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Mademuasel [1]
3 years ago
11

Please help ASAP

Mathematics
1 answer:
pochemuha3 years ago
5 0

Answer:5/4 and x

Step-by-step explanation:

dont trust me sorry

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3 years ago
Given three balls and four boxes. Assume the boxes are numbered 1, 2 ,3 and 4.Place the ball one by one into the boxes randomly.
KengaRu [80]

Answer:

The value of E(X) is 1.56

Step-by-step explanation:

Number of Boxes=4

Number of Balls =3

If X is the number of box in which the balls are than the number of balls in the box are given as ((4-X)+1)^3

Number of ways such that all the balls are in the box number 4, no balls in the first three boxes is given as n(x=4)=1^3=1

Number of ways such that all the balls are in the box number 3,  is given as n(x=3)=n(x=2)-n(x=4)=2^3-1^3=7

Number of ways such that all the balls are in the box number 2,  is given as n(x=2)=n(x=1)-n(x=3)=3^3-2^3=19

Number of ways such that all the balls are in the box number 1,  is given as n(x=1)=n(x=4)-n(x=3)=4^3-3^3=37

So the total number of ways are

n(total)=n(x=4)+n(x=3)+n(x=2)+n(x=1)

n(total)=1+7+19+37=64

So Probabilities are given as

p(x=1)=n(x=1)/n(total)=37/64

p(x=2)=n(x=2)/n(total)=19/64

p(x=3)=n(x=3)/n(total)=7/64

p(x=4)=n(x=4)/n(total)=1/64

So the E(x) is given as

E(X)=\sum^{n}_{i=1}p_ix_i\\E(X)=\sum^{4}_{i=1}p_ix_i\\E(X)=x_1p_1+x_2p_2+x_3p_3+x_4p_4\\E(X)=1\times\dfrac{37}{64}+2\times\dfrac{19}{64}+3\times\dfrac{7}{64}+4\times\dfrac{1}{64}\\E(X)=\dfrac{25}{16}=1.56

So the value of E(X) is 1.56

5 0
4 years ago
What is f (-4) = -2x+1
Vlad1618 [11]
The answer would be 5/2 (Decimal: 2.5) I hope this helped ^^
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