Forums / Miscellaneous Discussions / Math Tricks/Problems
Math Tricks/Problems | ||||
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1) 6210001000, the last digit would have to be a one binh 2) 1234759680, I had a good guess to start with.. psh...got them driving down I 65 ;), whilst eating my subway sandwich, and saving the world.... (to the rest of you reading this I'm not that arrogant, just brotherly taunting) | ||||
take 240 (look familiar argheyass) and switch the first 2 digits (sound familiar) ;) | ||||
1. Find a formula to add all integers from 5 to X where X >5. | ||||
(5+X)*(X-5+1)/2 | ||||
What number shows up most often when you roll 10 dice? | ||||
Seloc 10? :) Right sun. | ||||
10/3 hours (6000000/(20000-3000t)) dt. ~ 1386.3 miles give me a couple for a good one :) | ||||
argheyass cheats! | ||||
woman and seloc do you mean the sum of two dice? if its just each individual dice rolled I dont see how you can determine which one, as the probability is = for all 6.. maybe I'm just stupid tho, wouldn't've been the first time, can you PM the answer | ||||
Total of the 10 dice. | ||||
just a guess off the top of my head, 35. | ||||
oh now i get Seloc question. I thought it was a trick question since I didn't know dice is a plural term... "and seloc do you mean the sum of two dice? if its just each individual dice rolled I dont see how you can determine which one, as the probability is = for all 6.." Archias, the number of possible outcomes are 6^10. The probability of getting a 10 as a sum of all dice is the lowest since all dice need to roll on a 1 to get the sum of 10. So the probability of getting a sum of 10 is 1/(6^10) which is very small chance. now let say the sum is 11. Now one dice out of the 10 dice need to be a 2 while all other being a 1. This mean the probability of getting a sum of 11 is 10/(6^10). The question is to find out what sum have the most probability of getting it. I hope you understand more of this question now :) | ||||
Considering all have the same probability, | ||||
well, adding "6" 10 times is 60, and addnig "1" 10 times is 10 (the two sum with the lowest chance) the average is 70/2 = 35 which was my rough estimate. it serves the purpose of this question, but I did not calculate its actual probability ;o...
edit: oops i was editing some typo here, but i clicked on "REPORT" button on Lord Carnage's post by accident..... sowwiee..... | ||||
I believe sun is right. the middle number is the one that will get the most probability of getting it. 10 dice of 1-6 means minimum is 10 and maximum is 60. (10+60)/2 = 35. darn sun posted before me. :) | ||||
unless the dice are rigged. | ||||
The average sum of 10 n-sided dice is (n+1)*10/2 => 5n + 5. | ||||
Archias ...The question is to find out what sum have the most probability of getting it. I hope you understand more of this question now :) | ||||
Mr. Jenos | ||||
There's no such thing as a 0-sided or 1-sided dice... but I thought that was too trivial to mention. | ||||
and I also thought it was trivial to mention that a die with n-sides can only exist if n is a positive integer greater than one. | ||||
I haven't yet seen a two sided dice, or a three sided. Would you like to explain how these would work? | ||||
3 sided dice:
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Coins have more than one side 0.o And those dice are unfair! | ||||
0 sided dice is an imaginary nonexistent entity! | ||||
or a sphere ;) | ||||
i think a sphere is 1 sided ;o | ||||
yes its 35 probably sun | ||||
Actually Sun a sphere is a infinite sided dice if you want to look at it that way. But I'd love for you to explain how a sphere can be just 1 sides since it can stop anywhere. If you throw a ball will you be able to say which side it will be up? | ||||
it has one surface ;p it will always be that one side ;p | ||||
"it has one surface ;p it will always be that one side ;p" | ||||
but the round one has no edge on the surface :P (of course assuming its perfectly round!) | ||||
I think you mistake side with surface. They aren't the same, a ball has only one exterior surface but has an infinite number of sides. | ||||
2x^2-3x-4=0, find x. | ||||
x= 3+/- sqrt(9-4*2*4) ALL OVER 4 ... imaginary solution ... | ||||
It's not imaginary.....just follow through with your working... | ||||
a mistake on my part. 9-4*2*4 should be + instead of -. | ||||
The solution is 3+/- sqrt(41) like Sun said and please if you want to post a math problem post an interesting one not these 8th grade run of the mill ones that anybody can do, they should take a bit of thinking. | ||||
How'ld you guess.... | ||||
Only somebody that has these as homework would think they are by any means hard to do. | ||||
Can you do this one for me? (preferably before tommorow) [4^1/3(cos (7*pi/36) + isin (7*pi/36))]^12 i have to express the answer in the form a+bi too :( | ||||
I just learned that stuff last week.........and yet I can't remember it! I blame this on you!! Ugh, I shouldn't be staying up this late at all......................... | ||||
"Can you do this one for me? (preferably before tommorow)" | ||||
i forgot those classical algebra stuff, learned them in first year. i remember there was exponential representation involved somewhere... | ||||
The entire thing is to remember you need to remove the ^12 by multiplying it with the cos and sin arguments. | ||||
ya i vaguely remember some fundamental formula (i think we even proved it back then..) the exponent moving into the trig argument and stuff. but exatly what it was, i cant remember ;)
but those were easy if u know your stuffs.... | ||||
The forumula is: (cos(a*pi)+isin(a*pi))^b = cos(a*b*pi)+isin(a*b*pi). | ||||
:) that refreshed my memory :D | ||||
cant forget about the 4^1/3, raise it to the 12th | ||||
"cant forget about the 4^1/3, raise it to the 12th" | ||||
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