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Sergeeva-Olga [200]
4 years ago
12

A farm raises cows and chickens. The farmer has a total of 40 animals. One day he counts the legs of all his animals and realize

s he has a total of 118 legs. How many cows does the farmer have?
Mathematics
1 answer:
allsm [11]4 years ago
8 0

Answer: the farmer has 19 cows.

Step-by-step explanation:

Let x represent the number of cows that the farmer has.

Let y represent the number of chicken that the farmer has.

The farmer has a total of 40 animals. It means that

x + y = 40

A cow has 4 legs. A chicken has 2 legs. One day he counts the legs of all his animals and realizes he has a total of 118 legs. It means that

4x + 2y = 118- - - - - - - - 1

Substituting x = 40 - y into equation 1, it becomes

4(40 - y) + 2y = 118

160 - 4y + 2y = 118

- 4y + 2y = 118 - 160

- 2y = - 42

y = - 42/ - 2

y = 21

x = 40 - y = 40 - 21

x = 19

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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}

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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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Answer:

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