Answer:
C. Test for Goodness-of-fit.
Step-by-step explanation:
C. Test for Goodness-of-fit would be most appropriate for the given situation.
A. Test Of Homogeneity.
The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.
B. Test for Independence.
The chi square is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.
C. Test for Goodness-of-fit.
The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with their estimates and provided that one degree of freedom is deducted for each parameter estimated.
Answer:
w= -52
Step-by-step explanation:
We simplify the equation to the form, which is simple to understand
15-w/4=28
Simplifying:
15-0.25w=28
We move all terms containing w to the left and all other terms to the right.
-0.25w=+28-15
We simplify left and right side of the equation.
-0.25w=+13
We divide both sides of the equation by -0.25 to get w.
w= -52
Hope it Helps :) !!
X is greater than or equal to -200
