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Arlecino [84]
3 years ago
14

Need help asap will give brain

Mathematics
2 answers:
SCORPION-xisa [38]3 years ago
4 0
Question 1 is F question 2 is D
kogti [31]3 years ago
4 0
Question 1 is F
Question 2 is D
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Canadians who visit the United States often buy liquor and cigarettes, which are much cheaper in the United States. However, the
victus00 [196]

Answer:

Probability of bringing a bottle of liquor into the country that is, the probability of bringing 1 bottle liquor into the country = P(B) = 0.31

The probability of not bringing a bottle of liquor into the country, that is, the probability of bringing 0 bottle liquor into the country = P(B') = 0.69

Probability distribution of bottle liquor

Let X represent the random variable of the number of bottle liquor brought into the country by a person

X | P(X)

0 | 0.69

1 | 0.31

Step-by-step explanation:

The joint probability distribution for the number of bottles of liquor and the number of cartons of cigarettes imported by Canadians who have visited the United States for 2 or more days is given in the question as

V | B

C | 0 | 1

0 | 0.62 | 0.16

1 | 0.07 | 0.15

Note that B = bottle liquor

C = Carton cigarettes

V is each variable

Let the probability of bringing a bottle of liquor into the country be P(B), that is, the probability of bringing 1 bottle liquor into the country.

The probability of not bringing a bottle of liquor into the country is P(B'), that is, the probability of bringing 0 bottle liquor into the country.

Let the probability of bringing a carton of cigarettes into the country be P(C), that is, the probability of bringing 1 carton cigarettes into the country.

The probability of not bringing a carton of cigarettes into the country is P(C'), that is, the probability of bringing 0 carton cigarettes into the country.

From the joint probability table, we can tell that

P(B n C) = 0.15

P(B n C') = 0.16

P(B' n C) = 0.07

P(B' n C') = 0.62

Find the marginal probability distribution of the number of bottles imported.

Probability of bringing a bottle of liquor into the country that is, the probability of bringing 1 bottle liquor into the country = P(B)

P(B) = P(B n C) + P(B n C') = 0.15 + 0.16 = 0.31

The probability of not bringing a bottle of liquor into the country, that is, the probability of bringing 0 bottle liquor into the country = P(B')

P(B') = P(B' n C) + P(B' n C') = 0.07 + 0.62 = 0.69

Probability distribution of bottle liquor

Let X represent the random variable of the number of bottle liquor brought into the country by a person

X | P(X)

0 | 0.69

1 | 0.31

Hope this Helps!!!

8 0
3 years ago
The person who answers this will get brainliest, also a five start rating, and a thank you!
Sladkaya [172]
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8 0
3 years ago
Pls help!! i will give brainliest answer to the first right answer
Veronika [31]

Answer:

1600%

Step-by-step explanation:

Area = Side *Side

40%side * 40%side =1600% Area

40% *40%= 1600%

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finlep [7]
The answer is B:7!!!!!!!
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