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bearhunter [10]
2 years ago
8

Which expressions are equivalent to the model?

Mathematics
2 answers:
horsena [70]2 years ago
4 0
The correct answer is 2y(3+4x) and 6y + 8xy
you can distribute the 2y to both of the values inside the parentheses in order to simplify
Afina-wow [57]2 years ago
3 0
The correct answer is 2y(3+4x)
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postnew [5]

Answer:

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Step-by-step explanation:

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3 years ago
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A box has five items, three good and two defective. three items are selected at random without replacement. let x be the number
user100 [1]
This problem can be solved from first principles, case by case.  However, it can be solved systematically using the hypergeometric distribution, based on the characteristics of the problem:
- known number of defective and non-defective items.
- no replacement
- known number of items selected.

Let
a=number of defective items selected
A=total number of defective items
b=number of non-defective items selected
B=total number of non-defective items
Then 
P(a,b)=C(A,a)C(B,b)/C(A+B,a+b)
where 
C(n,r)=combination of r items selected from n,
A+B=total number of items
a+b=number of items selected

Given:
A=2
B=3
a+b=3
PMF:
P(0,3)=C(2,0)C(3,3)/C(5,3)=1*1/10=1/10
P(1,2)=C(2,1)C(3,2)/C(5,3)=2*3/10=6/10
P(2,0)=C(2,2)C(3,1)/C(5,3)=1*3/10=3/10
Check: (1+6+3)/10=1 ok
note: there are only two defectives, so the possible values of x are {0,1,2}

Therefore the 
PMF:
{(0, 0.1),(1, 0.6),(2, 0.3)}

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solniwko [45]

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Step-by-step explanation:

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Step-by-step explanation:

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