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n200080 [17]
1 year ago
7

Suppose CM and DM intersect AB as shownAlison states < AMC + < CMD+ < DMB = < AMB which of the following justifies A

lison's statement?A. Angle Addition PostulateB. Segment Addition PostulateС. Construction of an Angle BisectorD. Definition of Supplementary Angles

Mathematics
1 answer:
aivan3 [116]1 year ago
4 0
Answer: D. Definition of Supplementary Angles

Explanations:

Note from the diagram shown that:

AMB forms a straight line

This means that < AMB = 180 degrees (Because sum of angles on a straight line is 180 degrees)

Also note that the sum of angles on a straight line is 180 degrees

< AMC + < CMD + < DMB = 180 (sum of angles on a straight line)

Since < AMB = 180

< AMC + < CMD + < DMB = < AMB

Since the above indicates that the sum of the angles add up to 180 degrees, they are supplementary angles.

Note that supplementary angles are angles that add up to 180 degrees

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A bag contains 6 red marbles, 2 blue marbles, and 1 green marble. What is the
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3 years ago
In Applied Life Data Analysis (Wiley, 1982), Wayne Nelson presents the breakdown time of an insulating fluid between electrodes
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Answer:

The sample mean is \bar{x}=14.371 min.

The sample standard deviation is \sigma = 18.889 min.

Step-by-step explanation:

We have the following data set:

\begin{array}{cccccccc}0.15&0.82&0.81&1.44&2.70&3.28&4.00&4.70\\4.96&6.49&7.25&8.03&8.40&12.15&31.89&32.47\\33.79&36.80&72.92&&&&&\end{array}

The mean of a data set is commonly known as the average. You find the mean by taking the sum of all the data values and dividing that sum by the total number of data values.

The formula for the mean of a sample is

\bar{x} = \frac{{\sum}x}{n}

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\bar{x}=\frac{0.15+0.82+0.81+1.44+2.7+3.28+4+4.7+4.96+6.49+7.25+8.03+8.4+12.15+31.89+32.47+33.79+36.80+72.92}{19}\\\\\bar{x}=14.371

The standard deviation measures how close the set of data is to the mean value of the data set. If data set have high standard deviation than the values are spread out very much. If data set have small standard deviation the data points are very close to the mean.

To find standard deviation we use the following formula

\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }

The mean of a sample is  \bar{x}=14.371.

Create the below table.

Find the sum of numbers in the last column to get.

\sum{\left(x_i - \overline{X}\right)^2} = 6422.0982

\sigma = \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 6422.0982 }{ 19 - 1} } \approx 18.889

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