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adell [148]
3 years ago
7

Evaluate 13 - 0.75y + 8x when w =12 and x=1/2

Mathematics
1 answer:
JulijaS [17]3 years ago
8 0

<em>Note: I am assuming you meant to write the value of y = 12 instead of w = 12. </em>

<em>So, I am assuming y = 12</em>

Answer:

The value of 13 - 0.75y + 8x when we substitute y = 12 and x = 1/2 will be: 8

Step-by-step explanation:

Given the expression

13\:-\:0.75y\:+\:8x

Given

  • y = 12
  • x = 1/2

substituting y = 12 and x = 1/2 in the expresion

13\:-\:0.75y\:+\:8x=13\:-\:0.75\left(12\right)\:+\:8\left(\frac{1}{2}\right)

                            =13-0.75\cdot \:12+8\cdot \frac{1}{2}

                             =13-9+4

                             =8

Therefore, the value of 13 - 0.75y + 8x when we substitute y = 12 and x = 1/2 will be: 8

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Answer:

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Step-by-step explanation:

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In this case the system is:

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The following data are from an experiment designed to investigate the perception of corporate ethical values among individuals w
alukav5142 [94]

Answer:

Part 1

a. Sum of Squares, Treatment= 61

b. Sum of Squares, Error= 7.5

c. Mean Squares, Treatment = 30.5

d. Mean Squares, Error= 0.5

2. the F value lies in the rejection region > 3.6823

3. The value of the test statistic = 61

4. The p-value is < 0.00001

5. Conclusion

Since p-value < α, H0 is rejected.

6. Between x`2 and x`3

7. Fisher's Least Significant Difference value almost 0.869

8.There is a significant difference between the means

Step-by-step explanation:

Summary of Data

                       <u>   Treatments</u>

                       1             2             3                 Total

n                      6             6             6                   18

∑x                   42          57           30                  129

Mean              7            9.5           5                  7.167

<u>∑x2              298         543        152                    993</u>

<u>Sd.D       0.8944     0.5477     0.6325           2.0073</u>

ANOVA Table

<u>Source                                  SS              df                  MS </u>

Between-treatments           61              2                   30.5       F = 61

<u>Error                                     7.5           15                     0.5 </u>

<u>Total                                     6             8.5                     17 </u>

a. Sum of Squares, Treatment= 61

b. Sum of Squares, Error= 7.5

c. Mean Squares, Treatment = 30.5

d. Mean Squares, Error= 0.5

2. Using alpha= 0.05 the F value lies in the rejection region i.e F > 3.6823

x1` -x2`= 7-9.5= -2.5 Not significant as difference <3.68

x1`- x3`= 7-5= 2 Not significant as difference <3.68

x2` -x3`= 9.5-5= 4.5 Significant as difference > 3.68

3. The value of the F test statistic = 61

4. The p-value is < 0.00001

5. Conclusion

<em>Since p-value < α, H0 is rejected.</em>

6. Using alpha= .05, differences occurs between x2` and x3` as their difference is greater than 3.68

7. Fisher's Least Significant Difference value almost 0.869

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Least Significant Difference= 2.13 √ 2*0.50/ 6

=0.869

8.There is a significant difference between the means

x1` -x2`= 7-9.5= -2.5 Significant as difference > Least Significant Difference

x1`- x3`= 7-5= 2 Significant as difference > Least Significant Difference

x2` -x3`= 9.5-5= 4.5 Significant as difference > Least Significant Difference

6 0
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Answer:

your answer is

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Explain:

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