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grigory [225]
3 years ago
13

Divide (x^2+x-6) by (x+3)

Mathematics
1 answer:
seropon [69]3 years ago
3 0

Answer:

x-2

Step-by-step explanation  

\frac{x^{2} + x - 6}{x + 3} = \frac{(x-2)(x+3)}{(x + 3)} = (x-2)

eliminate (x+3) from top and bottom leaving (x-2)

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Please help I’ll mark brainliest as long as you tell me how you got that answer, also I apologize if it’s kinda blurry. I hope y
kolezko [41]

Answer:

B.

Step-by-step explanation:

Not another one! ;)

A can be eliminated because the numbers don‘t match. B seems like a possible answer, let’s first check the others out. C and D can be slim instead because the variable, g, should be next to the 1.5 and 2.5. C and D both change the variable location. So, B.

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What is 9.2 as a fraction
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VARVARA [1.3K]

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In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until
Juliette [100K]

Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

In this problem:

  • There are 12 bulbs, hence N = 12.
  • 3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

  • Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
  • The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

8 0
2 years ago
Please help i give brainly
ArbitrLikvidat [17]
What they said^ lolllllllllll
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