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tangare [24]
3 years ago
14

What is the sum of the polynomials? (m + n + 3) + (m + n + 4)?

Mathematics
2 answers:
kvv77 [185]3 years ago
8 0

Answer:

The sum of of the polynomials (m + n + 3) + (m + n + 4) is<u> </u><u>2m + 2n  +7</u>

Step-by-step explanation:

The given expression is (m + n + 3) + (m + n + 4)

we need  to find the sum of this expression

so, first we add the similar terms

m + n + 3 + m + n + 4

first write the like term together and then constant,

⇒ m + m + n + n +3 + 4

⇒ 2m + 2n  +3 + 4

⇒ 2m + 2n  +7

Therefore, The sum of of the polynomials (m + n + 3) + (m + n + 4) is<u> </u><u>2m + 2n  +7</u>

Margarita [4]3 years ago
6 0
2m+2n+7 is the answer to that
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Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 12
melamori03 [73]

Answer:

98.75% probability that every passenger who shows up can take the flight

Step-by-step explanation:

For each passenger who show up, there are only two possible outcomes. Either they can take the flight, or they do not. The probability of a passenger taking the flight is independent from other passenger. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples 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:

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

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)}.

The probability that a passenger does not show up is 0.10:

This means that the probability of showing up is 1-0.1 = 0.9. So p = 0.9

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets

This means that n = 125

Using the approximation:

\mu = E(X) = np = 125*0.9 = 112.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{125*0.9*0.1} = 3.354

(a) What is the probability that every passenger who shows up can take the flight

This is P(X \leq 120), so this is the pvalue of Z when X = 120.

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

Z = \frac{120 - 112.5}{3.354}

Z = 2.24

Z = 2.24 has a pvalue of 0.9875

98.75% probability that every passenger who shows up can take the flight

7 0
3 years ago
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