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3 years ago

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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kolezko [41]

Answer:

Step-by-step explanation:

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3 years ago

What is 9.2 as a fraction

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3 years ago

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

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2 years ago

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ArbitrLikvidat [17]

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3 years ago