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mars1129 [50]
2 years ago
8

Please help me ASAP ​

Mathematics
2 answers:
Sedaia [141]2 years ago
8 0

Answer:

  • 18°

Step-by-step explanation:

<u>The exterior angle of the polygon A is the difference of interior angles of pentagon and the square:</u>

  • 180(5 - 2)/5 - 180(4 - 2)/4 =
  • 540/5 - 360/4 =
  • 108 - 90 =
  • 18
Talja [164]2 years ago
6 0

Answer:

135 degrees

Step-by-step explanation:

360 - 90 = 270

270 / 2 = 135 degrees

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Factor the polynomial completely 6x^2-3x+2
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Answer:

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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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Step-by-step explanation:

The y intercept is where the line intercepts the y, or vertical axis.

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Hope this helps!

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1. What is the value of the expression f2 + 5 if f = 3?<br> A. 8<br> B. 2<br> C. 6<br> D. 14
zaharov [31]

Option D: 14 is the right answer

Step-by-step explanation:

In order to find the value of given expression we have to put the given value for f in the expression

Given expression is:

f^2+5

We have to find the value of expression on f=3

So,

putting f=3 in expression

= (3)^2+5\\= 9+5\\=14

The value of given expression at f=3 is 14

Hence,

Option D: 14 is the right answer

Keywords: expressions, variables

Learn more about expressions at:

  • brainly.com/question/10364988
  • brainly.com/question/10435816

#LearnwithBrainly

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