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1 year ago

5

WE ARE GONNA START A RIOT!! as you know, I wasted 310 points on an essay, counting, and all these bots do is put LINKS and do no

t help WHATSOEVER!!!!!!! I bet all of you people had one of the links too. Im tired, and i bet you are to. Me and others are fed up of these people disrespecting our learning, so we will drag them down the pit with us. WHO'S WITH ME?!?!

1 answer:

-Dominant- [34]1 year ago

6 0

Answer:

ME LOL

Step-by-step explanation:

XDHJCVEFBV,RW

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The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

2 years ago

X-5y=2 and 9x+8=15 solve by substitution

lozanna [386]

$x-5y=2$
Hence $x=2+5y$ (Equation I)
$9x+8=15$ (Equation II)
Substituting Equation I in Equation II
$9(2+5y)+8=15$
$18+45y+8=15$
$26+45y=15$
$45y=-11$
Hence $y= - \frac{11}{45}$

Substituting this in Equation in I to get X

$x-5(- \frac{11}{45} )=2$
$x+ \frac{11}{9}=2$
$x= 2- \frac{11}{9}$
$x= \frac{17}{9}$

7 0

2 years ago

A diagonal brace is to be placed in the wall of a room. the height of the wall is 6 ft, and the wall is 12 ft long. what is the

Varvara68 [4.7K]

We know that
Applying the Pythagoras Theorem
c=²=a²+b²
where c is the diagonal of the wall
a=12 ft
b=6 ft
c²=12²+6²-----> c²=144+36-----< c²=180-----> c=√180------> c=6√5 ft

the answer is
<span>the exact length of the brace is 6</span>√5 ft

8 0

2 years ago

Round 9991.55314546 to the nearest thousand

algol13

Answer:

9992.0.

Step-by-step explanation:

So eventually untill you get to the 2 you get 9991.5532 but, now since it is *thousand* it would be 9992.0.

Hope it helps.

8 0

2 years ago

Identify the number that is 9.5 units from 2 on a number line

dusya [7]

There are two such numbers.

One of them is 2 + 9.5 = 11.5.

The other one is 2 - 9.5 = -7.5.

_____

If we consider "9.5 units from 2" to be 9.5 units in the positive direction, then the appropriate choice is 11.5.

6 0

2 years ago