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Paha777 [63]
3 years ago
10

36 POINTS! Write a problem where the negative square root is not an integer. then graph the square root(tell me what to graph si

nce we cant on here)
Mathematics
1 answer:
Tcecarenko [31]3 years ago
7 0
<span>" - √5 "....and 'square root ' is statement and difficult to graph....you need a function</span>
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Which is a solution of x2 – x – StartFraction 3 Over 4 EndFraction = 0?
Anna [14]

Answer:

x = - \frac{1}{2}, x = \frac{3}{2}

Step-by-step explanation:

Given

x² - x - \frac{3}{4} = 0 ( add \frac{3}{4} to both sides )

x² - x = \frac{3}{4}

Solve using the method of completing the square

add ( half the coefficient of the x- term )² to both sides

x² + 2( - \frac{1}{2} )x + \frac{1}{4} = \frac{3}{4} + \frac{1}{4}

(x - \frac{1}{2} )² = 1 ( take the square root of both sides )

x - \frac{1}{2} = ± 1 ( add \frac{1}{2} to both sides )

x = \frac{1}{2} ± 1

Thus

x = \frac{1}{2} - 1 = - \frac{1}{2}

x = \frac{1}{2} + 1 = \frac{3}{2}

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Which of the following equations has the solution x= all real numbers?
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3 years ago
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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