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fgiga [73]
2 years ago
11

Do yall like my avatar

Mathematics
1 answer:
Talja [164]2 years ago
6 0

Answer:

Yeah

Step-by-step explanation:

lol

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Hey! Can somebody please help my friend with 5,211 ÷ 6? I need to have a picture of the work if somebody can please do that!
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868.5

Step-by-step explanation: If u devide by groups, you can find the answer.

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Tupac bought a new video game. While he learned how to play, he played 5 games and had a
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Solve this equation 4y-y+11=38
nordsb [41]

Answer:

y= 9

Step-by-step explanation:

4y-y+11=38

3y+11=38

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6 0
3 years ago
Answer and explain step by step.​
Vitek1552 [10]

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6 0
2 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

P(AE) = 500/2000

P(AE) = 0.25

Finally,

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
2 years ago
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