Please help Solve for n. 5/15 = n/21 A. N = 35 B. N = 7 C. N = 5 D. N = 90

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Combinatorial Enumeration. That whole class was a rollercoaster ride of mind-blowing generating functions to prove crazy things. The exam had ridiculous questions like 'count the number of cactus trees with n vertices such that etc etc etc' and you'd do three pages of terrible terrible sums and algebra. Then your final answer would be something beautiful like n/2 and you'd breath a sigh of relief and thank the math gods.

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Time zones and time offsets. A time zone is a geographical region in which residents observe the same standard time. A time offset is an amount of time subtracted from or added to Coordinated Universal Time (UTC) time to get the current civil time, whether it is standard time or daylight saving time (DST)

I hope my answer has come to your help. God bless you and have a nice day ahead!

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Anastasy [175]

This was what I got

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2 years ago

Read 2 more answers

Suppose that an airline overbooks seats on their flights. In particular, it sells 300 tickets for a flight when there are only 2

vladimir1956 [14]

Using the normal approximation to the binomial, it is found that there is a 0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

In this problem:

15% do not show up, so 100 - 15 = 85% show up, which means that $p = 0.85$ .
300 tickets are sold, hence $n = 300$ .

The mean and the standard deviation are given by:

$\mu = np = 300(0.85) = 255$

$\sigma = \sqrt{np(1-p)} = \sqrt{300(0.85)(0.15)} = 6.185$

The probability that we will have enough seats for everyone who shows up is the probability of at most 270 people showing up, which, using continuity correction, is $P(X \leq 270 + 0.5) = P(X \leq 270.5)$ , which is the p-value of Z when X = 270.5.

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

$Z = \frac{270.5 - 255}{6.185}$

$Z = 2.51$

$Z = 2.51$ has a p-value of 0.994.

0.994 = 99.4% probability that we will have enough seats for everyone who shows up.

A similar problem is given at brainly.com/question/24261244

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