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Anna [14]
3 years ago
9

Look at picture for the question I will make brainiest whoever has the correct answer!

Mathematics
1 answer:
aleksklad [387]3 years ago
6 0

Answer:

C

Step-by-step explanation:

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In the figure below, \overline{EO}=2x-5 <br> EO<br> =2x−5 and \overline{MO}=x+17 <br> MO<br> =x+17
jonny [76]

Answer: x = 69

Step-by-step explanation:

4 0
3 years ago
Functions math, please help.​
kozerog [31]
Y intercept is (0,-7). slope is 2.

Change the function to y=mx+b format first. m=slope and plug in 0 as x to get y intercept.

8 0
2 years ago
I need the answer for this question
Scrat [10]

Answer:

z=-13.9

Hope this helped!!

7 0
2 years ago
Read 2 more answers
If n = 2 + 6m and p = 2/n, what is the value of p when m = -1/2?
artcher [175]

Answer:

p = - 2

Step-by-step explanation:

Given

n = 2 + 6m ← substitute m = - \frac{1}{2} into the expression

n = 2 + 6(- \frac{1}{2} ) = 2 - 3 = - 1

Thus

p = \frac{2}{n} = \frac{2}{-1} = - 2

8 0
2 years ago
An aptitude test is designed to measure leadership abilities of the test subjects. Suppose that the scores on the test are norma
Komok [63]

Using the normal distribution, it is found that 0.0329 = 3.29% of the population are considered to be potential leaders.

In a <em>normal distribution</em> 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.

In this problem:

  • The mean is of 550, hence \mu = 550.
  • The standard deviation is of 125, hence \sigma = 125.

The proportion of the population considered to be potential leaders is <u>1 subtracted by the p-value of Z when X = 780</u>, hence:

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

Z = \frac{780 - 550}{125}

Z = 1.84

Z = 1.84 has a p-value of 0.9671.

1 - 0.9671 = 0.0329

0.0329 = 3.29% of the population are considered to be potential leaders.

To learn more about the normal distribution, you can take a look at brainly.com/question/24663213

8 0
2 years ago
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