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2 years ago

11

Which sets of angles listed are supplementary in this diagram?

1 answer:

AlladinOne [14]2 years ago

6 0

Supplementary angle = Angels which add up to 180 degrees
Angle CED

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3 and 1/3 + 4 and 5/7 =

klasskru [66]

The answer is 8 and 1/21 in fixed fraction.

and decimal form it would be 8.047619

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3 years ago

Which of the following is the quotient of the rational expressions shown

Natali5045456 [20]

Answer:

2x^3/x+5/x-5/ 3x-1

2x^3/ 3x-1

Step-by-step explanation:

we can cut plus and minus sign

3 0

3 years ago

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Ms. Carmell's art class is making a mosaic to cover an ugly concrete wall along a a highway off ramp the wall forms a right tria

GuDViN [60]

252 you multiply the 42 and 12 then devoid even by 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

determine the maximum or minimum of the quadratic function. express your answer in the form (x,y) and using decimals rounded to

kolezko [41]

We are given the following quadratic equation

$f(x)=2x^2+7x-10$

The vertex is the maximum/minimum point of the quadratic equation.

The x-coordinate of the vertex is given by

$h=-\frac{b}{2a}$

Comparing the given equation with the general form of the quadratic equation, the coefficients are

a = 2

b = 7

c = -10

$h=-\frac{b}{2a}=-\frac{7}{2(2)}=-\frac{7}{4}=-1.75$

The y-coordinate of the vertex is given by

$\begin{gathered} f(x)=2x^2+7x-10 \\ f(-1.75)=2(-1.75)^2+7(-1.75)-10 \\ f(-1.75)=2(3.0625)^{}-12.25-10 \\ f(-1.75)=6.125^{}-12.25-10 \\ f\mleft(-1.75\mright)=-16.13 \end{gathered}$

This means that we have a minimum point.

Therefore, the minimum point of the given quadratic equation is

$(-1.75,-16.13)$

7 0

1 year ago

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