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3 years ago

Does the function have an absolute maximum or minimum?

Mathematics

2 answers:

Dmitriy789 [7]3 years ago

8 0

A. Yes it has absolute maximum.

Alekssandra [29.7K]3 years ago

5 0

Answer:

kekkmwn.qmwmmwjehjrjdjrjjdndjdnrjjtjtjrk

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Keep the ratio simple.<br> 3/5 : 2/5

fredd [130]

Answer:

3:2

Step-by-step explanation:

I am assuming this means convert to whole numbers

3/5:2/5 is 6/10:4/10

6/10:4/10 is 0.6:0.4

0.6:0.4 is 6:4

6:4 is 3:2

8 0

1 year ago

Read 2 more answers

What could this answer be??

bekas [8.4K]

The number in the parentheses is the rate of change. Because this number is less than 1 it is a decrease, so it is a decay.

The percent decrease is 1 - 0.63 = 0.37 = 37% decrease

5 0

3 years ago

((6x-8)-2x)-((12x-7)-(4x-5))

KengaRu [80]

Answer:

-12x + 4

Step-by-step explanation:

(6x - 8 - 2x) - (12x - 7) - (4x - 5)

The subtraction sign before the parenthesis is basically multiplying by -1, so the subtraction signs in the parenthesis have to change to addition. After that, you can remove the parenthesis.

6x - 8 - 2x - 12x + 7 - 4x + 5

Simplified = -12x + 4

6 0

1 year ago

Which of these is not an equation?

timurjin [86]

Answer:

Step-by-step explanation:

an equation always shows that sum is equal

in this case 1825026 will subtract 17 and the remainder will be used to find A

3 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

2 years ago