Mathematics: Post your doubts here!

[Rutzaba]

can 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

here 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

 

[applepie1996]

u did a bad thing

LOL xD
why may i ask ??

 

[unseen95]

u did it?

part of it

 

[PANDA-]

I managed to get both M and A... but not C. Could anyone explain how to obtain the answer for C?

I'm reposting my query because it seems no one saw it and it got buried two pages back LOL.

 

[ahmed abdulla]

here 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

thanks ... just by readind ur first sentence i remembered the concept
i think ur a magician

 

[ahmed abdulla]

I'm reposting my query because it seems no one saw it and it got buried two pages back LOL.

check the guy who replyed to me ... i was having the same query !

 

[unseen95]

please help me with 10(b)(i) and (ii). Link to question http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_33.pdf

 

[salvatore]

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!

I'm reposting this.. anyone?

 

[Raiyan3]

I'm reposting my query because it seems no one saw it and it got buried two pages back LOL.

Okay once you have the co ordinates of A and M, you have done the toughest part.
Since you havent given the co ordinates of A and M, im gonna take the numbers that come firrst in my head, A and M in this situation be (4,0) and (2,4)
Okay now Let C be (p,q)
Now we use Mid point formulas.
(4+p/2 = 2), (0+q = 4)
Find p and q.
we used the mid point formulas because M is the mid point of A and C. We have the points of A and M. Fill in the missing spaces.
yet again mid point = (x1+x2/2 , y1+y2/2) Use the co ordinates of A(and let C be p,q), which equals to co ordinates of M. Hope it helps!

 

[syed1995]

I'm reposting this.. anyone?

Q7 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 0.9.

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...

 

[Rutzaba]

Q7 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...

wow such long answers

 

[Rutzaba]

LOL xD
why may i ask ??

cux i solved it -_-
but it really duxnt matter

 

[Rutzaba]

thanks ... just by readind ur first sentence i remembered the concept
i think ur a magician

magic be haram
wud appreciate duas cuming this way ^_^

 

[syed1995]

wow such long answers

haha yeah ... Took quite some time.. there were 3 parts and all 3 were long ones Akhir 14/50 marks khali isshi part kay thay

 

[syed1995]

magic be haram
wud appreciate duas cuming this way ^_^

Which maths papers you giving this year? P3 and S1?

 

[salvatore]

Q7 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 0.9.

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...

Thanks a lot for your help bruv.. much appreciated.
I don't understand a few things though:

For the first part, why can't I solve it this way: P(X=1) = (1/6) => (1/6)^9? (Since only one 3 out of 6 => 1/6)

For part (ii), is there any other way apart from log to solve for the value of n? I'm not really familiar with logarithm..

Finally, I totally don't understand how you solved part(b).. how did you get GY , GGGY, GGGGGY? Please help me understand this

I'm sorry if I'm bothering you with all these questions, but I really need your help.. I suck at statistics!

 

[Rutzaba]

Which maths papers you giving this year? P3 and S1?

im in uni bro

 

[syed1995]

im in uni bro

Nice! Which University?

Thanks a lot for your help bruv.. much appreciated.
I don't understand a few things though:

For the first part, why can't I solve it this way: P(X=1) = (1/6) => (1/6)^9? (Since only one 3 out of 6 => 1/6)

For part (ii), is there any other way apart from log to solve for the value of n? I'm not really familiar with logarithm..

Finally, I totally don't understand how you solved part(b).. how did you get GY , GGGY, GGGGGY? Please help me understand this

I'm sorry if I'm bothering you with all these questions, but I really need your help.. I suck at statistics!

First Part .. Well it is saying atleast one 3 .. not only one three .. so we need the probability of P(X≥1) .. it can be 1 three or 2 three or 3 three or even nine 3 it just can't be zero 3's... You need to solve it for X=1,2,3,4,5,6,7,8,9 and then add them all to get the correct answer .. or do what I did.

Second Part .. no other way that i know off .. I usually use my calculator to do these equations .. like i have a fx 991ES .. just do this

(1-(5/6)^x) then [ALPHA] the buttom below shift (It's SOLVE/CALC for me) then type the 0.9 and hit SOLVE [Shift and then press CALC]

so it becomes (1-(5/6)^x) = 0.9 then SOLVE .. It takes a little while .. but it gets the correct value of X.

Honestly there is no other way than log to resolve powers .. if there is then I don't know of the method.

Part b ..

Well it said that Julie was the first to play .. Yellow meant that the person won and if green was drawn another ball was drawn .. and we needed the probability of ronnie winning.

that is only possible is Julie lost every time (Got a green ball) and then Ronnie got a Yellow one (and won and game stops).. so GY .. but since there are 5 green balls .. until those end the game can continue.. so in the next one .. J gets a G and then R gets a G and then again J gets G and then R gets a Y to win and game stops.. and so on until all 5 green balls get used.. This part is tricky to understand .. read the question 2-3 times and then see my working .. you will hopefully understand it.

 

[salvatore]

Nice! Which University?



First Part .. Well it is saying atleast one 3 .. not only one three .. so we need the probability of P(X≥1) .. it can be 1 three or 2 three or 3 three or even nine 3 it just can't be zero 3's... You need to solve it for X=1,2,3,4,5,6,7,8,9 and then add them all to get the correct answer .. or do what I did.

Second Part .. no other way that i know off .. I usually use my calculator to do these equations .. like i have a fx 991ES .. just do this

(1-(5/6)^x) then [ALPHA] the buttom below shift (It's SOLVE/CALC for me) then type the 0.9 and hit SOLVE [Shift and then press CALC]

so it becomes (1-(5/6)^x) = 0.9 then SOLVE .. It takes a little while .. but it gets the correct value of X.

Honestly there is no other way than log to resolve powers .. if there is then I don't know of the method.

Part b ..

Well it said that Julie was the first to play .. Yellow meant that the person won and if green was drawn another ball was drawn .. and we needed the probability of ronnie winning.

that is only possible is Julie lost every time (Got a green ball) and then Ronnie got a Yellow one (and won and game stops).. so GY .. but since there are 5 green balls .. until those end the game can continue.. so in the next one .. J gets a G and then R gets a G and then again J gets G and then R gets a Y to win and game stops.. and so on until all 5 green balls get used.. This part is tricky to understand .. read the question 2-3 times and then see my working .. you will hopefully understand it.

I'm getting the concept now..

Thanks a lot for giving your time man

 

[Sanis]

I need urgent help please q1 how did we get a variance of 96 ??!!!
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s05_qp_1.pdf