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%

step-by-step explanation:

i could only see the first page. the other pages were blurry. ; -;

x hot dogs for 2$, y drinks for 3$. 29 items and total 70

2x+3y=70

x+y=29

if x=17and y=12

2(17)+3(12)=70

34+36=70

answer: the correct answer is -32

step-by-step explanation: