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%

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)

i don't know, but that looks pretty bad. my school passed us unless we failed in all the subjects, but idk abt ur school's rules.

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