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u did it?thanks it helped
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u did it?thanks it helped
its not tough.. its just that we had never seen this kind of questions ever beforei am soo glad i am not giving p3 till next year. s1 is killing me and i heard p3 is even tougher
ummmm.........the work done isn't equal to zerohttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_41.pdf
How to do question 5 (ii) ?
Is the work done = 0 ? if yes? then why ?
its okay i got it
u did a bad thingits okay i got it
no need to do it
here are two hints that id give and i want you to do it urselfcan any one help me with mj 2010 varient 13 ... p1 ... q 8(ii) ??????
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_13.pdf
LOL xDu did a bad thing
part of itu did it?
I managed to get both M and A... but not C. Could anyone explain how to obtain the answer for C?
thanks ... just by readind ur first sentence i remembered the concepthere are two hints that id give and i want you to do it urself
a) use the M1*M2 =-1 use this to take out the gradient of BD take midpoint
u have a grad u have coordinates make eq of line of BD
then to find b put x=0 as it is on the y axis
thenu have coords of B and M the midpoint
u can say that (coords of b + coords of c )/2 willgive you the coords of M the midpoint
here u can put midpoint and B to get the coords of C
check the guy who replyed to me ... i was having the same query !I'm reposting my query because it seems no one saw it and it got buried two pages back LOL.
I'm reposting this.. anyone?http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_61.pdf
Qn no. 7 - whole question.
Please help me solve the question above.. I'm totally confused!
Okay once you have the co ordinates of A and M, you have done the toughest part.I'm reposting my query because it seems no one saw it and it got buried two pages back LOL.
I'm reposting this.. anyone?
wow such long answersQ7 i) Atleast one 3 .. means that it is saying P(X≥1) or P(X>0) n = 9
Where X is the number of 3s.
So for it to be zero 3s the probability will be (5/6) and 0 3s at all 9 throws would be P(X=0)=(5/6)^9
Now it asked at least 1 3 .. meaning there are no zeros .. so 1 - P(X=0) = 1- (5/6)^9 = 0.806 Answer
ii) When n fair dice are thrown, the probability of getting at least one 3 is greater than 0.9.
Find the smallest possible value of n.
This is a real tricky question .. we know that for one dice which is thrown 9 times the probability of getting atleast one 3 is 1-(5/6)^9
so for n number of throws the probability will be 1- (5/6)^n
Now the question says .. the probability of n number of throws is 0.9 so
0.9 = 1-(5/6)^n
-0.1=-(5/6)^n
(5/6)^n=0.1
Now To get the value of n we need to use ln on both sides (a little something from my add maths).. using ln would get the power to the bottom
n* ln (5/6)= ln(0.1)
n = ln (0.1) / ln(5/6)
n = 12.62
so atleast 13 throws need to be made for the probability to be 13.
b)
This was a real tricky question .. I remember my friend asked me solve it..
5 Green Balls 3 Yellow Balls .. Player wins when a yellow ball is drawn.. Julie draws the first ball .. Probability of Ronnie Winning..
The ways in which ronnie can win are as follows
J'R , J'R'J'R , J'R'J'R'J'R (then the green balls end so yellow ball would come eventually..) To summarize these details in terms of balls
GY , GGGY, GGGGGY
Total = 9, 5 green and 3 yellow.. so probabilities become
P(GY) = (5/8 * 3/7) = 15/56
P(GGGY) = (5/8 * 4/7 * 3/6 * 3/5) = 3/28
P(GGGGGY) = (5/8 * 4/7 * 3/6 * 2/5 * 1/4 * 3/3) = 1 / 56
Total P = 15/56 + 3/28 + 1/56 = 11/28 Answer
Hope that helps...
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