All categories

3 years ago

∠2= ∠5 , because they are

Mathematics

1 answer:

Contact [7]3 years ago

3 0

$\sf \red {alternate \: angles}$

$\huge \purple{ \underbrace{ \overbrace{Hope \: it \: helps}}}$

You might be interested in

Kevin needs to find a cheap courier service to deliver birthday gift to his sister.the first courier he is considering charges $

Charra [1.4K]

Answer:

the cost is 21$, and it weighs 1 pound.

Step-by-step explanation:

7 0

3 years ago

A motel has a policy of booking as many as 150 guests in a building that holds 140. Past studies indicate that only 85% of booke

iren2701 [21]

Answer:

0.0015 = 0.15% probability that if the motel books 150 guests, not enough seats will be available.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x successes 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 distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ , if $np \geq 10$ and $n(1-p) \geq 10$ .

150 guests booked:

This means that $n = 150$

85% of booked guests show up for their room.

This means that $p = 0.85$

Is the normal approximation suitable:

$np = 150(0.85) = 127.5$

$n(1-p) = 150(0.15) = 22.5$

Both greater than 10, so yes.

Mean and standard deviation:

$\mu = E(X) = np = 150*0.85 = 127.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{150*0.85*0.15} = 4.3732$

Find the probability that if the motel books 150 guests, not enough seats will be available.

More than 140 show up, which, using continuity correction, is $P(X > 140 + 0.5) = P(X > 140.5)$ , which is 1 subtracted by the p-value of Z when X = 140.5. So