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bagirrra123 [75]
3 years ago
11

What are all the factors of 360?

Mathematics
1 answer:
Solnce55 [7]3 years ago
7 0
The factors of 360 are: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360

1 × 360 = 360
2 × 180 = 360
3 × 120 = 360
4 × 90 = 360
5 × 72 = 360
6 × 60 = 360
8 × 45 = 360
9 × 40 = 360
10 × 36 = 360
12 × 30 = 360
15 × 24 = 360
18 × 20 = 360

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ELEN [110]
46, 16 is a little less than 1/2 of 34, so and 46 is a little less than 1/2 of 100.
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What is the cube answer to 24
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Factor the following :
Otrada [13]

Answer:

(x - 3)(x² + 4)

Step-by-step explanation:

x³ - 3x² + 4x - 12 ( factor first/second and third/fourth terms )

= x²(x - 3) + 4(x - 3) ← factor out (x - 3 from each term

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8 0
2 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

8 0
2 years ago
What's the probability that the product of 2 rolls of a d6 is odd?
timofeeve [1]
My guess is 3/6 is your answer
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