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1) are our employees safe and sound, 2) are there any major customer issues or service problems, 3) What is new happening in the world and our industry today Less
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As CEO I don't need to micromanage. I'd check the news, weather, and email.
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1: How happy are your employees. 2: How happy are you customers. 3: Where are we wasting time and money and how can it be improved. Less
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Interview Candidate and Anonymous are right. This is also known as the Monte Hall problem Less
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Your choice splits the doors in two sets. Set A contains the door you selected, and the probability that is a prize behind this door is 1/3. The set B contains all remaining doors, and the probability that the winning door is somewhere in there is 2/3. By removing one door, which all have the success probability of zero because there's coal behind them, from set B, only one door remains in B, but the overall probability for success in set B is still 2/3. Therefore you must switch. Less
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Interview candidate is right. You got 1/3 chance that prize is behind door #1 and you lose if you switch. And you got 2/3 chance that prize is behind either door #2 or #3. Since the host will always eliminate the wrong one. 2/3 chance will be allocated on the left one. Less
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Using two digits in "base 6" you can count up to 35. Typically base 6 would use digits 0 to 5, but in this case you can adjust and use 1 through 6. Less
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How about... Dice 1: each pip equals 1 day: 1 pip =1 day, 2 pips = 2 days, etc. Dice 2 each pip equals 5 days: 1 pip = 5 days, 2 pips = 10 days 3 pips = 15 days etc. Die 2 is always to the left of Die 1. For days less than or equal to 5 use 1 die. When counting a day that is a multiple of 5 (days 5,10,15 etc.) count day by putting Die to 5 pips and Die 2 to the remaining number of days. Less
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@Rikesh: How will you indicate 17, 18, 19, 20, 27, 28, 29, 30. Here is what i think... D D 6*0+1=1 6*0+2=2 6*0+3=3 .. 6*0+6=6 6*1+1=7 6*1+2=8 .. 6*1+6=12 6*2+1=13 6*2+2=14 .. 6*2+6=18 6*3+1=19 6*3+2=20 Less
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It depends on the shape of the cake.
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3 First 0 degree Second 90 degree No I have 4 have equal parts. Third From the Centre height wise. (Z-axis) Now I have 8 equal parts. Less
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With 4 straight cuts through 0, 45, 90, and 135 degrees You can cut the cake into 8 equal pieces Less
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The address of a company I was interested in applying to; needed to determine if the commute was reasonable. It was, so I applied. Less
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The company I'm applying for. I wanted to find out their mission statement and general work activities to find out if it was a good fit for me. Less
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A Banana Split. I actually leered at it, more than goggled it...
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Just to clarify: the original reviewer's answer - C(10,3) - is indeed incorrect. More specifically, C(10,3) represents a number of ways to pick 3 DISTINCT desserts. The problem statement does not suggest that the desserts should be unique; moreover, it's very natural to order three identical desserts! I believe the correct answer to be a sum of C(10,3) /* number of ways to pick 3 distinct desserts */ + 10 * 9 /*number of ways to pick 2 matching and 1 distinct dessert */ + 10 /* number of ways to pick 3 matching desserts */ = 220 Less
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C (10, 3) = 120 - This is very simple
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If order matters (it would to me if I were eating them), and repetition is allowed, the answer would be 1,000 since you can choose any of the 10 for the first, second and third dessert resulting in 10 X 10 X 10 possible dessert sequences. Less
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Yes, ready to be placed in all the branches needed
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Yes I could. I could be ready to placed in all branches
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Yes, I'm ready
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Your talking about worst case only, but guess something to think about. Heard rumors that hbc might try to buy us but since their like macys, not high end fashion like us, they would need us to run Neimans. Didn't now they owned sfa. Want to be a buyer here but maybe should think about back up plan. Less
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Group project is about: a) demonstrating how you work as part of a team (sit back and let others do the work (not good), micro-manage and take over group to try to win (not good), motivate people, treat others with respect, TEAMWORK ATTITUDE (good). b) showing how you approach a problem, (critical) thinking skills. Attitude, ability to work well with others & apply yourself is what they want to see. Don't have to "win". Less
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No additional test on day 2. Focus is group project. Ability to work well with others, teamwork skills more important than coming in first. Less
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In fact the number is much larger than that. The average number of sparks per vehicle (in CA) is probably closer to 5.5. Four-cylinder vehicles are the norm in Europe, but in the US is common to see a higher percentage of V6 and V8 engines. For every gas (not diesel, as noted previously) vehicle in CA, working or not, you could estimate ~5.5 sparks. In addition, you have to account for all stored (i.e. not necessarily installed in vehicles) sparks. So if there are, say, 20k car shops in CA and each of them holds 100 sparks in stock, that's another 2M sparks. How about motorbikes? Probably 2.5 sparks on average? Any spark factories in CA? Finished product would count too. How many unsold/retired vehicles? I'd say the final number is well north of 100M. Less
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Cars, lawn mowers, edgers, and many other engines. Only going by cars is a huge underestimate. Less
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This is a fermi question. assuming there is one sparkplug per vehicle, and the number of sparkplugs used in other machinary pales in comparision to the number of vehicles in california, then you just need to get the ball park figure for the number of vehicles.. here's my shot at it.. CA population is roughly 40 million. I would estimate roughly 1 car per family.. so lets say about 15-20 million cars in CA (taking into account children). so ball park figure of 20 million. lets add 10 million more for commercial vehicles, motorcycles,trailers etc.. . I would estimate a total of 30 million spark plugs in the state of California. Less
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Use a decision tree.
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^That method won't work because it's possible that there are those on the email list that would respond without an email, so there would be some overlap, which means you can't just add the two together. Easiest way to approach it would be to draw some sort of Venn Diagram Less
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this is likely an applied probability question: My answer is to apply total probability formula here. Less