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saveliy_v [14]
3 years ago
5

Tell whether the ratios are proportions: 4/7,24/35

Mathematics
1 answer:
EastWind [94]3 years ago
7 0

To find whether or not the ratios are proportional, simplify. The first ratio is already simplified, so simplify the second one. to do this, find what factors 24 and 35 have in common.

Factors of 24: 1, 2, 3 ,4, 6, 8, 12, 24

Factors of 35: 1, 5, 7, 35

It looks like there are now common factors between 24 and 35. For the ratios to be proportional, they have to have the same value. If 24/35 can't be simplified to 4/7, the factors are not proportional.

So, to answer the question: No, the ratios are NOT proportions. I hope this helps and have a great rest of your day!

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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
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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.

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