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UkoKoshka [18]
3 years ago
7

El productor del números es 28. Sabiendo que el primer número es 3 unidades menor que el segundo, calcula los dos número

Mathematics
1 answer:
Rasek [7]3 years ago
8 0

Answer:

Tienes 3 números, "x", "y", "z"

=> El primero es 20 unidades menor que el segundo:

El primero es "x", y el segundo es "y", entonces:

x = y - 20

=> El tercero es igual a la suma de los dos primeros:

El tercero es "z", entonces:

z = x + y

z = (y - 20) + y

z =  2y - 20

=> Entre los tres suman 120

x + y + z = 120

y - 20 + y + 2y - 20 = 120

4y - 40 = 120

4y = 160

y = 40

Si y = 40, entonces:

x = y - 20

x = 20

Además, z = 2y - 20

z= 60

RPTA:

x=20

y=40

z=60

Step-by-step explanation:

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Answer:

1. 0.0910 = 9.10% probability that exactly 180 passengers show up for the flight.

2. 0.4522 = 45.22% probability that at most 180 passengers show up for the flight.

3. 0.5478 = 54.78% probability that more than 180 passengers show up for the flight.

Step-by-step explanation:

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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}

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Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

Assume that there is a 0.905 probability that a passenger with a ticket will show up for the  flight.

This means that p = 0.905

Also assume that the airline sells 200 tickets

This means that n = 200

Question 1:

Exactly, so we can use the P(X = x) formula, to find P(X = 180).

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

P(X = 180) = C_{200,180}.(0.905)^{180}.(0.095)^{20} = 0.0910

0.0910 = 9.10% probability that exactly 180 passengers show up for the flight.

2. When 200 tickets are sold, calculate the probability that at most 180 passengers show up for the flight.

Now we have to use the approximation.

Mean and standard deviation:

\mu = E(X) = np = 200*0.905 = 181

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.905*0.095} = 4.15

Using continuity correction, this is P(X \leq 180 + 0.5) = P(X \leq 180.5), which is the p-value of Z when X = 180.5. Thus

Z = \frac{X - \mu}{\sigma}

Z = \frac{180.5 - 181}{4.15}

Z = -0.12

Z = -0.12 has a p-value of 0.4522.

0.4522 = 45.22% probability that at most 180 passengers show up for the flight.

3. When 200 tickets are sold, calculate the probability that more than 180 passengers show up for the flight.

Complementary event with at most 180 passengers showing up, which means that the sum of these probabilities is 1. So

p + 0.4522 = 1

p = 1 - 0.4522 = 0.5478

0.5478 = 54.78% probability that more than 180 passengers show up for the flight.

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