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Sav [38]
3 years ago
6

A box contains three cards, labeled 1, 2, and 3. Two cards are chosen at random, with the first card being replaced before the s

econd card is drawn. Let X represent the number on the first card, and let Y represent the number on the second card.
A. Find the joint probability mass function of X and Y.
B. Find the marginal probability mass functions pX(x) and pY(y).
C. Find µX and µY.
D. Find µXY.
E. Find Cov(X,Y).
Mathematics
1 answer:
inysia [295]3 years ago
4 0

Answer:

Step-by-step explanation:

Given that a box ontains three cards, labeled 1, 2, and 3.

Two cards are chosen at random, with the first card being replaced before the second card is drawn.

Let X represent the number on the first card, and let Y represent the number on the second card.

Since drawing is done with replacement, we find that X and Y are independnet

A) Joint pdf of x and y are

     

y    x 0 1 2 3 Total

0 0.0625 0.0625 0.0625 0.0625 0.25

1 0.0625 0.0625 0.0625 0.0625 0.25

2 0.0625 0.0625 0.0625 0.0625 0.25

3 0.0625 0.0625 0.0625 0.0625 0.25

Total 0.25 0.25 0.25 0.25  

B) Pdf of X is

x        0     1       2       3

p    0.25  0.25 0.25 0.25

and Y also will have same pdf

C) \mu_x = 0.25(0+1+2+3) = 1.5\\\mu_y = 0.25(0+1+2+3) = 1.5

D) \mu_{xy } =E(x) E(y) = 2.25

E) Cov (x,y) =0 since X and Y are independent

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Step-by-step explanation:

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When simplifying 4x^3-10x^2+6x/2x^3+x^2-3x, what are the term(s) that can be cancelled
brilliants [131]

Answer:

x can be cancelled

Step-by-step explanation:

we are given

\frac{4x^3-10x^2+6x}{2x^3+x^2-3x}

Firstly, we will factor numerator and denominator

and then we can factor it

4x^3-10x^2+6x=2x(2x^2-5x+3)

4x^3-10x^2+6x=2x(2x+1)(x-3)

now, we can factor denominator

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now, we can replace it

\frac{4x^3-10x^2+6x}{2x^3+x^2-3x}=\frac{2x(2x+1)(x-3)}{x(x-1)(2x+3)}

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x can be cancelled

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3 years ago
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