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good luck to you as well! and thanks a lot for helping me out...Im sorry. Well. good uck. But my, p63 w11 is HARD
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good luck to you as well! and thanks a lot for helping me out...Im sorry. Well. good uck. But my, p63 w11 is HARD
yeah thats how we do it! thanks!Im not sure. but i think u can take probability at 63 to be 0.75 since it is upper quartile. u do know the mean.
P(63-mean/ S.D) =.75
try
yeah i tried that but couldnt get the answer..could you plz do it?Remember the fi thing i said before, use that formula
P( k-mean/standard devation <Z< 128-mean/s.d) = .7465
CNT GET UR WAYYY,,,,,,,,my stratergy was too find two Gs 2gether,,,,,,without emphaisizing exctly 2 meaning i ll have calculated both 2 and 3 and then subtract that from exactly 3....bt dount knw where's the problemWell, this was how I'd solved it before:
Simply 'choose' 2 Gs out of 3, and the remaining 7 letters can be completely random!
G G _ _ _ _ _ _ _...............[consider GG to be 'attached' together so that it can alternate in 2! ways]
n = 2! x 3c2 x 7!.................. [but don't forget that there are 3 recurring Es too!]
..............3!
...= 5040...... Q.E.D
P.S. If you're still uncertain, let me know. I'll work in it one more time.
E=summation easy calculate E(x) because its the same for both. E(x) = 245 +E(60)=4445
this question has been solved three to 4 times plz check from pages 170 onwardshelp???
thanks in advance
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_63.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_ms_63.pdf
question 3 part iii how to do?/ m unable to get
plz any one
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_63.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_ms_63.pdf
question 3 part iii how to do?/ m unable to get
plz any one
sorrY bt have askd fr the last part any ways thanx fr ur co-operation i gt ma answerSince probability of going to park is 0.6,hence not going is 0.4. The dog barks when goes to park = 0.6x 0.35 aand not going to park = 0.4x0.75.Sum it up =)
i know how to solve this i was just wondering when should we use this method and when do we just use the probability they've given?this question has been solved three to 4 times plz check from pages 170 onwards
i know how to solve this i was just wondering when should we use this method and when do we just use the probability they've given?
what do u mean x2?
ohh okthe probability given will be still using it.. and when they say "within", we hv to find the lower and upper boundary.
dude i did it my way first taking P(X>9.2) then adding the probability found to P(X<7.2) ul get the same answer
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