A un Technology Analyst Summer Intern le preguntaron...21 de marzo de 2013

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It's simple. Take 1 coin from the first stack, 2 coins from the second, 3 from the thirds - you see where this is going - finally 10 from tenth stack. put them on weighing scale. If in result you get 56 - it was first stack, 57 - second.... 65 - tenth stack. Menos

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1-If all the coins are made from the same material and same diameter, then the heavier coin stack should be slightly taller. 2- Weight all the stacks at once, and remove each stack until you see a dip in the tota weight . Menos

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I didn't quite understand what you meant. So there are 10 coins, 9 weigh 1 gram, and one weighs 2 grams? The second sentence seems contradictory. If you can only make one measurement, then you would divide the stacks into two. Place each stack on either scale, and whichever one is heavier contains the 2g coin. Menos

A un Trading Analyst Intern le preguntaron...29 de enero de 2014

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Expected value of the first game is (2/36)($1) + (1/36)($2) = 1/9 of $1. The expected value of the second game is (1/6)($1) = 1/6 of a dollar. Therefore, the second game is preferred and you would pay (1/6 of $1) - (1/9 of $1) = 1/18 of $1 to play the second game instead of the first game. Menos

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I believe the interview candidate's answer is incorrect. The expected value of game #1 is (3/36)($1) + (33/36)($0) = 1/12 of $1. The expected value of game #2 is (1/6)($1) = 1/6 of $1. The second game is preferred, and you'd pay 1/12 of $1 to play the second over the first. Menos

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The interview candidate's answer is correct. The wording states that in the first game, you receive $1 for every number by which the sum of the dice exceeds 10. This means that rolling a sum of 12 gives you $2 and rolling a sum of 11 gives you $1, so the expected value of the first game is indeed 1/9 of $1. Menos

A un Trading Analyst Intern le preguntaron...29 de enero de 2014

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Each coin is independent; let p(H) denote the probability of the coin being heads and p(T) denote the probability of the coin being tails in the long run. To get a head, note that we must be at a tail, and transition with 0.5 probability. So, p(H) = 0.5 * p(T). To get a tail, note that we can (1) be at a tail and transition to tail with 0.5 probability, or (2) be at a head and transition to tail with 1 probability. So, p(T) = 0.5 * p(T) + p(H). Lastly, p(H) + p(T) = 1. Note that the first two equations give the same thing: p(T) = 2 p(H). Substituting into p(H) + p(T) = 1, we get p(H) = 1/3, p(T) = 2/3. So, over time, we'll end up with 1/3's of the coins as heads and 2/3's of the coins as tails. Menos

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It will eventually reach and remain at 1/3 heads and 2/3 tails.

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The answer above is good. Another way to phrase it is that at round n, the prob for a single coin to be H is p_H (n) = 1/2 p_T (n-1). Parenthesis indicate functional dependence. So we can say p_H(n) = 1/2 (1-p_H(n-1)). If we assume a stable solution exists at infinity, then it must be independent of n. So guess p_H(n) = const. Substitute and solve for const to get 1/3. Menos

A un Technology Developer and Business Analyst Intern le preguntaron...14 de febrero de 2013

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how do you convey a technical banking transaction to a street man? Can someone answer this?! Menos

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so what are the classes and methods required for a transaction?

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I think every one knew it.

A un Financial Analyst Intern le preguntaron...30 de marzo de 2011

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An insurance that you pay to hedge against price changes in the future. buy put to hedge against price going down. Menos

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Where you “put” options on high or low before a date

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A put option is a right but not obligation to sell the underlying asset at the strike price at maturity. One put option contains 100 shares of the underlying asset. A put option can be an effective tool for managing risk and uncertainty. Using a protective put one can hedge the potential downside risk of the portfolio, with only giving up a small portion of the upside potential, which is the premium paid for the option(s). This is known as the breakeven point. Menos

A un Business Analyst Intern le preguntaron...22 de febrero de 2021

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Hey can you help me with assumptions for these?

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Do justify your calculations including the assumptions.

A un Technology Analyst Summer Intern le preguntaron...16 de febrero de 2010

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Oval would be the super class and circle would be the sub class which extends oval. As every circle is an oval whereas every oval is not a circle Menos

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I dunno they're both pretty class. but circles would be my fave, like when ur on a roundabout and you keep going round with a big spliff in ur hand. it's sooo good Menos

A un Business Analyst Intern le preguntaron...6 de octubre de 2013

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Yeahhhh...

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do you have more details about set up of the 1st case?

A un Financial Analyst Intern le preguntaron...6 de septiembre de 2011

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Anybody have an answer for this?

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ROI, percent change, Futures

A un Financial Analyst Intern le preguntaron...2 de noviembre de 2011

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Cash Flow, ROI, Profit Margin,

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Percent change, Futures, ROI