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3 years ago

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2 answers:

3241004551 [841]3 years ago

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-19 49/50

Ber [7]3 years ago

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You might be interested in

How many zeros does this polynomial function, y=(x-8)(x+3)^2

lidiya [134]

Answer:

3

Step-by-step explanation:

The factor (x - 8) has 1 zero at x = 8

The factor (x + 3)² has a zero at x = - 3 of multiplicity 2

In total there are 3 zeros

3 0

3 years ago

Read 2 more answers

Just some math :> and meh needs halp

vesna_86 [32]

Answer:16

Step-by-step explanation:

5 times 2 is 10. 10 times 0 is 0. 1 times 6 is six

8 0

3 years ago

If sin tetha=root3/2, what is cos tetha?

Minchanka [31]

Answer:

The value of cos Ф is ± $\frac{1}{2}$

Step-by-step explanation:

There are important rules for sin Ф and cos Ф

sin²Ф + cos²Ф = 1

sin²Ф = 1 - cos²Ф

cos²Ф = 1 - sin²Ф

∵ sin Ф = $\frac{\sqrt{3}}{2}$

∴ sin²Ф = $(\frac{\sqrt{3}}{2})^{2}$

∴ sin²Ф = $\frac{3}{4}$

→ By using the third rule above

∵ cos²Ф = 1 - sin²Ф

∴ cos²Ф = 1 - $\frac{3}{4}$

∴ cos²Ф = $\frac{1}{4}$

→ Take square root for both sides

∴ cos Ф = ± $\frac{1}{2}$

∴ The value of cos Ф is ± $\frac{1}{2}$

6 0

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

If using the method of completing the square to solve the quadratic equation x^2 + 9 x + 7 = 0 , which number would have to be a

FinnZ [79.3K]

Answer:

(x+9)^2=53/4

Step-by-step explanation:

6 0

3 years ago