To work this problem I simulated 70,000 poker hands and noted the ones that had 3 of a kind (including ones that were full house and 4 of a kind). I counted the number of hands between instances of this, and quit after I had counted 1000 instances. (There were generally more than 1000 instances in 70,000 hands.)
In 10 such simulations, the average number of hands that had to be dealt before 3 of the same kind were seen ranged from 40.507 to 46.064. The mean of those 10 simulations was 43.359.
The published numbers for frequency of occurrence of the hands of interest are
54,912 . . . . for 3 of a kind
3,744 . . . . . for full house
624 . . . . . . .for 4 of a kind
for a total of 59,280 out of 2,598,960 possible poker hands. Then you'd expect a hand containing 3 of the same number approximately one every 43.842 hands.
This number is pretty close to the simulated value.
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The problem isn't clear as to the number of cards that are dealt for one hand. If it is only 3, then the odds change somewhat, as do the simulation results. I used a 5-card hand for my simulation.
First, let x be the amount of 20% acid solution in liters. With this, 60 - x is the amount of 50% solution. The acid balance before and after mixing becomes,
(0.2)(x) + (0.5)(60 - x) = (0.40)(60)
The value of x is 20. Therefore, Gabe needs 20 liters of 20% acid solution and 40 liters of 50% acid solution.
Answer:
Step-by-step explanation:
So we have the inequality:
Definition of Absolute Value:
Note that the sign is flipped in the second case because we multiplied by a negative.
Add 5 to both sides to both equations:
Merge:
And we're done!
4.33 or 4.333 it really doesn't matter you just have to have at least 2 "3's" because the "3's" just go on forever and ever
Answer:
Tip = $4.34
Total = $18.34
Step-by-step explanation:
Dinner = $14
Tip is 31% of dinner
14 x 0.31 = 4.34
14 + 4.34 = 18.34