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3 years ago

I don’t know how to do number 9. Can some one please help? URGENT.

Mathematics

1 answer:

vfiekz [6]3 years ago

6 0

Here you are !! Please mark BRAINLEST

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the sum of ages of father and his son is 60 years.if the difference between their ages is 40 years then find the age of father a

cluponka [151]

Answer:

Father's age is 50, Son's age is 10

Step-by-step explanation:

X+Y=60

X-Y= 40

Let X represent the Father's age while Y represents the Son's age;

therefore, from the two equations, the value of X is

X= 60-Y

X= 40+Y then putting in the value of x

(40+Y) + Y= 60,

40+2Y=60

2Y=60-40 =20

Y= 10,

Now that we know the value of Y, we can get the value of x

If X+Y=60, then

X+10=60,

X=60-10,

X= 50

8 0

3 years ago

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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

2 years ago

In ΔABC , angle C is a right angle.

guajiro [1.7K]

Answer: Choice A. sin(A) = cos(B)

============================================================

Explanation:

The rule is that sin(A) = cos(B) if and only if A+B = 90.

Note how

sin(A) = opposite/hypotenuse = BC/AB
cos(B) = adjacent/hypotenuse = BC/AB

Since both result in the same fraction BC/AB, this helps us see why sin(A) = cos(B). Similarly, we can find that cos(A) = sin(B).

In the diagram below, the angles A and B are complementary, meaning they add to 90 degrees. So this trick only applies to right triangles.

The side lengths can be anything you want, as long as you're dealing with a right triangle.

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2 years ago

15 is 60% of what number? please answer really soon!

Anna35 [415]

Answer:25 hope this helps

Step-by-step explanation:

5 0

2 years ago

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314,207 in standard form expanded form and word form

Gennadij [26K]

Three hundred fourteen thousand two hundread seven

Hope it helps!!

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3 years ago

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