pound chocolate must be <span>20−x
so it will be</span>.
8x + 3(20 - x) = 5(20)
so correct option is C
hope it helps
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Answer:

D. 108 liters

Step-by-step explanation:

Given that the leak is ;

5 drops = 5 minutes

?= 4 weeks

Find number of hours in 4 weeks

1 week=7days

4 weeks=?--------------cross multiply

4×7=28 days

1 day=24 hours

28 days=?--------------cross multiply

28×24=672 hours

Hour= 60 minutes

672 hours=?

672×60=40320 minutes

Finding the leakage in 4 weeks

80 drops= 5 minutes

?=40320 minutes------------cross multiplication

(40320×80)÷5 =645120 drops

Finding the liters of water wasted in 4 weeks

60 drops of water= 10 milliliters

645120 drops of water=?-------------cross multiply

(645120×10)÷60 =107520 milliliters

But you know

1 milliliter=0.001 liters

107520 milliliters=?-----------cross multiply

107520×0.001

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3 years ago
What is the sum of the polynomials?
Vaselesa [24]

Answer:

<u>The answer is the option B</u>

4m-24n+3

Step-by-step explanation:

we have

17m-12n-1+4-13m-12n

Group terms that contain the same variable

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vladimir1956 [14]

Answer:

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Step-by-step explanation:

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The formula for the area of a circle is A=pi r^2 how does the value for a change when r increases
marin [14]
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A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement
dsp73

Answer:

P(First\ White\ and\ Second\ Blue) = \frac{3}{28}

P(Same) = \frac{67}{210}

Step-by-step explanation:

Given (Omitted from the question)

Red = 7

White = 9

Blue = 5

Solving (a): P(First\ White\ and\ Second\ Blue)

This is calculated using:

P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)

P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}

P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}

P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}

P(First\ White\ and\ Second\ Blue) = \frac{45}{420}

P(First\ White\ and\ Second\ Blue) = \frac{3}{28}

Solving (b) P(Same)

This is calculated as:

P(Same) = P(First\ Blue\ and Second\ Blue)\or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)

P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})

P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})

P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}

P(Same) = \frac{20+42+72}{420}

P(Same) = \frac{134}{420}

P(Same) = \frac{67}{210}

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