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3 years ago

Write the following equation in slope-intercept form: 3x+4y=12

1 answer:

8 0

Hello, MissMisty here to help you.

• The answer •
Y = -3/4 + 3

Thanks! I hope it helped you, I'm glad I could help if it did. Have a good morning.

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Find the values of x and y - i will give brainliest if right

kherson [118]

Answer:

itd x=68 and y=5 I apolgize if its wrong

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2 years ago

2x + 4y = 22 2x - 2y = -8

NISA [10]

Answer:

2x+4y=22

-2x+2y=+8

6y=30

y=5

now

replace the value of y in equ i

2x +4y =22

2x+20=22

x=1

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3 years ago

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Evaluate (a + b)^2 for a = 2 and b = 3.

shepuryov [24]

Hope this helps you!!!

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3 years ago

Read 2 more answers

Use the commutative property to rewrite the following expression. -16 · y

Y_Kistochka [10]

Commutative property of multiplication is just mixing up the order but the answer stays the same.
Examples: 4 + 5 + 6 = 6 + 4 + 5 ; (a)(d) = (d)(a) ;

Answer:
(-16)(y) or/and (y)(-16)

7 0

3 years ago

Given a joint PDF, f subscript X Y end subscript (x comma y )equals c x y comma space 0 less than y less than x less than 4, (1)

ioda

(1) Looks like the joint density is

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

In order for this to be a proper density function, integrating it over its support should evaluate to 1. The support is a triangle with vertices at (0, 0), (4, 0), and (4, 4) (see attached shaded region), so the integral is

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

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

(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:

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

6 0

3 years ago