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I dont think so!Can't we do it like: No.of arrangements - No G together - All G together
[9!/(3!*3!)] - [(7C3*6!)/3!] - 7!/3!
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I dont think so!Can't we do it like: No.of arrangements - No G together - All G together
[9!/(3!*3!)] - [(7C3*6!)/3!] - 7!/3!
Why?I dont think so!
its w11 qp 62
I got it hassam:Can't we do it like: No.of arrangements - No G together - All G together
[9!/(3!*3!)] - [(7C3*6!)/3!] - 7!/3!
The answer is coming negative!Why?
i get 5040The answer is coming negative!
* If i did the calculation right!
Well then my calculation must have been wrong but i cant seem to get the calculations right!i get 5040
its very easy bro just think of it like thiswell this is so damn cnfusing
i had the same question!! this one's bugging me since yesterdayum in this question, im finding the z value for 0.94, but the answer's wrong, somehow they are giving the z value for 0.97..... can sm1 pls explain whats going on here?View attachment 10654View attachment 10655
i had the same question!! this one's bugging me since yesterday
The probability of a letter being >12g above the mean is 0.94+(0.06/2)=0.94+0.03=0.97um in this question, im finding the z value for 0.94, but the answer's wrong, somehow they are giving the z value for 0.97..... can sm1 pls explain whats going on here?View attachment 10654View attachment 10655
how do u get 0.06??The probability of a letter being >12g above the mean is 0.94+(0.06/2)=0.94+0.03=0.97
What u are taking is 0.94 which is the probablity of the letters being within 12g of the mean but what u have to take is the probabality of the letters being MORE than 12g above the mean!
oh finally! THANKU!The probability of a letter being >12g above the mean is 0.94+(0.06/2)=0.94+0.03=0.97
What u are taking is 0.94 which is the probablity of the letters being within 12g of the mean but what u have to take is the probabality of the letters being MORE than 12g above the mean!
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