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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 partSince 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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