A fair die is rolled 4 times. What is the probability of having no 3 and no 4 among the rolls? Round your answer to three decimal places

66.667%

Step-by-step explanation:

In a fair die, there are 6 possible rolls. Out of the 6, 1 number is chosen each time. Since order does not matter in this problem, we will be applying combination, so this would be written as 6C1 x 4. In order to exclude 3 and 4, from 6, the choices must be narrowed down to 4. Written out as an equation, this would be 4C1 x 4. To get the percentage, simply divide the two equations.

(4C1 x 4)/(6C1 x 4) = 0.66667 = 66.667%

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )