Conditional probablility P(A/B) = P(A and B) / P(B). Here, A is sum of two dice being greater than or equal to 9 and B is at least one of the dice showing 6. Number of ways two dice faces can sum up to 9 = (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 10 ways. Number of ways that at least one of the dice must show 6 = (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2), (6, 1) = 11 ways. Number of ways of rolling a number greater than or equal to 9 and at least one of the dice showing 6 = (3, 6), (4, 6), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6) = 7 ways. Probability of rolling a number greater than or equal to 9 given that at least one of the dice must show a 6 = 7 / 11
The answer is a.
Substitute (-2,1) and (1,10) into each answer. Only a is correct because you will get -7 from both points.
So all you need to do if find the min of the function:
f(x)=0.04x^3-4x^2-176x
f'(x)=0.12x^2-8x-176
The only zero for this is x=84.105 <---first
Plug this into f(x) to get f(84.105)=-19300 <---second
Answer:
16.83
Step-by-step explanation:
7.36
+ 9.47
----------
16.83
