You roll a number cube twice. what is the probability of rolling a 2 first and then rolling an odd number? 1/36,1/12,1/9,1/4

A = event of rolling a 2 on the first roll

P(A) = 1/6 because there is only one "2" out of six labels total on the number cube.

B = event of rolling an odd number on the second roll

P(B) = 3/6 = 1/2 since there are three odd numbers {1,3,5} out of six total

The events A and B are independent, so we can multiply the probabilities

P(A and B) = probability of both events happening simultaenously

P(A and B) = P(A)*P(B)

P(A and B) = (1/6)*(1/2)

P(A and B) = (1*1)/(6*2)

P(A and B) = 1/12

Note: The fraction 1/12 is approximately equal to 0.0833

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

idk

step-by-step explanation: