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2 years ago

3. Scatter plots are used to look for a pattern of association, or trend, between

Mathematics

1 answer:

Marina86 [1]2 years ago

4 0

I would say b because its 2 or more points on a graph otherwise its just a point

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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

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

5 0

3 years ago

Write the product of 7/20 and 3 as a decimal

posledela

7.8
is the answer :) because you have to turn the fraction into a decimal then multiply by 3.

7 0

3 years ago

Can someone explain how to do this

alex41 [277]

When any angle bisects with other angle then both the angles are equal.

Then, m of angle ABD = angle CBD

ABD = 70

ABD = CBD = 70°

ABC = 70 + 70 = 140°

3 0

2 years ago

Read 2 more answers

Find the solution(s) of the following equation.

kobusy [5.1K]

Answer:

The answer is A: y=7

Step-by-step explanation:

If you multiply 7 x 7 x 7 the answer would be 343.

7 0

2 years ago

Find the x-coordinate of the point on the graph of y=x^2 where the tangent line is parallel to the secant line that cuts the cur

jek_recluse [69]

First find the secant line. The slope of the secant line through $(-1,1)$ (when $x=-1$ ) and $(2,4)$ (when $x=2$ ) is the average rate of change of $y=x^2$ over the interval $[-1,2]$ :

$\text{slope}_{\text{secant}}=\dfrac{2^2-(-1)^2}{2-(-1)}=\dfrac33=1$

The tangent line to $y=x^2$ will have a slope determined by the derivative:

$y=x^2\implies y'=2x$

Both the secant and tangent will have the same slope when $2x=1$ , or when $x=\dfrac12$ .

4 0

2 years ago