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3 years ago

THE RIGHT ANSWER WILL RECEIVE A BRAINLEST AND POINTS AND THANXS!!!

Mathematics

1 answer:

trapecia [35]3 years ago

6 0

Answer:

1. F

2. G

3. H

4. E

5. C

6. B

7. A

8. D

Step-by-step explanation:

1. For a horizontal line, this is zero. F. Slope

2. These lines have the same slope. G Parallel Lines

3. These lines meet at 90°. H Perpendicular Lines

4. This is where two lines meet. E. Point of Intersection

5. For the line 3 2 6 x y   , this is −3. C. Y-intercept

6. The numbers 10 and 1 /10 are examples. B Reciprocals

7. This is the name for an equation of a line in the form Ax By C    0. A. Standard Form

8. For a vertical line, the value of x is constant and equal to this D. x-intercept

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Answer:

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3 years ago

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2 years ago

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ioda

(1) Looks like the joint density is

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

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$\displaystyle\int_0^4\int_y^4 cxy\,\mathrm dx\,\mathrm dy=\int_0^4\frac{cy}2(4^2-y^2)=32c=1$

$\implies\boxed{c=\dfrac1{32}}$

(2) The region in which X > 2 and Y < 1 corresponds to a 2x1 rectangle (see second attached shaded region), so the desired probability is

$P(X>2,Y$

(3) Are you supposed to find the marginal density of X, or the conditional density of X given Y?

In the first case, you simply integrate the joint density with respect to y:

$f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_0^x\frac{xy}{32}\,\mathrm dy=\begin{cases}\frac{x^3}{64}&\text{for }0$

In the second case, we instead first find the marginal density of Y:

$f_Y(y)=\displaystyle\int_y^4\frac{xy}{32}\,\mathrm dx=\begin{cases}\frac{16y-y^3}{64}&\text{for }0$

Then use the marginal density to compute the conditional density of X given Y:

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}\frac{2xy}{16y-y^3}&\text{for }y$

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3 years ago

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I can answer you in the comments below but I’m not sure what you’re asking

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