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@qq11, I don't think your answer is right either. The question specifically says the balance beam can only be used twice, so I assume the question is asking how to find the heaviest marble by using the beam twice, not 3 times. You simply just split the 9 marbles into groups of 3. Put two sets on each side of the balance beam, and this will tell you which set has the heavy marble. After that, just take the heavy set of 3 marbles and put 2 of them on the balance beam, and you'll know which marble is the heavy marble. Less
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Ooo and ooo and ooo. First try: Ooo vs ooo => Ooo is heavier. Now take two of those and put one on each scale. If equal than the third is heaviest. If not equal, heavier one is heaviest. OR try ooo vs. ooo. Equal. Repeat step two for the pile that is established to be Ooo. Less
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@qq : go kill yourself.....are you really thinking about sayng to put the marble in your hand?? c mon bro.......alex is right btw Less
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Isn't it 90 + 15 = 105 degrees?
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Yea its 105 degrees
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It's 90 degrees between 9 and 6 o'clock but the hour hand will move half way towards 10. Since there's 3 hours between the 9 and 6 we can say 1 hour = 90/3 = 30. The initial 90 degrees + the additional (1/2)*30 = 105 degrees. Less
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20
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correction to the above, if you're writing the numbers with letters as opposed to numerically the correct answer is 10. Less
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9,29,39,49,59,69,79,89,99...19 isn't one..so 9
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5/36
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5/36 is right.... Combinations that add up to six: 1 and 5, 2 and 4, 3 and 3, 4 and 2, 5 and 1. So we have 5 successes. There are 36 individual outcomes because there are 6 sides on each die and 6*6 = 36 So we have 5/36. Less
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1/6
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Answer is 7, the solution can be found online as this is somewhat common in interviews. Less
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Wouldnt it be 5 races? just time each of them
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Wouldn't it be 6? 5 inital races, then you take the winners of each race and race them once more Less
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Chance of 1 Heads = .5 Chance of 7 heads in a row = (.5)^7 to do this in your head think of fractions... (1/2)^7 = (1/2*1/2=1/4, * 1/2=1/8, * 1/2 = 1/16....if you do it seven times you get 1/128. so you have a better chance to pick out the penny with two heads (1/100) than to get 7 heads in a row (1/128) Less
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99 times out of 100 you take the normal penny. prob that normal penny give 7 heads in a row is 1/128. so, prob that he has special penny is 1/100 / (1/100 + 0.99*1/128) = 56.2% Less
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~56.4% that you have the 2-headed penny.
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If margins remain the same, earnings should come in closer to last year's .26 vs estimates of .20. Less
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NSSC is the little stock that should and will at some point. It has produced a double digit earnings surprise in each of the last 4 quarters. It will in the coming quarter as well, as management assures that they will produce flat revenues for the coming quarter of Less
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NSSC is the little stock that should and will at some point. It has produced a double digit earnings surprise in each of the last 4 quarters. It will in the xoming quarter Less
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total market cap of sector x market share
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market share price * nb of shares
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I was at my second year of university, and I didn't thought about the possibility of tricky questions, so I spent my time trying to find a solution and at the end I answered: "I don't know!" (How stupid I was) Less
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Expected Payoff: ((1/2)^4)*10 = 0.625 Cost = 1 ==> No
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I will consider 2 cases. First case: we pay a buck to flip the coin 4 times and if we don't get 4 heads, we again pay a dollar to flip the coin 4 times and the results of the first round do not count (i.e getting THHH for the first round and HTTT for the second won't do).So here we pay 1 dollar for 4 flips. In this case we would expect HHHH to happen once in 16 rounds (since the probability of having HHHH is 1/16). Then expected value of this game = -16 (the cost of playing 16 rounds) + 10 = -6. Hence, we shouldn't play. Second case: we pay a dollar to flip the coin 4 times and if we do not get 4 heads we pay another buck for ONE more flip until we get 4 heads in a row. Let us work out the expected number of flips to get 4 heads in a row. Let Ei = expected number of flips to get i heads in a row. Then E4 = 0.5 * E3 + 0.5 * E4 + 1 (since if we get a head then the problem is reduced to finding the expected number of flips to get 3 heads in a row (ie E3), if we get a tail, then we start from scratch (ie E4)). Similarly, E3 = 0.5 * E2 + 0.5 * E4 + 1; E2 = 0.5 * E1 + 0.5 * E4 + 1; E1 = 0.5 * E0 + 0.5 * E4 + 1; E0 = 0 since the expected number of flips required to get 0 heads is simply zero. Solving these equations, we get that E4 = 30, ie we expect to flip the coin 30 times until we get 4 heads in a row. Remember that the first 4 flips cost us a dollar and EACH subsequent flip is an extra dollar. So the cost of this game would be 1 + 26 = 27. So the epected value of this game = - 27 + 10 = - 17. Therefore, we shouldn't play this game. Less
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EV is -3/8