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3 years ago

Find the coordinates of the vertices of the triangle after a 270° rotation

Mathematics

1 answer:

Minchanka [31]3 years ago

8 0

Answer:

Step-by-step explanation:

(a). 270° clockwise or 90° counterclockwise:

(x, y) --------> ( - y, x )

(b). 270° counterclockwise or 90° clockwise:

(x, y) ---------> ( y, - x )

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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

3 years ago

Plzz helppp it's due today

juin [17]

Answer:

Step-by-step explanation:

use trig functions

2 sides and 1 angle given

Sides given

hypotenuse -> x
opposite side to 30˚ -> 4

Angle

30˚

trig function associated with hyp and opp is sin

sin(x) = opp/hyp

substitute

sin 30˚ = 4/x

solve for x

0.5 = 4/x

0.5x = 4

x = 8

7 0

3 years ago

Read 2 more answers

Plz help 10 points ;-;

Oksana_A [137]

Answer:

See explanation.

Step-by-step explanation:

Coordinates of point P is (2.5, 0).

Coordinates of point T are (-1.5, -1).

Coordinates of point R are (0.5, -2.5).

hope this helps.

5 0

3 years ago

Read 2 more answers

4x10^4 in standard notation

lyudmila [28]

Answer:

40000

Step-by-step explanation:

4×10^4 = 4×10000 = 40000

6 0

3 years ago

Read 2 more answers

Which of the following best describes the equation below?

Reika [66]

4(x + 9) + 5x = 9x - 36

=> 4x + 36 + 5x = 9x - 36

=> 9x + 36 = 9x - 36

=> 9x - 9x = -36 - 36

=> 0 = -72

You can see that the equation doesn't hold true.

So there is no solution for this equation.

Answer is C. no real solutions

5 0

3 years ago