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
Answer:
OPTION B - 2
Step-by-step explanation:
What is the upper limit for the zeros of the function P(x) = 4x^4 + 8x^3 - 7x^2 - 21x - 9. Ans: 2 is an upper limit. Use synthetic division. and the remainder are all positive, 2 is an upper limit.
Answer:
If a graph is an Euler Circuit that mean that it can be traversed and begins and has all even verticies. This allows you to start and stop at the same verticie.
Step-by-step explanation:
A. -(a+5)
Because the negative sign is outside the parenthesis, multiplying by -1 just removes the negative sign:
-(a+5) * -1 = a+5
B. -(-x+31)
Apply the distributive property:
-(-x+31) becomes (- -x +31) which simplifies to (x+31)
multiply that by -1 to get -x+31
C. -(4x+12)
Because the negative sign is outside the parenthesis, multiplying by -1 just removes the negative sign:
-(4x+12) * -1 = 4x+12
