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harkovskaia [24]

2 years ago

11

John needs to wrap the edge of his rectangular desk with duct tape. He knows the width is 10/8 of a meter and the length is 3/8

of a meter. How much duct tape does he need to wrap the desk’s perimeter?

1 answer:

Zielflug [23.3K]2 years ago

7 0

Answer: 1 and 5/8

Step-by-step explanation:

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YOOOOOOO HELLLPPPPPP PT.2 <br> HERE 100 PLUS BRAINLIST

pshichka [43]

I believe it is the second one hope that helped

4 0

1 year ago

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Can someone help me and explain? I will mark brainlest. ♡

Sedbober [7]

Answer:

$p'(4) = -3$

$q'(8) = \frac{1}{4}\\$

Step-by-step explanation:

For p'(4):

$p(x) = f(x)g(x) \\ p'(x) = \frac{d}{dx}(f(x)g(x)) \\ p'(x) = f'(x)g(x) +f(x)g'(x)$

$p'(4) = f'(4)g(4) + f(4)g'(4) \\ p'(4) = (-1)(3) +(7)(0) \\ p'(4) = -3$

For q'(8):

$q(x) = \frac{f(x)}{g(x)} \\ q'(x)= \frac{d}{dx}(\frac{f(x)}{g(x)}) \\ q'(x) = \frac{f'(x)g(x) -f(x)g'(x)}{{g(x)}^2}$

$q'(8) = \frac{f'(8)g(8) -f(8)g'(8)}{{g(8)}^2} \\ q'(8) = \frac{(2)(2) -(6)(\frac{1}{2})}{{2}^2} \\ q'(8) = \frac{4 -3}{4} \\ q'(8) = \frac{1}{4}$

4 0

3 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

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

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2 years ago

DJ has 16 movies, and Three-eighths of them are comedies. How many movies are comedies?

IrinaVladis [17]

Answer:

16 * 3/8 = 6

6 movies are comedies

7 0

3 years ago

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Which of the following is the inverse of the given function? <br><br> y= 3x^5-4

ziro4ka [17]

I hope this helps you

y+4/3=x^5

x= (y+4/3)^1/5

f^-1(x)=(x+4/3)^1/5

8 0

2 years ago