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2 years ago

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PLEASE HELP 20 POINTS!!

1 answer:

Harman [31]2 years ago

7 0

Answer:

<u>D) x² - y = 10x + 25 + y</u>

Step-by-step explanation:

A relation only becomes a function if each x-value only has a single y-value as the answer.

<u>Evaluating</u>

A and B cannot be the answers as each value will get 2 answers
C will also not be the answer as y² means at least 2 answers will be given
<u>D</u> is the best option as each inputted x-value will get an answer of a single y-value

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Figure ABCD is dilated by a scale factor of 1/3. If point B is at (6,-3), where will point B’ be located

Lesechka [4]

Answer:

Step-by-step explanation:

Multiply both numbers by 1/3. That is your new coordinates. (2, -1)

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3 years ago

Line "AB" and line "CD" are perpendicular and intersect at point P. What is the measure of angle APC?

Step2247 [10]

Answer:

APC is 90 degrees

Step-by-step explanation:

When lines intersect and are perpendicular. that means the angle of intersection is 90 degrees.

APC is 90 degrees

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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

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

7 0

3 years ago

After taxes, Liza makes $2,185.76 each month. From that, she has to pay three bills. For the last month, one bill is for $73.49,

Taya2010 [7]

Answer:

the amount left to spend is $740.77

Step-by-step explanation:

The computation of the amount left to spend is as follows:

= Each month payment - last month payment bills

= $2,185.76 - $73.49 - $897.19 - $474.31

= $740.77

Hence, the amount left to spend is $740.77

We simply deduct three amount bills from the each month payment

3 0

2 years ago

216 as a power number

I am Lyosha [343]

Hey There,

216 is equal to $6^{3}$

Solution:

6*6*6=213

Thank You!

5 0

3 years ago

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