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3 years ago

PLEASE HELP ME QUICK!!!20 POINTS!!!!!!!!

Mathematics

1 answer:

Blizzard [7]3 years ago

5 0

the volume scales by (1/10)^3. so it would be 48.875*10^-3.

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Help<br><br> (5a-3a^3)*(4a-1)

ivann1987 [24]

Answer:

$-12a^{4} + 3a^{3} + 20a^{2} -5a$

Step-by-step explanation:

$(5a-3a^{3} )$ • $(4a-1)$

Let's use FOIL (first, outer, inner, last) to solve this. We'll multiply the terms by one another in that fashion.

$20a^{2} - 5a - 12a^{4} + 3a^{3}$

Rearrange in decreasing exponents.

$-12a^{4} + 3a^{3} + 20a^{2} -5a$

3 0

3 years ago

Write 68,127,000,000,000,000 in scientific notation. (x10^

Alexeev081 [22]

Answer:

6.8127 × 10^16

Step-by-step explanation:

Hope this helps:)

7 0

2 years ago

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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

What is this please help or else

Mademuasel [1]

Answer:

8:45

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Please I need the answer

Vedmedyk [2.9K]

Answer:

x=4

Step-by-step explanation:

Simplifying

2(6x + 4) + -6 + 2x = 3(4x + 3) + 1

Reorder the terms:

2(4 + 6x) + -6 + 2x = 3(4x + 3) + 1

(4 * 2 + 6x * 2) + -6 + 2x = 3(4x + 3) + 1

(8 + 12x) + -6 + 2x = 3(4x + 3) + 1

Reorder the terms:

8 + -6 + 12x + 2x = 3(4x + 3) + 1

Combine like terms: 8 + -6 = 2

2 + 12x + 2x = 3(4x + 3) + 1

Combine like terms: 12x + 2x = 14x

2 + 14x = 3(4x + 3) + 1

Reorder the terms:

2 + 14x = 3(3 + 4x) + 1

2 + 14x = (3 * 3 + 4x * 3) + 1

2 + 14x = (9 + 12x) + 1

Reorder the terms:

2 + 14x = 9 + 1 + 12x

Combine like terms: 9 + 1 = 10

2 + 14x = 10 + 12x

Solving

2 + 14x = 10 + 12x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-12x' to each side of the equation.

2 + 14x + -12x = 10 + 12x + -12x

Combine like terms: 14x + -12x = 2x

2 + 2x = 10 + 12x + -12x

Combine like terms: 12x + -12x = 0

2 + 2x = 10 + 0

2 + 2x = 10

Add '-2' to each side of the equation.

2 + -2 + 2x = 10 + -2

Combine like terms: 2 + -2 = 0

0 + 2x = 10 + -2

2x = 10 + -2

Combine like terms: 10 + -2 = 8

2x = 8

Divide each side by '2'.

x = 4

Simplifying

x = 4

Hope it helps!

7 0

2 years ago

Read 2 more answers