All categories

3 years ago

What's the answer? Idk it bc I'm stupid and I don't feel like doing it

Mathematics

1 answer:

aliya0001 [1]3 years ago

4 0

H=-16x to the second power +136

You might be interested in

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

3 years ago

Please!!! Follow me _/(•_•)\_

Vikki [24]

Answer:

Why

Step-by-step explanation:

How do I follow I'm new to dis

7 0

3 years ago

Read 2 more answers

A book is 9 inches wide and 13.5 inches tall. What will the area of the front of the book?

Nina [5.8K]

Answer:

121.5

Step-by-step explanation: so what u are supost to do is multiply the sides which will give you your answer of 121.5.

8 0

3 years ago

Read 2 more answers

4 5/6+3/5= my daughter is needing help she's in the 5th grade

V125BC [204]

Answer:5 13/30

Step-by-step explanation:

4+5/6 + 3/5

5/6 +3/5

Find LCD of 5/6 and 3/5

25/30 + 18/30

43/30= 1 13/30

4+1+13/30= 5 13/30

7 0

3 years ago

Read 2 more answers

Evaluate. (1/4)^-2 - (5^0x2) x 1^-1

blondinia [14]

Answer:

Step-by-step explanation:

Arithmetic expressions are best evaluated using a reliable calculator. The Google calculator always follows the Order of Operations, so can give you a good answer to such questions.

(1/4)^-2 - (5^0)(2)(1^-1)

= 16 - (1)(2)(1) . . . . evaluate the exponents first

= 16 -2 . . . . . . . . . evaluate the product next

= 14 . . . . . . . . . . . finally, evaluate the sum

___

The rule of exponents is ...

a^-b = 1/(a^b)

4 0

3 years ago