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rodikova [14]
3 years ago
11

The product of 8 and x is greater than 50.

Mathematics
1 answer:
Vsevolod [243]3 years ago
8 0

Answer:

X would have to be any number greater than 6.25

Step-by-step explanation:

8*7=56, 8*8=64, and so on.

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Use the Distributive Property to expand 6(-4x + 3y).​
Anna11 [10]

Answer:

-24x + 18y

Step-by-step explanation:

6 x -4x = -24x

6 x 3y = 18y

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A recursive rule for an arithmetic sequence is a1=4;an=an−1−3.
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Read 2 more answers
John had 77 apples he gave 20 to alex and 15 to jack how many apples does john have left
Diano4ka-milaya [45]

Answer:

John has 42 apples left

Step-by-step explanation:

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Hope this helps

6 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 <em>X</em> > 2 and <em>Y</em> < 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 <em>X</em>, or the conditional density of <em>X</em> given <em>Y</em>?

In the first case, you simply integrate the joint density with respect to <em>y</em>:

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 <em>Y</em>:

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 <em>X</em> given <em>Y</em>:

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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