If the rank correlation test is used with paired sample data, and the conclusion is that there is sufficient evidence to support the claim of a correlation between the two variables, can linear regression methods be use to find the regression equation that can be used for predictions? why or why not? fill in the blanks to complete the paragraph. the methods of linear regression ▼ can be used should not be used for predictions. the regression equation is based on ▼ any correlation a linear correlation between the two variables, but the methods of rank correlation ▼ do not require require a linear relationship.

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3

step-by-step explanation:

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6   = 0.00000180

1.80 x 10-4   = 0.000180 (bigger, or "less small", than the former one)

so

1.8 x 10-2       = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.