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Nadya [2.5K]
3 years ago
5

Austin is filling shelves at a grocery store. With every carton of 12 cereal boxes, he can fill 60 cm of shelving. The rack has

10 m of shelving, but 2 m is already occupied with other products. Austin must use enough cereal boxes to fill the rack.
Mathematics
1 answer:
avanturin [10]3 years ago
8 0

Answer:

<h2>Austing needs 160 boxes to complete the rack.</h2>

Step-by-step explanation:

Givens

  • 12 cereral boxes occupies 60 centimeters of shelving.
  • The rack has 10 meters of shelving, equivalent to 1000 centimeters.
  • 2 meters are already used.
  • There are 800 centimeters of shelving remaining.

To find the total number of cereal boxes needed to use the whole rack, we just need to use the rule of three: If 12 cereal boxes occupies 60 centimeters of shelving, 800 centimeters would require how many cereal boxes

n=800cm \times \frac{12 \ boxes}{60cm} =160 \ boxes

Therefore, Austing needs 160 boxes to complete the rack.

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Travelers who fail to cancel their hotel reservations when they have no intention of showing up are commonly referred to as no-s
notsponge [240]

Answer:

a) 0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) 0 is the most likely value for X.

Step-by-step explanation:

For each traveler who made a reservation, there are only two possible outcomes. Either they show up, or they do not. The probability of a traveler showing up is independent of other travelers. This means that the binomial probability distribution is used to solve this question.

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.

No-show rate of 10%.

This means that p = 0.1

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This means that n = 4

a) What is the probability that at least two of the four selected will turn to be no-shows?

This is P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)

In which

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

P(X = 2) = C_{4,2}.(0.1)^{2}.(0.9)^{2} = 0.0486

P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036

P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001

P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0486 + 0.0036 + 0.0001 = 0.0523

0.0523 = 5.23% probability that at least two of the four selected will turn to be no-shows.

b) What is the most likely value for X?

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

P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.6561

P(X = 1) = C_{4,1}.(0.1)^{1}.(0.9)^{3} = 0.2916

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P(X = 3) = C_{4,3}.(0.1)^{3}.(0.9)^{1} = 0.0036

P(X = 4) = C_{4,4}.(0.1)^{4}.(0.9)^{0} = 0.0001

X = 0 has the highest probability, which means that 0 is the most likely value for X.

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