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For the circle x^2 + y^2 = 100, a) State the radius at 6,8

alexandr1967 [171]

Answer:

add multiply divide or subtract make sure to use Pemdas then and you should have your answer

Step-by-step explanation:

8 0

3 years ago

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For which survey is a sample not necessary?

BARSIC [14]

Answer:

<h2>How many car accidents involve airbag malfunctions?</h2><h2 />

This is because there's no need to survey an entire population for accurate results.

The other 3, regarding Colorado residents, middle school students, and classmates are considered a population.

For testing the amount of car accidents, a population is not needed.

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<em>I hope this helped you!</em>

7 0

3 years ago

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The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo

dusya [7]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

$\to f_X (x) \ \ xe^{-x} \ , \ x>0$

For point a:

Moment generating function of X=?

Using formula:

$\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx$

$M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx$

integrating the values by parts:

$u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0} -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\$

$= \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\$

Therefore, the moment value generating by the function is $=\frac{1}{(t-1)^2}$

In point b:

$E(X^n)=?$

Using formula: $E(X^n)= M^{n}_{X}(0)$

form point (a):

$\to M_{X}(t)=\frac{1}{(t-1)^2}$

Differentiating the value with respect of t

$M'_{X}(t)=\frac{-2}{(t-1)^3}$

when $t=0$

$M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\$