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3 years ago

WILL GIVE 97 POINTS AND MARK BRAINLIEST

Mathematics

1 answer:

julia-pushkina [17]3 years ago

3 0

Answer:

There is no fundamental difference between a “rotation” and an “orbit” and or "spin". The key distinction is simply where the axis of the rotation lies, either within or outside of a body in question. This distinction can be demonstrated for both “rigid” and “non rigid” bodies.

Step-by-step explanation:

thats what google says, hope it helps

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Solve the exponent equation for x 7(4x-6)= 1/49

tatiyna

X=1. Chose C. Hoped that will help

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3 years ago

I just have one question left you don’t need to simplify if you give the answer

Ugo [173]

It is either 400 or negative 400 because if you devide 12 by .03 it is 400

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3 years ago

Read 2 more answers

The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 < x < 1, 0 < y < 2

fredd [130]

I'm going to assume the joint density function is

$f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0$

a. In order for $f_{X,Y}$ to be a proper probability density function, the integral over its support must be 1.

$\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1$

b. You get the marginal density $f_X$ by integrating the joint density over all possible values of $Y$ :

$f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0$

c. We have

$P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}$

d. We have

$\displaystyle P\left(X$

and by definition of conditional probability,

$P\left(Y>\dfrac12\mid X\frac12\text{ and }X$

$\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}$

e. We can find the expectation of $X$ using the marginal distribution found earlier.

$E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}$

f. This part is cut off, but if you're supposed to find the expectation of $Y$ , there are several ways to do so.

Compute the marginal density of $Y$ , then directly compute the expected value.

$f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0$

$\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87$

Compute the conditional density of $Y$ given $X=x$ , then use the law of total expectation.

$f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0$

The law of total expectation says

$E[Y]=E[E[Y\mid X]]$

We have

$E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}$

$\implies E[Y\mid X]=1+\dfrac1{6X+3}$

This random variable is undefined only when $X=-\frac12$ which is outside the support of $f_X$ , so we have

$E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87$

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3 years ago

Analyze the asymptotes of the function.

koban [17]

Answer:

vertical asymptote at x=-1

horizontal asymptote at y=0

Step-by-step explanation:

$y=\frac{1}{x+1}$

To find vertical asymptote we set the denominator =0 and solve for x

x+1=0 (subtract 1 from both sides)

x=-1

So, vertical asymptote at x=-1

To find horizontal asymptote we look at the degree of both numerator and denominator

there is no variable at the numerator , so degree of numerator =0

degree of denominator =1

When the degree of numerator is less than the degree of denominator

then horizontal asymptote at y=0

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3 years ago

I'm not very good with special right triangles, can someone please help?

Arturiano [62]

All 45 45 90 triangles sides are equal to x, x, and xsqrt2. which is the biggest side. So here x(11) and x(11) are both given. the answer to the x shown on the triangle is xsqrt2

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