2fair dice are rolled. what is the probability that the sum is even given that the first die is rolled is a 2?

wouldnt you add 6+6=12 then divide that with the probability of getting a 2?

Explanation:

"given that the first die rolled is a 2" means we know 100% that the first die shows a 2. Either we can see the die or a friend is telling us the status. Since we know the first die is a 2, this means we can effectively ignore it. Everything will hinge on the second die. If the second die shows an odd number, something like 1, then 2+1 = 3 is the result which is also odd.

The general rule is odd+even = odd and even+even = even.

Therefore, the two dice must together be even for the sum to be even.

Of the six possible ways to roll a die {1,2,3,4,5,6}, there are 3 even values {2,4,6} so the chances of rolling an even number on the second die is 3/6 = 1/2. Again we dont need to consider the first die at all since everything practically hinges on this second die.

50%

Step-by-step explanation:

For the total to be even, when the first number is even, the second number also has to be even.

2, 4, and 6 are even, out of 6 numbers. That is 3 numbers out of 6, which gives us that the probability is 1 out of 2, or 50%.

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~

x-3y=4

step-by-step e