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3 years ago

What is the solution to the following equation? -6(x-2) + 3x = -3(x + 3) + 21

Mathematics

2 answers:

NeX [460]3 years ago

4 0

9514 1404 393

Answer:

no solution

Step-by-step explanation:

Simplifying the given equation, you have ...

-6(x -2) +3x = -3(x +3) +21 . . . . . given

-6x +12 +3x = -3x -9 +21 . . . . . . eliminate parentheses

-3x +12 = -3x +13 . . . . . . . . . . . . collect terms

0 = 1 . . . . . . . . . . . . . add (3x-12) to both sides

There is no value of x that will make this true.

The equation has NO SOLUTION.

storchak [24]3 years ago

3 0

-6(x+2) + 3x = -3(x+3) + 21

-6x - 12 + 3x = -3x - 9 + 21

-3x - 12 = -3x + 12

-3x + 3x = 12 + 12

0 = 24

Sorry but this equation has no solution.
-3x + 3x = 0, x is cancelled out so we can’t find ‘x’.

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Answer:

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2 years ago

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4 years ago

What is the area of the figure? Round to the nearest tenth.

nikdorinn [45]

Answer:

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Step-by-step explanation:

This is made up of 4 triangular segments. If we assume the cross section is perpendicular then we can use the 1/2b*h formula for each piece with the given values. So we have 2 triangles with the formula: 1/2(3)*(4.5)=6.75 * 2 triangles = 13.5.

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3 years ago

The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.

Rainbow [258]

Answer:

(a) 0.2061

(b) 0.2514

Step-by-step explanation:

Let <em>X</em> denote the heights of women in the USA.

It is provided that <em>X</em> follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

$P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z$

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

$P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z$

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

$P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0$

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

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3 years ago