If a single six-sided number cube is rolled twice, what is the probability that the sum is greater than 5, given that the first roll was a 3? a) 1/9 b) 13/18 c) 2/3 d)cannot be determined

-2/

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116