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3 years ago

A’B’C’ is a dilation image if ABC which is the correct description of the dilation

Mathematics

2 answers:

irakobra [83]3 years ago

8 0

Answer: center a scale factor 4

Step-by-step explanation:

Sav [38]3 years ago

3 0

Answer:

The dilation is an enlargement

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is equal and this ratio is called the scale factor

In this problem Triangles A'B'C' and ABC are similar by AA Similarity Postulate

Let

z------> the scale factor

$z=\frac{A'C'}{AC}$

substitute the values

$z=\frac{2+8}{2}$

$z=\frac{10}{2}=5$

The scale factor is greater than 1

therefore

The dilation is an enlargement

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Dahasolnce [82]

Step-by-step explanation:

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y-4/x-8= -2/5

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3 years ago

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3 years ago

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Write a function describing the relationship of the given variables.

zhuklara [117]

Answer:

The function is A = 10√r

Step-by-step explanation:

* Lets explain the meaning of direct variation

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- If Y is in direct variation with x (y ∝ x), then y = kx, where k is the

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* Now lets solve the problem

# A is varies directly with the square root of r

- Change the statement above to a mathematical relation

∴ A ∝ √r

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∴ A = k√r

- To find the value of the constant of variation k substitute A and r

by the given values

∵ r = 16 when A = 40

∵ A = k√r

∴ 40 = k√16 ⇒ simplify the square root

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3 years ago

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Rainbow [258]

Answer:

(a) 0.2061

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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

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telo118 [61]

Answer:

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