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Alinara [238K]
3 years ago
6

chelsea takes the bus to school,but she walks home. the bus travels to the school at an average speed of 45 miles per hour. whil

e going home from school, chelsea walks at a speed of 2.5 miles per hour. in total, she takes 1 hour to travel to school and back. if (t) is the time it takes to go to school by bus, the equation that will help find the time it takes chelsea to get to school on the bus is?

Mathematics
1 answer:
jekas [21]3 years ago
4 0
For this case, the first thing you should know is:
 d: v * t
 Where,
 d: distance
 v: speed
 t: time
 To go to school by bus we have:
 d = 45 * t
 To return from school we have:
 d = 2.5 * (1-t)
 how the distance is the same:45 * t = 2.5 * (1-t)
 Answer:
 the equation that will help find the time it takes chelsea to get to school on the bus is:
 45 * t = 2.5 * (1-t)
 Note: the equation will have one solution. 
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d. (-14) + (-8)

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notsponge [240]

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

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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.

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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}

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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

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 = 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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