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Say 2 indentical bags contain different numbers of and
counters.
A bag is chosen AT RANDOM and a single counter removed from it without looking.
bag A : 2 3
Bag B : 3 4
If the counter was what was the probability that the bag was Bag A??
I got it it's 2499! 10000! = [10000/3125]+[10000/625]+[10000/125]+[10000/25]+[10000/5]=a number with 2499 0s
Thank you
I think i get it, thanks again for your attention and time, and i presume this will work on fractorial? what is this formula called?
Hey!
I searched floor funtion, so basically it just "round it down to the whole number" as in my words
Is that somewhat correct?
Thanks for the help, but why 125, 25, and 5 as denominators?
i am guessing about 17?
So i am encounter with the question of:
How many 0s are there at the end of the # that is the product of the first 150 POSITIVE intergers (i.e. 1,2,3,.....150)?
anyone know?
or have a method that probably work?
'Cause I don't want to calculate for a week or so
May I ask how did you got the answer, we probably used different methods?
Yep, you are correct, Thank you again!!
thank u!!
i will double check but i think i m on the right track
Here is the question:
what is the difference for the equation of "square of 1 - square of 2 + square of 3 - square of 4........+ square of 99 - square of 100"?
I got -5500, is that correct?
Please give any assitance!
Thank you for helping!!!:D
Can anybody tell me more about the Deep Blue ? like how does it work, i heard that's the most intelligent chess computer
Also I have win quite a couple times where I played against other people and won by black.
Well I think the computer is smarter than the most of us here. The software programmers programmed the chess software to try their best and win us
I agree with Xyz, once the infinite decimal (0.9999) gets multiplied, it's not infinite any more, so
0.9999 X 10= 9.999 ( 1 "9" less)≠ 9.9999
9.999-0.9999=8.9991
8.9991÷9= 0.9999
P.S. This works for any amount of nines