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ludmilkaskok [199]
3 years ago
9

One institute mandates that for every 100 items scanned through the electronic checkout scanner at a retail​ store, no more than

2 should have an inaccurate price. A recent study of the accuracy of checkout scanners at the stores of a certain company showed​ that, of the 70 company stores​ investigated, 56 violated the​ institute's scanner accuracy standard. If 1 of the 70 company stores is randomly​ selected, what is the probability that that store does not violate the institute scanner accuracy​ standard?
Mathematics
1 answer:
ozzi3 years ago
4 0

Answer:

There is a 20% probability that the store selected does not violate the institute scanner accuracy​ standard

Step-by-step explanation:

70 company stores were investigared.

56 violated the​ institute's scanner accuracy standard.

70-56 = 14 did not violate the institute scanner accuracy​ standard.

If 1 of the 70 company stores is randomly​ selected, what is the probability that that store does not violate the institute scanner accuracy​ standard?

This is 14 divided by 70, so:

P = \frac{14}{70} = 0.2

There is a 20% probability that the store selected does not violate the institute scanner accuracy​ standard

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Answer:

An event with a probability of 0 is impossible.  

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Probability is typically expressed in terms of a fraction between 0 and 1 where the denominator is the total number of outcomes and the numerator is the number of desired outcomes.  Since probability is expressed as a fraction, if the probability is 0, that means it is impossible, or there is no chance that the event can happen.  However, if the probability is 1, that means that the event is certain to happen and the odds are completely in your favor that the event will happen.  

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Hi, does anyone know the answer to this question? I’m bad at geometry and I’m struggling to answer it.
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Answer:

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3 years ago
In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa
Elden [556K]

Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

^nC_r=\dfrac{n!}{r!(n-r)!}

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434

Hence, the required number of ways =27434

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3 years ago
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