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arlik [135]
3 years ago
14

HEADS UP there is a bot going around saying to put in a link to get your answers dont do it i didnt click it but it could be a v

irus because all there answers lead to that site and it answers every minute so dont click on any links in the answer area
Mathematics
1 answer:
blsea [12.9K]3 years ago
6 0

Answer:

Okay! Thx for letting us know!

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b) 8.5 feet

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How does the volume of a cylinder change when its diameter is tripled
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Police estimate that​ 25% of drivers drive without their seat belts. If they stop 6 drivers at​ random, find the probability tha
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17.80% probability that all of them are wearing their seat belts.

Step-by-step explanation:

For each driver stopped, there are only two possible outcomes. Either they are wearing their seatbelts, or they are not. The drivers are chosen at random, which mean that the probability of a driver wearing their seatbelts is independent from other drivers. So we use the normal probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Police estimate that​ 25% of drivers drive without their seat belts.

This means that 75% wear their seatbelts, so p = 0.75

If they stop 6 drivers at​ random, find the probability that all of them are wearing their seat belts.

This is P(X = 6).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{6,6}.(0.75)^{6}.(0.25)^{0} = 0.1780

17.80% probability that all of them are wearing their seat belts.

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3 years ago
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