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NISA [10]
3 years ago
8

What is the geometric mean of 50 and 1/8

Mathematics
1 answer:
lesya692 [45]3 years ago
8 0
2.5 is the geometric mean
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makvit [3.9K]

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2 years ago
512, 256, 128, 64, __, __, __, ...​
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3 years ago
Need help on homework <br><br> #8
irakobra [83]
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7 0
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
2 years ago
A day-care center has 2 baskets of dolls. One basket has 8 dolls, and the other basket has an unknown number of dolls in it. Wha
xenn [34]

Answer:

d+8

Step-by-step explanation:

Let d represent number of dolls in 2nd basket.

We have been given that a day-care center has 2 baskets of dolls. One basket has 8 dolls, and the other basket has an unknown number of dolls in it. We are asked to represent this situation in an expression.

The number in both baskets would be equal to dolls in 1st basket plus dolls in 2nd basket.

We can represent this information in an expression as:

d+8

Therefore, our required expression would be d+8, where d represents number of dolls in 2nd basket.

3 0
3 years ago
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