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2 years ago

Find the slope of the line. [?] Give your answer as a fraction in simplest form.

Mathematics

1 answer:

anyanavicka [17]2 years ago

3 0

Answer:

7/14

I am sure is that

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Which system has infinitely many solutions?

GREYUIT [131]

The answer is "b"
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8 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

3 years ago

QF Q7.) Use properties of logarithms to expand the logarithmic expression as much as possible. Evaluate logarithmic expressions

Softa [21]

Attached is the solution.
The following log properties are used:
$log(x^n) = nlog(x)$
$log(\frac{a}{b}) = log(a) - log(b) \\ \\ log(ab) = log(a) + log(b) \\ \\ log_9 (81) = 2 \rightarrow 9^2 = 81$

5 0

3 years ago

Read 2 more answers

If Angle P and Angle Q are supplementary,

Ede4ka [16]

Answer: 313% 79% 3%

Step-by-step explanation:

3 0

2 years ago

Wale was twice as old as Ajay 10 years back. How old is Ajay today if Wale will be 40 years old 10. years hence?

vagabundo [1.1K]

Using a system of equations, it is found that Ajay is 25 years old today.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are:

Variable x: Wale's age.

Variable y: Ajay's age.

Wale was twice as old as Ajay 10 years back, hence:

$x - 10 = 2(y - 10)$

$x - 10 = 2y - 20$

$x = 2y - 10$

Wale will be 40 years old in 10 years, hence:

$x + 10 = 40$

$x = 40$

Then, Ajay's present age is found as follows.

$2y = x + 10$

$2y = 40 + 10$

$y = \frac{50}{2}$

$y = 25$

You can learn more about system of equations at brainly.com/question/14183076

4 0

2 years ago

Read 2 more answers