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sineoko [7]
3 years ago
12

4 intersecting lines are shown. Line m contains points Y, K, and B. Another line contains points C, Y, and W. Another line conta

ins points B, A, and C. The fourth line contains points J, A, K, and W. Points P and X are not on the lines. Points W, K, and are collinear. Line CA and line YK intersect at . Point is not contained on line m.
Mathematics
1 answer:
Ilya [14]3 years ago
5 0

Answer:

Points W, K, and

✔ J

are collinear.

Line CA and line YK intersect at

✔ point B

.

Point

✔ W

is not contained on line m.

Step-by-step explanation:

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3 years ago
1500 customers hold a VISA card; 500 hold an American Express card; and, 75 hold a VISA and an American Express. What is the pro
alex41 [277]

Answer:

There is 15% probability that a customer chosen at random holds a VISA card, given that the customer has an American Express card.

P(VISA \:| \:AE) = 15\%\\

Step-by-step explanation:

Number of customers having a Visa card = 1,500

Number of customers having an American Express card = 500

Number of customers having Visa and American Express card = 75

Total number of customers = 1,500 + 500 = 2,000

We are asked to find the probability that a customer chosen at random holds a VISA card, given that the customer has an American Express card.

This problem is related to conditional probability which is given by

P(A \:| \:B) = \frac{P(A \:and \:B)}{P(B)}

For the given problem it becomes

P(VISA \:| \:AE) = \frac{P(VISA \:and \:AE)}{P(AE)}

The probability P(VISA and AE) is given by

P(VISA and AE) = 75/2000

P(VISA and AE) = 0.0375

The probability P(AE) is given by

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P(AE) = 0.25

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P(VISA \:| \:AE) = \frac{P(VISA \:and \:AE)}{P(AE)}\\\\P(VISA \:| \:AE) = \frac{0.0375}{0.25}\\\\P(VISA \:| \:AE) = 0.15\\\\P(VISA \:| \:AE) = 15\%\\

Therefore, there is 15% probability that a customer chosen at random holds a VISA card, given that the customer has an American Express card.

8 0
3 years ago
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