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3 years ago

Please help, need to get this done quick!

Mathematics

1 answer:

Murljashka [212]3 years ago

4 0

our X is 30 because 3x = 90 and if we divide that by 3 then we are left with 30, I knew it was equal to 90 degrees because of the little red box that symbolizes a 90 degree angle.

So now that we know X = 30 we can plug it in for (2x + 3y)

2(30) + 3y is equal to (60 + 3y) and if we equal that to the other side 3x and plug in for x on that side, we have a new equation of 90 = 60 + 3y so then from here we subtract 60 from both sides and are left with 30 = 3y and finally we divide 3

and our final answer is Y = 10 (((((((((((((((((((((((((((((:

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Answer:

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3 years ago

A video game store allows customers to rent games for 4.75 each. Customers can also buy a membership for $54 annually, and video

ankoles [38]

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Step-by-step explanation:

Let x represent number of video games.
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We are also told that customers can also buys a membership for $54 annually, and video games would only cost $2.50 each. The cost of renting x video games after membership would be 2.50x + 54

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3 years ago

What is the solution to the equation4x-6=10x-3 ?

shepuryov [24]

Answer:

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Step-by-step explanation:

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5 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

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DH, CG, AE are parallel to BF. They are not skew to BF.

CD, AC, GH, EG are neither parallel nor do they intersect BF, so they are skew to BF

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3 years ago