8 Team Bracket Form

In absolute terms pump A takes 8 hours to finish.If it’s the time between pump B stops and A finish, it’s 30 min. Pump B takes 10 hours and 3/4 of that is 7 hours and 30 min. Since pump A takes 8 hours, it takes 30 more minutes to finish its pool.If you mean how much it takes pump A to finish pump B’s pool, it’s 2 hours. Pump B filled 3/4 pool, so 1/4 remain “empty” and it takes 2 hours for pump A (8 hours/4) to fill that much.If it’s fill both pools, it’s 2 and 1/2 hours. Half an hour more to finish A’s pool plus 2 hours to finish pump B’s work.All of this is assuming both pools are of the same size!EDIT:After some clarification from the OP, the right answer should be 2 hours. It’s the time pump A takes to finish 1/4 pool.

17I arrived at this number in a somewhat unusual way, so I’ll explain. I didn’t so much calculate it as intuit it.A 3:2 ratio means 3 of 5 games were won and 2 of 5 were lost. But what was important to me was that we can work with multiples of 5. This is important because the total number of games is also a multiple of 5.85 games is 15 short of 100 (which is 20 times 5), or 17 times 5. If I wanted to know how many games were won or lost, I can just multiply 17 by 3 (51 wins) or by 2 (34 losses) and confirm that 51+34=85.But that’s not what was asked. The question asked how many more games were won than lost, which I knew based on my deduction of the multiple of 17, and the difference of the 3:2 ratio, would be one multiple of 17, or… 17.

No it didn’t… It never has… But it was Beautiful - Read This - https://www.quora.com/What-didnt...

Jacques Laskar has done large scale numerical simulations of the solar system, seeking to answer a number of different questions regarding the stability of the solar system.On the topic of this question, his results show that it’s possible that Mercury’s orbit could become so eccentric at times (with about a 1% probability over several thousand million years) that it suffers a close encounter with Venus, and in turn Earth’s orbit is catastrophically altered by a subsequent close encounter with Venus. That’s the bad news (which isn’t really that bad on the scale of things).The good news is that another question on the solar system’s stability is “over what time scale can we consider our predictions to be accurate?” The answer to that is apparently some 65 million years. That is, we can predict with quite reasonable accuracy and certainty where all the major bodies will be for the next 65 million years.What happens at 65 million years? It’s merely a consequence of chaotic motion: 65 million years is the time scale for small errors in current observations to build up to sufficiently large errors in our predictions that we can no longer be very confident about them. This is the same as with the weather: we can make some pretty good observations and even interpolations to fill in missing data, and we have great models, but the time scale for predictions to become uselessly uncertain is about a week for the weather in many places.Coming back to the question at hand, yes, Earth could be ejected from its current quasiperiodic orbit of the sun. Thankfully, no, it won’t happen in our lifetimes (or a long time hence) unless something totally unforeseen shoots through our solar system with a strong enough disruptive influence.

Problem is to choose 6 out of 8 boys and 3 out of 6 girls. Since 2 boys will be “unchosen” we could restate as choose 2 boys out of 8 and 3 girls out of 6.Start with boys. 8 choices for first boy, but only 7 to choose from for 2nd. So number of choices is 8×7=56. But since either of the 2 boys could have been chosen first, we have to divide 56 by 2 giving 28.6 choices for 1st girl, 5 for 2nd and 4 for 3rd. So 6x5x4=120. But again the order of the girls is not relevant. We can order 3 girls in 3x2x1=6 ways. So ways of choosing girls reduces to 20 (120/6).Since any of the boy choices can work with any of the girl choices, the overall answer is 560 (28×20).Stated mathematically the answer is 8!/(6!×2!) × 6!/(3!×3!), where 8! (8 factorial) = 8×7×6×5×4×3×2×1.

Let’s say there’s Jools, Jops, Stoo, Rj, Ziggy and Cheese. If we start forming a first group we have six students to choose from. The second person of that group could then be filled by any of the 5 remaining students. Which leaves us with 4 individuals eligable for the third position. The number of possible combinations of this group would thus be 6*5*4 = 120.Since the question does not state an outline of specific positions or roles in the group(s) formed, a group consisting of Jops, Ziggy and Cheese should be considered equal to the combination Cheese, Jops and Ziggy. The internal ordering of the group(s) should therefore not matter for the answer to this question. Whether we are forming one or two groups is therefore not really relevant: since all six students have been considered in the forming of the first group and count towards the number of possible combinations.In short, forming a group of three out of six people can be done in 120 ways, all of which leaving an arbitrary remainder of three students.

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