Mathematics: Post your doubts here!

smzimran

smzimran

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hassam said:
its w11 qp 62
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alphabravocharlie said:
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 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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hassam

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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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mukki

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hassam said:
well this is so damn cnfusing
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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
 
usmiunique

usmiunique

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Diamona151294 said:
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
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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!
 
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sweetyrox

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usmiunique said:
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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how do u get 0.06??
 
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hassam

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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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Diamona151294

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usmiunique said:
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!
Click to expand...
oh finally! THANKU!
 
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