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Mathematics: Post your doubts here!

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its w11 qp 62
Can't we do it like: No.of arrangements - No G together - All G together
[9!/(3!*3!)] - [(7C3*6!)/3!] - 7!/3!
I got it hassam:
They are first taking 8! / 3! to find arrangements in which two Gs are grouped together and the third is placed randomly
Then they are subtracting 2 occasions when this third G may come besides the group of Gs
First possibility :
G (GG) ............
(GG) G ............
That is why the 2 is present,
So it is (8! / 3!) - 2 * (7! / 3!) [grouping the third G with the rest two in subtraction]
 
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"Then they are subtracting 2 occasions when this third G may come besides the group of Gs
First possibility :
G (GG) ............
(GG) G ............"
now this bit i cant understand...@smzimran wat i was thinking that.....i got 840 arrangements in which the 3Gs were 2gether and 6720 arrangements in which 2Gs and 3Gs were 2 gether......so just thought of subtracting them
 
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well this is so damn cnfusing
its very easy bro just think of it like this
32 teams play the first match but 16 get knocked out so hence the teams who have played only A SINGLE MATCH IS 16. then these 16 teams who survived play a second match of which 8 got knocked out hence 8 teams have played exactly two match. the remaining 8 teams play a third match in which 4 of them get knocked out hence 4 teams have played exactly three matches. Then out of the remaining 4 teams 2 get knocked out in the 4th match hence 2 teams have played exactly 4 matches. For the 5th match there are only two teams remaining hence only two teams have played exactly 5 matches
 
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um 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
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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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!
how do u get 0.06??
 
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well this is wat i read smwhere on yahoo answers The class boundaries are found by taking the average of the highest value of one class and the lowest value of the next class
 
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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!
 
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