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3 years ago

Brainliestto who helps me plz I need it

Mathematics

1 answer:

Mashutka [201]3 years ago

3 0

AB || CD and AC is oblique ;

So : Angle ( BAC ) = Angle ( DCA )

_________________________________

∆ ABC - ∆ ACD :

(1) AB = CD

(2) Angle ( BAC ) = Angle ( DCA )

(3) AC = AC

So According to (1) , (2) , (3) :

∆ ABC =~ ∆ ACD

Well, if two triangles are deposited, all their components are equal peer to peer.

So : AD = BC

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What is the value of 3x +2 when x = 5

Bingel [31]

Answer:

Step-by-step explanation:

Always put questions like these with an equal sign

3x + 2 = ?

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3 years ago

Read 2 more answers

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3 years ago

Mathematical Statistics with Applications Homework Help

photoshop1234 [79]

7.37:

a. W 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 n d.f.. Here we have $\mu=0$ and $\sigma=1$ .

b. U 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 n - 1 d.f..

c. Y₆² is chi-square distributed for the same reason as W, 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 U + Y₆² 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 t distribution. Since Y₆ is normally distributed with mean 0 and s.d. 1, it follows that this random variable is t distributed with 5 degrees of freedom.

b. Similar manipulation gives

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