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nirvana33 [79]
4 years ago
14

Plot C(-1,–7), D (6,–7), E(5,3), and F(5,-5) in a coordinate plane. Are CD and EF congruent?

Mathematics
1 answer:
N76 [4]4 years ago
7 0

Answer:

No, they are not congruent

Step-by-step explanation:

By CD and EF being congruent, we want to know if they are equal.

What we can simply do here is to find the distance between the points present.

That is, the distance between C and D, the distance between E and F

To get the distances between two points , we use the following equation

D = √(x2-x1)^2 + (y2-y1)^2

For CD, we have

D = √(6 -(-1)^2 + (-7 + 7)^2

D = √(7)^2 = 7 units

For EF, we have

D = √(5-5)^2 + (-5-3)^2

D = √(-8)^2

D = √(64) = 8 units

This clearly shows that they are both not congruent

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Answer:

The mean and standard deviation changed to 23.5 and 14.62 respectively, based on all 12 samples.

Step-by-step explanation:

We are given that the Six samples are collected of the number of scanning errors: 36, 14, 21, 39, 11, and 2 errors, per 1,000 scans each.

Representing the data in tabular form;

           X                            X - \bar X                            (X - \bar X)^{2}

          36                     36 - 20.5 = 15.5                   240.25

          14                      14 - 20.5 = -6.5                     42.25  

          21                       21 - 20.5 = 0.5                      0.25

          39                      39 - 20.5 = 18.5                   342.25

           11                        11 - 20.5 = -9.5                     90.25

           2                         2 - 20.5 = -18.5               <u>    342.25    </u>

        Total                                                              <u>   1057.5      </u>

Now, the mean of these value is given by;

        Mean, \bar X  =  \frac{\sum X}{n}

                         =  \frac{36+14+21+39+11+2}{6}

                         =  \frac{123}{6}  =  20.5

Standard deviation formula for discrete distribution is given by;

           Standard deviation, \sigma  =  \sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }

                                                 =  \sqrt{\frac{1057.5 }{6-1} } = 14.54

Now, the manager has six more samples taken:

33, 45, 34, 17, 1, and 29 errors, per 1,000 scans each

So, the modified table would be;

           X                            X - \bar X                            (X - \bar X)^{2}

          36                     36 - 23.5 = 12.5                   156.25

          14                      14 - 23.5 = -9.5                     90.25  

          21                       21 - 23.5 = -2.5                     6.25

          39                      39 - 23.5 = 15.5                   240.25

           11                        11 - 23.5 = -12.5                   156.25

           2                         2 - 23.5 = -21.5                   462.25    

          33                       33 - 23.5 = 9.5                     90.25

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         34                        34 - 23.5 = 10.5                   110.25

          17                         17 - 23.5 = -6.5                    42.25

           1                           1 - 23.5 = -22.5                   506.25

          29                        29 - 23.5 = 5.5               <u>     30.25      </u>

        Total                                                              <u>      2353      </u>

<u />

Now, the mean of these value is given by;

        Mean, \bar X  =  \frac{\sum X}{n}

                         =  \frac{36+14+21+39+11+2+33+45+34+17+1+29}{12}

                         =  \frac{282}{12}  =  23.5

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           Standard deviation, \sigma  =  \sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }

                                                 =  \sqrt{\frac{2353 }{12-1} } = 14.62

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