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Irina-Kira [14]
3 years ago
5

PLEASE HELP ASAP! BRAINLIEST TO BEST/RIGHT ANSWER

Mathematics
2 answers:
Papessa [141]3 years ago
7 0
I would go with B becuase when dividing by powers you really subtract the numbers
Ne4ueva [31]3 years ago
5 0
The right answer I think is B, hope it helps

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isted below are amounts​ (in millions of​ dollars) collected from parking meters by a security service company and other compani
AlladinOne [14]

Answer:

Means:

1.55

1.73

Standard Deviation:

0.1434

0.1338

Coefficient of variation:

9.2

7.7

the limited data listed here shows evidence of stealing by the security service​ company's employees.

Step-by-step explanation:

Given data:

security Service Company          Other Companies    

               x₁                                                 x₂

               1.5                                              1.8

               1.7                                               1.9

               1.6                                               1.6

               1.4                                                1.7

               1.7                                                 1.8

               1.5                                                 1.9

               1.8                                                 1.6

               1.4                                                  1.5

               1.4                                                  1.7

               1.5                                                  1.8

      n₁ = 10                                                 n₂ = 10

To find:

coefficient of variation for each of the two​ samples

Solution:

The formula for calculating coefficient of variation of sample is:

Coefficient of Variation (CV) = (Standard Deviation / Mean) * 100%

Calculate Mean for Security Service Company data:

Mean =  (Σ x₁) / n₁

         = (1.5 + 1.7 + 1.6 + 1.4 + 1.7 + 1.5 + 1.8 + 1.4 + 1.4 + 1.5) / 10

         = 15.5 / 10

Mean = 1.55

Calculate Standard Deviation for Security Service Company data:

Standard Deviation = √∑(x₁ - Mean)²/n₁-1

                                = √∑(1.5-1.55)² + (1.7-1.55)² + (1.6-1.55)² + (1.4-1.55)² + (1.7-1.55)² + (1.5-1.55)² + (1.8-1.55)² + (1.4-1.55)² + (1.4-1.55)² + (1.5-1.55)² / 10-1

=√∑ (−0.05)² + (0.15)² + (0.05)² + (−0.15)² + (0.15)² + (−0.05)² + (0.25)² +  (−0.15)² +  (−0.15)² +  (−0.05)² / 10 - 1

= √∑0.0025 + 0.0225 + 0.0025 + 0.0225 + 0.0225 + 0.0025 + 0.0625 +        0.0225 + 0.0225 + 0.0025 / 9

= √0.185 / 9

= √0.020555555555556

= 0.14337208778404

= 0.143374

Standard Deviation = 0.143374

Coefficient of Variation for Security Service Company:

CV = (Standard Deviation / Mean) * 100%

     = (0.143374 /  1.55) * 100

     = 0.09249935 * 100

     = 9.249935

CV  = 9.2

CV  = 9.2%

Calculate Mean for Other Companies data:

Mean =  (Σ x₂) / n₂

          =  (1.8 + 1.9 + 1.6 + 1.7 + 1.8 + 1.9 + 1.6 + 1.5 + 1.7 + 1.8) / 10

          =   17.3 / 10

Mean  = 1.73

Calculate Standard Deviation for Other Companies data:

Standard Deviation = √∑(x₂-Mean)²/n₂-1

                                = √∑[(1.8-1.73)² + (1.9-1.73)² + (1.6-1.73)² + (1.7-1.73)² + (1.8-1.73)² + (1.9-1.73)² + (1.6-1.73)² + (1.5-1.73)² + (1.7-1.73)² + (1.8-1.73)²] / 10 - 1

                                = √∑ [(0.07)² + (0.17)² + (-0.13)² + (-0.03)² + (0.07)² + (0.17)² + (-0.13)² + (-0.23)² + (-0.03)² + (0.07)²] / 9

                                = √∑ (0.0049 + 0.0289 + 0.0169 + 0.0009 + 0.0049 + 0.0289 + 0.0169 + 0.0529 + 0.0009 + 0.0049) / 9

                                = √(0.161 / 9)

                                = √0.017888888888889                                

                                = 0.13374935098493

                                =  0.13375

Standard Deviation = 0.13375

Coefficient of Variation for Other Companies:

CV = (Standard Deviation / Mean) * 100%

     = (0.13375 / 1.73) * 100

     = 0.077312 * 100

     = 7.7312

CV = 7.7

CV = 7.7%

Yes, the limited data listed here shows evidence of stealing by the security service​ company's employees because there is a significant difference in the variation.

6 0
3 years ago
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