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3 years ago

Which classification best describes the following system of equations x=5 y=8 -x-y+z=0

Mathematics

1 answer:

vichka [17]3 years ago

7 0

What are the answer choices?

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11. Which is an equation of the circle whose center is

VladimirAG [237]

I think that it is b sorry if I’m wrong

4 0

2 years ago

What is the center of a circle?<br> What is the radius of the circle?

lutik1710 [3]

Answer:

Center: (-9, -7)

Radius: 5

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What is the value of x in the right triangle below? Show your work and round your answer to the nearest hundredth.

navik [9.2K]

The length of the hypothenuse or the value of x is equal to 19.53

Data;

hypothenuse = x
adjacent = 11.2
opposite = 16

<h3>Pythagoras's Theorem</h3>

To solve this problem, we have to use Pythagoras's theorem which is used to find the missing side in a right angle triangle if we have at least two sides.

The formula for this is

$x^2 = y^2 + z^2\\$

Let's substitute the values and solve for the missing side

$x^2 = 16^2 + 11.2^2 \\x^2 = 256 + 125.44\\x^2 = 381.44\\x = \sqrt{381.44} \\x = 19.53$

The length of the hypothenuse or the value of x is equal to 19.53

Learn more Pythagoras theorem here;

brainly.com/question/3317136

6 0

2 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

In the spinner below the large wedges are twice the size of the smaller ones. What is true about the probability of landing on 6

mr_godi [17]

Answer:

Step-by-step explanation:

If we take each small section to be "1" unit, we can say the large sections (for "2" and "5") are "2" units each. So in total there will be 8 sections.

Since 5 is "2" sections, we can say:

P(5) = 2/8 = 1/4

And 6 is "1" section, so we can say:

P(6) = 1/8

Definitely, Probability of landing a 6 is HALF that of probability of landing a 5. Also we can see this is the picture.

So, from the answer choices, A is right.

6 0

3 years ago