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4 years ago

13

Mr. Hillman is buying boxes of colored pencils for his classroom. They regularly cost $1.80 each but are on sale for 30% off. If

the sales tax is 6% and he has $40 budget, how many boxes can he buy.

1 answer:

alexira [117]4 years ago

4 0

Answer:

joe

Step-by-step explanation:

joe momma

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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 $p = 0.1$

Four travelers who have made hotel reservations in this study.

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?

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

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

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

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

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

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

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

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