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Reil [10]
3 years ago
6

Simplify. y 5 · y 3 ÷ y 2

Mathematics
1 answer:
Gnesinka [82]3 years ago
3 0
Y^5 x y^3 /y^2
y^8/y^2
y^4
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Suppose you just received a shipment of twelve televisions. Three of the televisions are defective. If two televisions are rando
qwelly [4]

Answer: The probability that both televisions work is 0.5625 .

The probability at least one of the two televisions does not​ work is 0.4375.

Step-by-step explanation:

Given : Number of televisions received in shipment= 12

Number of defective televisions received in shipment= 3

Then proportion of defective televisions =\dfrac{3}{12}=0.25

Using binomial probability formula, the probability of getting success in x trials is given by :-

P(x)=^nC_xp^x(1-p)^{n-x}

If two televisions are randomly​ selected, then the probability that both televisions work will be :-

P(0)=^2C_0(0.25)^0(0.75)^2=(0.75)^2=0.5625

The probability at least one of the two televisions does not​ work will be :-

P(x\geq1)=1-P(0)=1-0.5625=0.4375

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Answer:

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Step-by-step explanation:

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3 years ago
Are these all correct ? If not please let me know
Licemer1 [7]
They are all correct except in 5), 4+8 is 12, not 13.
8 0
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Step-by-step explanation:

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