Answer:
Step-by-step explanation:
Refer to the attachment below
This may be a bit awkward to explain in writing, so please bear with me:)
You are given the equations. Begin by focusing on ad = 11.6. Treating variables normally, this reads as a times d = 11.6.
From that, d = 11.6/a by dividing both sides by a.
With d expressed, substitute (11.6/a) into cd = 6.7. Then isolate c by multiplying both sides by a/11.6, yielding c = (6.7a)/11.6.
Now that c is known, insert (6.7a)/11.6 for c in bc = 8.3. The algebra becomes a bit messy, but solving for b gives approximately 14.3705 / a. Since you need ab, multiply both sides by a, and rounding to one decimal place produces ab = 14.4
Answer:
51%.
Step-by-step explanation:
Given that the probability of at least one win is 0.3, it indicates that the chance of none winning is:
1 - 0.3 = 0.7 represents the probability of losing in the first two games. Since these are independent events, the chance of not winning in all four games is:
(0.7) * (0.7) = 0.49
Consequently, the likelihood of winning at least one game equals the complement of not winning any, hence:
1 - 0.49 = 0.51
This implies a 51% probability of winning at least one out of four games.