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Lisa [10]
3 years ago
12

How to solve 4(7-3x)=7(4-2x)

Mathematics
2 answers:
Colt1911 [192]3 years ago
6 0

Answer:

Step-by-step explanation:

First, you distribute the 4 and 7 to the stuff inside the parentheses 28 - 12x = 28 - 14x then add 28 to both sides -12x = -14x then add 14x to both sides 2x = 0 then divide both sides by 2 x = 0

Marina CMI [18]3 years ago
4 0

Answer:

x=0

Step-by-step explanation:

4(7-3x)=7(4-2x)

Distribute

28 -12x = 28 -14x

Subtract 28 from each side

-12x = -14x

Add 14x to each side

-12x+14x = -14x+14x

2x=0

x =0

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An equation for that would be.

x => 67
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The gallons of water used to take a shower varies directly with the number of minutes in the shower. If a 6 minute shower uses 3
Lapatulllka [165]
Answer:

You use 6 gallons of water per minute.

Explanation:

If you use 36 gallons in 6 minutes then you need to divide 6 by 36. 36/6=6. That would mean you use 6 gallons per minute.
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Does someone know the correct answer?
Ksenya-84 [330]

Answer:

C

Step-by-step explanation:

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Read 2 more answers
Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of
IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  • np ≥ 10
  • n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

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3 years ago
In order for the data in the table to represent a linear function with a rate of change of –8, what must be the value of m? m =
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19 because it's being subtracted by 8
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