Answer:
There is a 0.2785 probability of your opponent rolling two even numbers.
Step-by-step explanation:
Since we are looking for the probability of rolling even numbers, we must consider the odds of rolling a 2, a 4 or a 6.
A fair dice would yield in a 6/36 chance of rolling each number, but since these dice are loaded, there is a 7/36 chance of rolling a 4 and a 5/36 chance of rolling a 3 (opposite side of 4); the probabilities for rest remain the same.
Therefore, the probability (P1) of rolling and even number in one die is:
![P_{1} = P(2) + P (4) + P (6)\\P_{1} = \frac{6}{36} + \frac{7}{36} +\frac{6}{36} \\P_{1} = \frac{19}{36}](https://tex.z-dn.net/?f=P_%7B1%7D%20%3D%20P%282%29%20%2B%20P%20%284%29%20%2B%20P%20%286%29%5C%5CP_%7B1%7D%20%3D%20%5Cfrac%7B6%7D%7B36%7D%20%2B%20%5Cfrac%7B7%7D%7B36%7D%20%2B%5Cfrac%7B6%7D%7B36%7D%20%5C%5CP_%7B1%7D%20%3D%20%5Cfrac%7B19%7D%7B36%7D)
Since the game involves two dice, the probability of rolling even numbers on both is:
![P_{2} = P_{1} ^{2}\\P_{2} = (\frac{19}{36}) ^{2}\\P_{2} =0.2785](https://tex.z-dn.net/?f=P_%7B2%7D%20%3D%20P_%7B1%7D%20%5E%7B2%7D%5C%5CP_%7B2%7D%20%3D%20%28%5Cfrac%7B19%7D%7B36%7D%29%20%5E%7B2%7D%5C%5CP_%7B2%7D%20%3D0.2785)
There is a 0.2785 probability of your opponent rolling two even numbers.
The answer is 12x^3+20xy
Hope this helped :)
if you use the desmos graphing calculator - since you didn't provide answer choices - it should be able to help you.
Answer:
That would be 3, 10 and 4.
Step-by-step explanation:
Is the tenths position is 4 or less you round down Otherwise round up.
Answer:
It's either B or C.
Step-by-step explanation:
Sorry, I'm in a rush, no time to explain . . .