For independent-measures anova, individual differences contribute to the variance in the numerator and in the denominator of the f-ratio. what happens to the individual differences in the denominator of the f-ratio for repeated-measures anova?

where is abdul's work? don't see that anywhere posted : /

step-by-step explanation:

its a, i had the same answer, its from study island right

step-by-step explanation:

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube

proportions are fractions that equal to each other. so if two fractions are the same, they are proportional.

let's check the answers to see if they are proportional.

question a is not proportional because you can't turn 1.5/3 to 3/9. if you try to multiply 1.5/3 by 2 or 3, you wont get 3/9. it's impossible to make the fractions the same.

question b is not proportional because you can't turn 22/33 to 4/6. 33 can't go into 6.

question c is proportional because when you multiply 2 and 3 by 71, you would get 142/213, and those fractions equal to each other

question d is proportional because when you divide 14 and 7 by 2, you would end up with 7/3.5, and those fractions equal to each other.