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4 years ago

PLS solve THIS ASAP!!

Mathematics

2 answers:

statuscvo [17]4 years ago

6 0

Answer choices???? Idk if this is right but I calculated and I got 16.2

nalin [4]4 years ago

6 0

The answer is 20 because C is the square root of 20. Using Pythagorean theorem, we can prove this. Please let me know if you need more help. :)

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Which tables display linear functions? Check all that apply. A 2-column table with 4 rows. Column 1 is labeled x with entries ne

DIA [1.3K]

I think it’s A because

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3 years ago

Read 2 more answers

Divide. Write your answer in simplest form. 6/7 ÷ 5/21

Serjik [45]

The answer is 8/5.
To divide fractions, flip the second fraction and multiply.
For example,
= 6/7 • 21/5
= 126/35
Simplify by dividing both sides by 7 to get your answer.
= 8/5

5 0

4 years ago

Let X be a random variable with mean X = 25 and X = 6 and let Y be a random variable with mean Y = 30 and Y = 4. It is known

LenaWriter [7]

I'm guessing you intended to say $X$ has mean $\mu_X=E[X]=25$ and standard deviation $\sigma_x=\sqrt{\mathrm{Var}[X]}=6$ , and $Y$ has means $\mu_Y=E[Y]=30$ and standard deviation $\sigma_Y=\sqrt{\mathrm{Var}[Y]}=4$ .

If $W=X+Y$ , then $W$ has mean

$E[W]=E[X+Y]=E[X]+E[Y]=55$

and variance

$\mathrm{Var}[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2$

Given that $\mathrm{Var}[X]=36$ and $\mathrm{Var}[Y]=16$ , we have

$\mathrm{Var}[X]=E[X^2]-E[X]^2\implies E[X^2]=36+25^2=661$

$\mathrm{Var}[Y]=E[Y^2]-E[Y]^2\implies E[Y^2]=16+30^2=916$

Then

$E[W^2]=E[(X+Y)^2]=E[X^2]+2E[XY]+E[Y^2]$

$X$ and $Y$ are independent, so $E[XY]=E[X]E[Y]$ , and

$E[W^2]=E[X^2]+2E[X]E[Y]+E[Y^2]=661+2\cdot25\cdot30+916=3077$

so that the variance, and hence standard deviation, are

$\mathrm{Var}[W]=3077-55^2=52$

$\implies\sqrt{\mathrm{Var}[W]}=\sqrt{52}=\boxed{2\sqrt{13}}$

# # #

Alternatively, if you've already learned about the variance of linear combinations of random variables, that is

$\mathrm{Var}[aX+bY]=a^2\mathrm{Var}[X]+b^2\mathrm{Var}[Y]$

then the variance of $W$ is simply the sum of the variances of $X$ and $Y$ , $\mathrm{Var}[W]=36+16=52$ , and so the standard deviation is again $\sqrt{52}$ .