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Elenna [48]
3 years ago
15

Could someone help me please ? :)

Mathematics
1 answer:
Degger [83]3 years ago
8 0

ANSWER = 17.5% of 360 = 63

EXPLANATION:

Find 10% of 360 = 36

Then 5% of 360 = 18

Then 2.5% of 360 = 9

Add them together

so 17.5% of 360 = 63

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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 =
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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
Please help me w the answer
evablogger [386]

Answer:

\frac{2(x-6)(x-10)}{(x-4)(x-5)}

Step-by-step explanation:

In essence, one needs to work their way backwards to solve this problem. Use the information to construct the function.

The function has verticle asymptotes at (x = 4) and (x = 5). This means that the denominator must have (x - 4) and (x - 5) in it. This is because a verticle asymptote indicates that the function cannot have a value at these points, the function jumps at these points. This is because the denominator of a fraction cannot be (0), the values (x - 4) and (x - 5) ensure this. Since if (x) equals (4) or (5) in this situation, the denominator would be (0) because of the zero product property (this states that any number times zero equals zero). So far we have assembled the function as the following:

\frac{()}{(x-4)(x-5)}

The function has x-intercepts at (6, 0), and (0, 10). This means that the numerator must equal (0) when (x) is (6) or (10). Using similar logic that was applied to find the denominator, one can conclude that the numerator must be ((x - 6)(x-10)). Now one has this much of the function assembled

\frac{(x-6)(x-10)}{(x-4)(x-5)}

Finally one has to include the y-intercept of (0, 120). Currently, the y-intercept is (60). This is found by multiplying the constants together. (6 * 10) equals (60). One has to multiply this by (2) to get (120). Therefore, one must multiply the numerator by (2) in order to make the y-intercept (120). Thus the final function is the following:

\frac{2(x-6)(x-10)}{(x-4)(x-5)}

6 0
3 years ago
Charlotte has 30 beads. She has 8 more red beads than blue beads. Which equation represents this situation? Please help :(
Darina [25.2K]

Answer:

The second answer choice is right

Step-by-step explanation:

8 0
3 years ago
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