A single die is rolled twice. the set of 36 equally likely outcomes is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),
Maksim231197 [3]
<span>We need to find the rolls whose sum is greater than 10. By looking at the outcomes, we see that (5,6), (6,5), and (6,6) all have a sum greater than 10. Therefore, there are 3 chances to get a sum greater than 10. Since there are 36 chances overall, the probability of rolling greater than 10 are 3/36 = 1/12.</span>
(2x^2 -5y)/(3x-y)
((2*2)^2-(5*-4))/((3*-4)-(-4))
(4^2+20)/(-12+4)
36/-8
-4.5
180, is the answer because when I had a question like that you have to add
Answer:
40.9 - 5y
Step-by-step explanation:
25.6 - 5y + 15.3
40.9 - 5y ## ^ combine like-terms
