a standard six-sided die is rolled. what is the probability that the value of the second die is higher than the first

The probability that the value of the second die is higher than the first

Step-by-step explanation:

Explanation:-

In a single throw with two dice total number of sums = 6 X 6 = 36

Let 'E' be the event of the second die is higher than the first

Total number of cases

= {( 1,1), (1,2 )( 1,3), (1,4) , (1,5) ,(1,6)

  ( 2,1), (2,2 )( 2,3), (2,4) , (2,5),(2,6)

   ( 3,1), (3,2 )( 3,3), (3,4) , (3,5),(3,6)

( 4,1), (4,2 )( 4,3), (4,4) , (4,5),(4,6)

( 5,1), (5,2 )( 5,3), (5,4) , (5,5),(5,6)

( 6,1), (6,2 )( 6,3), (6,4) , (6,5),(6,6)}

The favourable cases

      n(E) = {(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)} = 15

The probability that the value of the second die is higher than the first

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer

4.

step-by-step explanation:

it have the same area.brainliest