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3 years ago

(x+f)+2=j solve for x ...?

2 answers:

8 0

The equation given in the question has to be studied minutely. Only thing that needs to be kept in mind are the signs.

(x+f) + 2 = j
x + f + 2 = j
x + f = j - 2
x = j - 2 - f
x = j - (2 + f)

I hope the procedure is clear enough for you to understand. I also hope that this is the answer that you were looking for and the answer has actually come to your desired help.

Murljashka [212]3 years ago

7 0

(x + f) + 2 = j subtract 2 from both sides

x + f = j - 2 subtract f from both sides

x = j - f - 2

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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

Treatments

1 2 3 Total

n 6 6 6 18

∑x 42 57 30 129

Mean 7 9.5 5 7.167

∑x2 298 543 152 993

Sd.D 0.8944 0.5477 0.6325 2.0073

ANOVA Table

Source SS df MS

Between-treatments 61 2 30.5 F = 61

Error 7.5 15 0.5

Total 6 8.5 17

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

Since p-value < α, H0 is rejected.

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

Least Significant Difference= t( 0.025,15) √2s²/r s²= 0.50 r= 6 =n1=n2=n3

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

3 years ago

True or False? Each point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.

Bogdan [553]

True

The Perpendicular Bisector Theorem states, If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment

3 0

3 years ago

Solve the inequality and graph the solution on the line provided. 3 + 2x > 5

ElenaW [278]

Answer:

x>1

Step-by-step explanation:

In order to do this, we can take the inequality:

3+2x>5

now there are constants on both sides meaning we can combine them like the following:

3+2x>5

-3 -3

------------

2x>2

Now we can simplify

2x/2>2/2

therefore, the answer is x>1

In order to graph this on a line you would need to put in a circle which is not filled in because x isn't equal to and greater than 1 so we put in a blank circle and since x is always going to be greater than 1, we would put a line that keeps going on.

You would put this line to the right of 1

o----------->

1 2 3 4 5 6

that is how the line would look like

8 0

2 years ago

How would I solve this, converting the left side to a square root? -16 = r^3

ZanzabumX [31]

$-16=r^3\implies \sqrt[3]{-16}=\sqrt[3]{r^3}\implies \sqrt[3]{-16}=r\implies \sqrt[3]{-2^4}=r \\\\\\ \sqrt[3]{(-2)^3\cdot 2^1}=r\implies -2\sqrt[3]{2^1}=r\implies -2\sqrt[3]{2}=r$

7 0

2 years ago

Can someone help me with a math problem please

Nina [5.8K]

Where is the math problem?

8 0

3 years ago