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TEA [102]
3 years ago
10

simply combining like teams 3a-2b-ab-(a-b+ab)-4ab+2b-a if l write tge answers as 1a- 6ab is it correct ?​

Mathematics
1 answer:
Semenov [28]3 years ago
8 0

Answer:

-6 a b + b + a

Step-by-step explanation:

Simplify the following:

-(a - b + a b) - a b - 4 a b + 2 b - 2 b + 3 a - a

Hint: | Group like terms in 3 a - 2 b - a b - (a b - b + a) - 4 a b + 2 b - a.

Grouping like terms, 3 a - 2 b - a b - (a b - b + a) - 4 a b + 2 b - a = -(a b - b + a) + (-a b - 4 a b) + (3 a - a) + (2 b - 2 b):

-(a b - b + a) + (-a b - 4 a b) + (3 a - a) + (2 b - 2 b)

Hint: | Combine like terms in -a b - 4 a b.

a b (-1) + a b (-4) = -5 a b:

-(a b - b + a) + -5 a b + (3 a - a) + (2 b - 2 b)

Hint: | Combine like terms in 3 a - a.

3 a - a = 2 a:

-(a b - b + a) - 5 a b + 2 a + (2 b - 2 b)

Hint: | Look for the difference of two identical terms.

2 b - 2 b = 0:

-(a b - b + a) - 5 a b + 2 a

Hint: | Distribute -1 over a b - b + a.

-(a b - b + a) = -a b + b - a:

-a b + b - a - 5 a b + 2 a

Hint: | Group like terms in -a b - 5 a b + b + 2 a - a.

Grouping like terms, -a b - 5 a b + b + 2 a - a = (-a b - 5 a b) + b + (-a + 2 a):

(-a b - 5 a b) + b + (-a + 2 a)

Hint: | Combine like terms in -a b - 5 a b.

a b (-1) + a b (-5) = -6 a b:

-6 a b + b + (2 a - a)

Hint: | Combine like terms in 2 a - a.

2 a - a = a:

Answer: -6 a b + b + a

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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

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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.

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Vedmedyk [2.9K]
The best method here is to plug in so plug in -1 (-1)+5+3(-1)=1 so the answer is A
3 0
2 years ago
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