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3 years ago

Can someone help me thanks you get 15 points!

Mathematics

2 answers:

stira [4]3 years ago

4 0

Answer:

your awnser would be B //////////////

d1i1m1o1n [39]3 years ago

3 0

Letter D is the answer

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In a linear Pair angles are?

tatiyna

Answer:

angles in linear pair are 90° hope it helps u ♥♥♥♥♥♥♥

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3 years ago

Read 2 more answers

Carl, Gilda, Conroy, and Kyla are the four candidates in a school election. Carl received 1/5 of the votes, Gilda received 5% of

Juliette [100K]

Firstly, we can convert all of the fractions into percentages. To do this, we need to make the denominator of the fraction 100, and whatever we do to the denominator we must also do to the numerator.

5 x 20 = 100
1 x 20 = 20.

So Carl recieves 20/100 or 20% of the votes.

4 x 25 = 100
1 x 15 = 25

So Conroy receives 25/100 or 25% of the votes.

If we add these together and Gilda's 5%, we get 50%. Since there are 100% votes overall, we need to do 100 - 50 = 50.

Kyla receives 50% of the votes.

8 0

3 years ago

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Find the absolute value -5/6|

Nadya [2.5K]

Answer:

5/6 OR 0.83 (with the three repeating)

Hope that helps!

Step-by-step explanation:

5 0

3 years ago

Begin by graphing the standard absolute value function f(x)=|x|. Then use transromations of this graph to describe the graph the

Len [333]

The +2 would cause a shift of up two on the y-axis. the 2 infront of the x, would cause the graph to narrow. using the point 0 f(x) would be 2 or (0,0) and on g(x) it would be (0,2) using x=3, you would have (3,3) and (3,8)

4 0

3 years ago

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

8 0

3 years ago