Can someone pleaseee help!!!

Mathematics

1 answer:

krok68 [10]3 years ago

3 0

Answer:

Step-by-step explanation:

use known angles and rules to find unknown values

vertical angles (pairs of angles on opposite sides of an intersection) are equivalent

supplementary angles (angles found next to each other on a line) add up to 180 degrees

opposite interior angles are equivalent

And I don't remember the name of this kind of angle, but angle 1 and 10 are equivalent because they are found on the same "location"

using these rules you can find all missing angles.

1= 75

2=180-75=105

let me know if you need more help than this

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An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0

zmey [24]

Answer:

96.33% probability that everyone who appears for the flight will get a seat

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .