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3 years ago

Mr. Brown can grade 4 exams in 15 minutes. How many exams can he grade in 1 hour? How do you know?

Mathematics

2 answers:

Savatey [412]3 years ago

4 0

16 because you divide i hour by 15 and multiply that by 4:)

mel-nik [20]3 years ago

4 0

4:15
1:.15
X:1H
16:1H

This is because 15x4=1hour and 4x4=16

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Step-by-step explanation: Because.

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3 years ago

Read 2 more answers

Need help on this question Plzzz

stepladder [879]

Answer:

B. -9, -5, -1

Step-by-step explanation:

<u>Solve the inequality:</u>

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a < 1/3 - 1/4
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2 years ago

What is the slope of the line that passes through the points (8,-4) and (6,3) in simplest form

Nesterboy [21]

Step-by-step explanation:

Rise of y-value = (-4) - (3) = -7.

Run of x-value = (8) - (6) = 2.

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3 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

3 years ago

The sale bin in a clothing store contains an assortment of t-shirts in different sizes. There are 7 small, 8 medium, and 4 large

umka21 [38]

Answer:

P(at least 1 large) = 0.648

P(at least 1 large) = 64.8%

Step-by-step explanation:

We have 7 small shirts, 8 medium shirts and 4 large shirts

Total number of shirts = 7 + 8 + 4 = 19 shirts

The probability that at least one of the first four shirts he checks is a large is given by

P(at least 1 large) = 1 - P(no large)

So first we need to find the probability that the none of the first four shirts he checks are large.

For the first check, there are 15 small and medium shirts and total 19 shirts so,

15/19

For the second check, there are 14 small and medium shirts and total 18 shirts left so,

14/18

For the third check, there are 13 small and medium shirts and total 17 shirts left so,

13/17

For the forth check, there are 12 small and medium shirts and total 16 shirts left so,

12/16

the probability of not finding the large shirt is,

P(no large) = 15/19*14/18*13/17*12/16

P(no large) = 0.352

Therefore, the probability of finding at least one large shirt is,

P(at least 1 large) = 1 - P(no large)

P(at least 1 large) = 1 - 0.352

P(at least 1 large) = 0.648

P(at least 1 large) = 64.8%

4 0

3 years ago