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olga2289 [7]
2 years ago
6

Select the correct answer. Triangle is reflected across the y-axis and then dilated by a factor of centered at the origin. Which

statement correctly describes the resulting image, triangle ? A. The dilation preserves the side lengths and angles of triangle . The reflection does not preserve side lengths and angles. B. The reflection preserves the side lengths and angles of triangle . The dilation preserves angles but not side lengths. C. Neither the reflection nor the dilation preserves the side lengths and angles of triangle . D. Both the reflection and dilation preserve the side lengths and angles of triangle .
Mathematics
1 answer:
DanielleElmas [232]2 years ago
8 0

Answer:

B. The reflection preserves the side lengths and angles of triangle. The dilation preserves angles but not side lengths.

Step-by-step explanation:

A reflection is when the position of an object is shifted and does not affect the side lengths or angle measures of the object. (this eliminates options A. and C.)

A dilation is when you change the size of an object a=and changes the side lengths but does not affect angle measures (this eliminates option D.)

The correct answer is:

B. The reflection preserves the side lengths and angles of triangle. The dilation preserves angles but not side lengths.

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m>14  \wedge  m\leq -4

There's no solution. It is the first one already. There is no number that is both greater than 14 and less than or equal to -4. That is no solution because there's no m that satisfy the compound inequality.

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Mathematical Statistics with Applications Homework Help
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7.37:

a. <em>W</em> follows a chi-squared distribution with 5 degrees of freedom. See theorem 7.2 from the same chapter, which says

\displaystyle \sum_{i=1}^n\left(\frac{Y_i-\mu}{\sigma}\right)^2

is chi-squared distributed with <em>n</em> d.f.. Here we have \mu=0 and \sigma=1.

b. <em>U</em> follows a chi-squared distribution with 4 degrees of freedom. See theorem 7.3:

\displaystyle \frac1{\sigma^2}\sum_{i=1}^n (Y_i-\overline Y)^2

is chi-squared distributed with <em>n</em> - 1 d.f..

c. <em>Y₆</em>² is chi-square distributed for the same reason as <em>W</em>, but with d.f. = 1. The sum of chi-squared distributed random variables is itself chi-squared distributed, with d.f. equal to the sum of the individual random variables' d.f.s. Then <em>U</em> + <em>Y₆</em>² is chi-squared distributed with 5 + 1 = 6 degrees of freedom.

7.38:

a. Notice that

\dfrac{\sqrt 5 Y_6}{\sqrt W} = \dfrac{Y_6}{\sqrt{\frac W5}}

and see definition 7.2 for the <em>t</em> distribution. Since <em>Y₆</em> is normally distributed with mean 0 and s.d. 1, it follows that this random variable is <em>t</em> distributed with 5 degrees of freedom.

b. Similar manipulation gives

\dfrac{2Y_6}{\sqrt U} = \dfrac{\sqrt4 Y_6}{\sqrt U} = \dfrac{Y_6}{\sqrt{\frac U4}}

so this r.v. is <em>t</em> distributed with 4 degrees of freedom.

4 0
3 years ago
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