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>
I'm pretty sure it's -4x.
Answer: c=5.4
The two roots are 3/5 and 9/5
Step-by-step explanation:
assume 5x2−12x c=0 is supposed to be 5x^2 - 12x + c = 0
p = (12 + sqrt(144-20c))/10
q = (12 - sqrt(144-20c))/10
p-3q=0,
1.2 + 0.1sqrt(144-20c) +
-3.6 + 0.3sqrt(144-20c) = 0
-2.4 + 0.4sqrt(144-20c) +2.4 = 2.4
sqrt(144-20c) = 2.4/0.4 = 6
144-20c=36
144-36=20c
c = 108/20 = 5.4
5x^2-12x+5.4=0
x = 3/5 or x = 9/5
