A chi-square test of independence is considered to be valid if each of the expected cell frequencies are:

12*c

step-by-step explanation:

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.

(5x-1)^2= (5x)^2-2(5x)×1+(1)^2=25x^2-10x+1