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2 years ago

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Which expression is equivalent to StartRoot negative 80 EndRoot? Negative 4 StartRoot 5 EndRoot Negative 4 StartRoot 5 EndRoot i

4 StartRoot 5 EndRoot i 4 StartRoot 5 EndRoot
or

Which expression is equivalent to √-80

A. -4√5

B. -4√5i

C. 4√5i

D. 4√5

2 answers:

fomenos2 years ago

8 1

The answer is D. 4√5

hichkok12 [17]2 years ago

4 1

Answer:

C. 4√5i

Step-by-step explanation:

on edge

please vote brainliest i have never gotten it before

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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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$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

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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3 years ago

An aircraft factory manufactures airplane engines. The unit cost C (the cost in dollars to make each airplane engine) depends on

butalik [34]

Answer:

90 engines must be made to minimize the unit cost.

Step-by-step explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

$f(x) = ax^{2} + bx + c$

It's vertex is the point $(x_{v}, y_{v})$

In which

$x_{v} = -\frac{b}{2a}$

$y_{v} = -\frac{\Delta}{4a}$

Where

$\Delta = b^2-4ac$

If a>0, the minimum value of the function will happen for $x = x_v$

C(x)=x²-180x+20,482

This means that $a = 1, b = -180, c = 20482$

How many engines must be made to minimize the unit cost?

x value of the vertex. So

$x_{v} = -\frac{b}{2a} = -\frac{-180}{2(1)} = 90$

90 engines must be made to minimize the unit cost.

3 0

3 years ago