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3 years ago

PLEASE HELP ME THIS IS URGENT I AM BEING TIMED!!!

Mathematics

2 answers:

valina [46]3 years ago

4 0

Answer:

1/4

Step-by-step explanation:

printed part = 210

total pages = 280

probability to open printed page 210/280 =3/4

probability to not open printed page = 1 - probability to open printed page = 1 - 3/4 = 1/4

hichkok12 [17]3 years ago

3 0

Answer:

1/4

Step-by-step explanation:

280-210=70

P=70/280=1/4

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den301095 [7]

<h3>Answer: Choice A) -2</h3>

=========================================================

Explanation:

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2 years ago

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3 years ago

Find two consecutive odd integers such that the sum of their squares is 34

AfilCa [17]

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4 0

3 years ago

Suppose a shipment of 120 electronic components contains 3 defective components. to determine whether the shipment should be ac

Likurg_2 [28]

For this problem, the most accurate is to use combinations

Because the order in which it was selected in the components does not matter to us, we use combinations

Then the combinations are $nC_r = \frac{ n! }{r! (n-r)!}$

n represents the amount of things you can choose and choose r from them

You need the probability that the 3 selected components at least one are defective.

That is the same as:

(1 - probability that no component of the selection is defective).

The probability that none of the 3 selected components are defective is:

$P = \frac{_{117}C_3}{_{120}C_3}$

Where $_{117}C_3$ is the number of ways to select 3 non-defective components from 117 non-defective components and $_{120}C_3$ is the number of ways to select 3 components from 120.

$_{117}C_3 = 260130$

$_{120}C_3 = 280840$

So:

$P = \frac{260130}{280840} = 0.927$

Finally, the probability that at least one of the selected components is defective is:

$P = 1-0.927 = 0.0737$

P = 7.4%

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4 years ago