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PSYCHO15rus [73]
3 years ago
11

Which data set is the farthest from a normal distribution?

Mathematics
2 answers:
viktelen [127]3 years ago
8 0

Answer:

(C)0.9, 1.0, 1.0, 1.1, 1.1, 1.1, 1.2, 1.2, 1.3

Step-by-step explanation:

In (A), (B), (D) and (E) options, all have similar set of numbers and ranges between 2 to 10. Also, these data sets contains the whole numbers, thus when we find mode, median and mean of these data sets, they will follow the same path.

But in (C), the data values of the given data set involves decimals and are not the part of the other data values and it goes from 0.9-1.3, thus making them farthest from a normal distribution when mean, mode and median are calculated.

Thus, option (C) is correct.

Trava [24]3 years ago
5 0

Answer:

i belive its c

Step-by-step explanation:

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Step-by-step explanation:

Trying to factor as a Difference of Squares :

1.1      Factoring:  r2-96

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =

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Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 96 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares.

Equation at the end of step  1  :

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Step  2  :

Solving a Single Variable Equation :

2.1      Solve  :    r2-96 = 0

Add  96  to both sides of the equation :

                     r2 = 96

When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  

                     r  =  ± √ 96  

Can  √ 96 be simplified ?

Yes!   The prime factorization of  96   is

  2•2•2•2•2•3

To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 96   =  √ 2•2•2•2•2•3   =2•2•√ 6   =

               ±  4 • √ 6

The equation has two real solutions  

These solutions are  r = 4 • ± √6 = ± 9.7980  

Two solutions were found :

                  r = 4 • ± √6 = ± 9.7980

Processing ends successfully

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