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3 years ago

14

Log38–log37=log3 8 7

1 answer:

Pachacha [2.7K]3 years ago

6 0

? It looks like u answered ur self

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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

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

3 years ago

Camille stopped at the grocery store to buy milk. While at the store, she spotted cans of soup on sale for $2.25 each. If the mi

Mrrafil [7]

$2.25x+$3.50=$35

so, $35-$3.50=$31.50

Then you divide $31.50 by $2.25

That equals 14

So, x=14

8 0

3 years ago

Find the comthon ratio of the geometric sequence 10,60, 360,...

Makovka662 [10]

Answer:

10

Step-by-step explanation:

5 0

3 years ago

Please answer! It would be appreciated if work is also shown!

borishaifa [10]

Quantity of paint left = $\frac{389}{480}$ gallons

Solution:

Quantity of yellow paint = $1\frac{1}{5}=\frac{6}{5}$ gallons

Quantity of green paint = $1\frac{1}{6}=\frac{7}{6}$ gallons

Quantity of blue paint = $\frac{7}{8}$ gallons

Quantity of paint used = $\frac{3}{4}$ gallons of each paint

Quantity of paint left = $1-\frac{3}{4}=\frac{1}{4}$ gallons of each paint

Quantity of paint left = $(\frac{6}{5}+\frac{7}{6}+\frac{7}{8})\times\frac{1}{4}$

Taking LCM for 5, 6, 8, we get LCM = 120

$(\frac{6}{5}+\frac{7}{6}+\frac{7}{8})\times\frac{1}{4}=(\frac{144}{120}+\frac{140}{120}+\frac{105}{120})\times\frac{1}{4}$

$=(\frac{389}{120})\times\frac{1}{4}$

$=\frac{389}{480}$

Hence, Mark will have $\frac{389}{480}$ gallons of paint left after painting the mural.

3 0

3 years ago

Plz help thanks sm!!!!

Snowcat [4.5K]

n = a number

2n = twice a number

2n - 7 = seven less than twice a number

4 0

3 years ago