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2 years ago

I need help with this plz.

Mathematics

2 answers:

aalyn [17]2 years ago

8 0

<h3>Answers:</h3>

T ' = (3,0)
Q ' = (1, 0)
M ' = (0,2)
Y ' = (0,3)

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

Explanation:

T = (0,0) moves to T ' = (3,0) since we shift 3 units to the right. We add 3 to the x coordinate. The y coordinate does not change since we're not shifting up or down.

Q = (-2,0) moves to Q ' = (1,0) after applying the same rule we did to point T.

M = (-3,2) moves to M ' = (0,2)

Y = (-3,3) moves to Y ' = (0,3)

See the diagram below

pogonyaev2 years ago

4 0

Move each point 3 spaces to the right and connect the dots

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2) X and Y are jointly continuous with joint pdf

Nady [450]

From what I gather from your latest comments, the PDF is given to be

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}$

and in particular, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)

(a) Find c such that f is a proper density function. This would require

$\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}$

(b) Get the marginal density of X by integrating the joint density with respect to y :

$f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(d) The conditional distribution of X given Y can obtained by dividing the joint density by the marginal density of Y (which follows directly from the definition of conditional probability):

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

(e) From the definition of expectation:

$E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}$

$E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}$

$E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}$

(f) Note that the density of X | Y in part (d) identical to the marginal density of X found in (b), so yes, X and Y are indeed independent.

The result in (e) agrees with this conclusion, since E[XY] = E[X] E[Y] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)

8 0

2 years ago

Can somebody explain what the problem is explaining to me

Olin [163]

Answer:

The problem basically is saying that find two numbers that add up to 8, where one is three times the other.

Step-by-step explanation:

The best thing to do here is to write a simple system of equations.

b+y=8

b=3y

The first equation means that the total blue paint plus the total yellow paint is equal to 8 ounces.

The second equation means that the blue paint is equal to 3 times the yellow paint.

We can isolate b in the first equation to get b=-y+8

-y+8=3y

8=4y

y=2.

So we have that she needs two ounces of yellow paint.

Therefore she needs six ounces of blue paint.

Drag the yellow bar to 2, and the blue bar to 6.

Hope this helps!

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2 years ago

Put the following numbers in order from least to greatest -2/3,-1/5,.04,2

julsineya [31]

-1/5, -2/3, .04, 2

Hope this helps :)

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