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RUDIKE [14]
3 years ago
14

It is well-defined collection of distinct objects A.class B.elements C.Group D.Set

Mathematics
1 answer:
Elena L [17]3 years ago
4 0
Answer

hope correct

D
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Write the point slope form of the equation of the line through the given point with the given slope.
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Point slope form is y - y1 = m( x - x2)
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What are your strength and weaknesses? List three of each.
Nat2105 [25]

This question appears to be asking what concepts in math you are good at and what concepts you are not good at. I’m not sure how to answer this since it seems very dependent on your own knowledge levels. Just be honest! :)

4 0
3 years ago
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In a two week period it was sunny 3/7 of the days. how many days were sunny?
Dafna11 [192]
Assuming two weeks is 14 days
3/7*14=6
therefore it was sunny for 6 days
6 0
4 years ago
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A manufacturing company with 450 employees begins a new product line and must increase their number of employees by 18 percent.
Tcecarenko [31]

Answer:

531 employees

Step-by-step explanation:

A manufacturing company with 450 employees begins a new product line and must increase their number of employees by 18 percent. Note that

450 employees - 100%

x employees - 18%.

Mathematically, you can write proportion:

\dfrac{450}{x}=\dfrac{100}{18},\\ \\x=\dfrac{450\cdot 18}{100}=81.

The new number of employees is

450+81=531.

3 0
4 years ago
I rlly need help with this :(
kobusy [5.1K]

Using the normal distribution, it is found that:

a) The pilot is at the 72th percentile.

b) 19.13% of pilots are unable to fly.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

\mu = 72.6, \sigma = 2.7.

Item a:

The percentile is the <u>p-value of Z when X = 74.2</u>, hence:

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

Z = \frac{74.2 - 72.6}{2.7}

Z = 0.59

Z = 0.59 has a p-value of 0.7224.

72th percentile.

Item b:

The proportion that is able to fly is the <u>p-value of Z when X = 78 subtracted by the p-value of Z when X = 70</u>, hence:

X = 78:

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

Z = \frac{78 - 72.6}{2.7}

Z = 2

Z = 2 has a p-value of 0.9772.

X = 70:

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

Z = \frac{70 - 72.6}{2.7}

Z = -0.96

Z = -0.96 has a p-value of 0.1685.

0.9772 - 0.1685 = 0.8087 = 80.87%.

Hence the percentage that is unable to fly is:

100 - 80.87 = 19.13%.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

3 0
2 years ago
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