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2 years ago

I NEED HELP OR I FAIL THIS TEST PLEASE

Mathematics

2 answers:

VladimirAG [237]2 years ago

7 0

Answer:

Step-by-step explanation:

You start with 15 dollars then you can use up to x amount of tickets.

Lilit [14]2 years ago

7 0

Answer: c

Step-by-step explanation:

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Please help!!!

V125BC [204]

Answer: 7 π r² + π r³

Explanation:

1) The general formula for the volumen, V, of a cylinder with radius r and height h is:

V = π r² h

2) Now you have the condition that the height is 7 inches greater than the radius, that is:

h = 7 + r

3) Replace h with its expression

V = π r² (7 + r)

4) Expand the product (use distributive property):

V = π (7r²) + π r³ = 7 π r² + π r³

Answer: 7 π r² + π r³

3 0

3 years ago

Mia discovered that if she takes half her age and adds 3, it produces the same results as taking one-fourth of her age and addin

Vikentia [17]

See https://web2.0calc.com/questions/help_29833.

5 0

2 years ago

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

Please help!! thank you.

Zina [86]

(A) Product of $-2x^{3}+x-5$ and $x^{3}-3x-4$ is:

$(-2x^{3}+x-5)(x^{3}-3x-4)$

$=-2x^{3}(x^{3}-3x-4)+x(x^{3}-3x-4)-5(x^{3}-3x-4)$

$=(-2x^{3})(x^{3})+(-2x^{3})(-3x)+(-2x^{3})(-4)+x(x^{3})+(x)(-3x)+(x)(-4)+(-5)(x^{3})+(-5)(-3x)+(-5)(-4)$

$=-2x^{6}+6x^{4}+8x^{3}+x^{4}-3x^{2}-4x-5x^{3}+15x+20$

$=-2x^{6}+6x^{4}+x^{4}+ 8x^{3}-5x^{3}-3x^{2}-4x+15x+20$

$=-2x^{6}+7x^{4}+3x^{3}-3x^{2}+11x+20$

(B) Yes.

Product of $-2x^{3}+x-5$ and $x^{3}-3x-4$ = Product of $x^{3}-3x-4$ and $-2x^{3}+x-5$ because multiplication is commutative.

Commutative Property of multiplication says that a.b = b.a.

Thus, multiplication is same irrespective of the order of two numbers.

5 0

3 years ago

What is the rational number equivalent to 3.12 repeating? 3 4/33, 3 8/33 ,3 10/33 ,3 5/33?

Schach [20]

3.12 = 3 4/33 (<span>3.12 repeating)</span>
hope it helps

5 0

2 years ago