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Georgia [21]
3 years ago
10

Which expression uses the greatest common factor and the distributive property to write the sum 36 1 54 as a product? A 6(6 1 9)

B 9(4 1 6) C 18(2 1 3) D 27(9 1 27)
Mathematics
1 answer:
Eddi Din [679]3 years ago
8 0

9514 1404 393

Answer:

  C  18(2 + 3)

Step-by-step explanation:

The GCD of 36 and 54 is their difference, 18. Factoring that out of the sum, you have ...

  36 + 54 = (18·2 + 18·3) = 18(2 + 3) . . . . matches choice C

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jekas [21]

Answer: The required expected value is $3.46.

Step-by-step explanation:

Since we have given that

Number of value of 4 or less are

2,3,4

So, there are 12 in numbers.

So, probability would be

\dfrac{12}{52}

Since Aces are considered as the highest card in the deck.

So, remaining probability would be

\dfrac{40}{12}

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So, the expected value would be

\dfrac{12}{52}\times 165-\dfrac{40}{52}\times 45\\\\=\dfrac{1980}{52}-\dfrac{1800}{52}\\\\=\dfrac{180}{52}\\\\=3.46

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7 0
3 years ago
The joint probability mass function of XX and YY is given by p(1,1)=0p(2,1)=0.1p(3,1)=0.05p(1,2)=0.05p(2,2)=0.3p(3,2)=0.1p(1,3)=
inessss [21]

Answer:

P(Y=1|X=3)=0.125

Step-by-step explanation:

Given :

p(1,1)=0  

p(2,1)=0.1

p(3,1)=0.05

p(1,2)=0.05

p(2,2)=0.3

p(3,2)=0.1

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p(2,3)=0.1

p(3,3)=0.25

Now we are supposed to find the conditional mass function of Y given X=3 :  P(Y=1|X=3)

P(X=3) = P(X=3,Y=1)+P(X=3,Y=2) +P(X=3,Y=3)

P(X=3)=p(3,1) +p(3,2) +p(3,3)

P(X=3)=0.05+0.1+0.25=0.4

P(Y=1|X=3)=\frac{P(X=3,Y=1)}{P(X=3)} =\frac{p(3,1)}{P(X=3)}=\frac{0.05}{0.4}= 0.125

Hence P(Y=1|X=3)=0.125

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Answer

Step-by-step explanation:

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