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VashaNatasha [74]

2 years ago

6

Help me me please https://brainly.com/question/22734415

1 answer:

matrenka [14]2 years ago

7 0

Answer:

2) 1

3) 9

4) 78

Step-by-step explanation:

2)

1 showed up the most

3)

range = max value - min value

range = 9 - 0

range = 9

4)

mode means which showed up the most,

78 showed up the most

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What’s 10x - 4 - 7/2x - 3/4(2x - 3) in its simplest form?

lyudmila [28]

10x-4-7/2x-3/4(2x-3)
6 1/2 -4 - 1 1/2x - 2 1/4
2 1/2- 1 1/2x- 2 1/4
1/4-1 1/2x
decimal form:
6.5-4-1.5x-2.25
2.5-1.5x-2.25
0.25-1.5x

7 0

3 years ago

Find the surface area

ki77a [65]

Step-by-step explanation:

so for this one u do

$15 \times 9 + 9 \times 4 + 15 \times 4$

231 the times it by 2 for the both sides

231×2=462 cm2

hope this make sense :)

7 0

2 years ago

Sammy Sales had the following receipts for his tree trimming business for the first six months of the year. What is the average

fredd [130]

$average = \frac{sum \: \: of \: \:all \: \: obeservations}{no. \: \: of \: observaions}$

$= \frac{1205 + 1450 + 1530 + 1655 + 1610 + 1710}{6}$

$= \frac{9160}{6}$

$= 1526.66$

___________________________________________

1526.66 is the average of his sales for this period.

6 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

Jan has a piece of yarn that is 6 1/2 feet long. She cuts off a piece that is 2 5/12 feet long. How much yarn does she have left

icang [17]

The answer is b because all you have to do is just subtract if I am wrong so sorry

5 0

2 years ago