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igor_vitrenko [27]
3 years ago
13

What is the answer to 9x(6+x)

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
5 0

You use distributive property. You first multiply 9x to x and 6. Which gives you 9x^2+54x

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Answer:

a) P=0.8

b) P=0.67

c) P=0.05

d) P=0.33

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Step-by-step explanation:

a. What is the probability that the household has only a cell phone and has high-speed Internet?

This probability is stated in the question: "Suppose of U.S. households having only a cell phone, 80% have high-speed Internet", so the probability is P=0.8.

P(C\&I)=0.8

b. What is the probability that the household has only a cell phone or has high-speed Internet?

This probability is equal to the sum of the probability of having only a cell phone and the probability of having high-speed internet, less the probability of having both (to avoid counting this household twice).

P(C\,orI\,)=P(C)+P(I)-P(C)*P(I|C)=0.25+0.65-(0.25*0.8)=0.25+0.62-0.20=0.67

c. What is the probability that the household has only a cell phone and does not have high-speed Internet?

This is equal to the probability of not having high-speed internet given that it has a cell phone (complementaty of the proability of Point (a)) multiplied by the probability of having a cell phone.

P(C\&\bar I)=P(\bar I|C)*P(C)=(1-P(\I|C))*P(C)=(1-0.8)*0.25 = 0.2*0.25 = 0.05

d. What is the probability that the household does not have just a cell phone and does not have high-speed Internet?

This probability is complementary of the one calculated in Point (c).

P(\bar C \,or\, \bar I)=1-P(C\, or\,I)=1-0.67=0.33

e. What is the probability that the household does not have just a cell phone and does have high-speed Internet?

This is equal to the probability of having high-speed internet less the probability it has both (cell phone and internet).

P(\bar C\&I)=P(I)-P(I|C)*P(C)=0.65-0.8*0.2=0.65-0.2=0.45

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