Mathematics: Post your doubts here!

[angelicsuccubus]

Where is this question? because the way u wrote it, looks unusual.

uh I donno how else to type it .. but here


and a=v(dv/dx) .. I need to find a.

 

[Khan_971]

yeah i know that much.. that leaves us with -a//root21 <Z < a/root21 = 0.5
what to do after that! i dont understand how ms got 0.75..

I mentioned somewhere a few minutes ago that this will lead to
2fi(a/root21)-1 = .5
fi(a/root21) = 1.5/2
fi(a/root 21) = .75

 

[Khan_971]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_63.pdf
q2 Khan_971

Already cleared before. check the previous pages.

 

[smartangel]

I mentioned somewhere a few minutes ago that this will lead to
2fi(a/root21)-1 = .5
fi(a/root21) = 1.5/2
fi(a/root 21) = .75

OH right. thanksss

 

[hassam]

Already cleared before. check the previous pages.

cnt find. sorry

 

[Khan_971]

uh I donno how else to type it .. but here
View attachment 10557

and a=v(dv/dx) .. I need to find a.

attached a file. see if that's what you need.

 

[Khan_971]

cnt find. sorry

Here. Comment #3264

 

[angelicsuccubus]

attached a file. see if that's what you need.

yes thankyou =]

 

[sunnyexo]

2^12 implies each coin in 2 ways. It wont matter if its heads or tails. but we do require a specific number

Thx~~~Finally I get it

 

[shan5674]

thnx.. but in the markscheme, the median is given as 19(11th value). the LQ is given as 10(mean of the 5th and 6th values) and the UQ is given as 24(the mean of the 16th and 17th value)...
WHY??

which paper is this? :S

 

[Khan_971]

View attachment 10552

how about this? I'm so confused when it comes to probability distribution T.T

Ok so the table will be like this, 5 girls and 3 boys , and a team of 4. so at minimum we can have 3 guys and 1 girl.
B- boy. G- girl
BBBG = 5C1
BBGG= 3C2*5C2
BGGG= 3C1*5C3
GGGG= 5C4
Add them up to get total possibilities. Divide each value above by this Total possibilities to get its probability. Hope this is it.
X..........1...2....3....4....
P(X)....

 

[Khan_971]

Eight cards are selected with replacement from a standard pack of playing cards
with 12 pic cards, 20 odd cards
and 20even cards.
a) how many different sequences of cards are possible?
b) how many of the sequencees in part (a) will contain three picture cards, three odd cars and two even cards??

Part a is either 52C8 or 52P8. If each picture is different then its P.
b) 12C3 * 20C3 * 20C2 OR P as im not sure about the pictures being different.

 

[Wanzi21]

Ok so the table will be like this, 5 girls and 3 boys , and a team of 4. so at minimum we can have 3 guys and 1 girl.
B- boy. G- girl
BBBG = 5C1
BBGG= 3C2*5C2
BGGG= 3C1*5C3
GGGG= 5C4
Add them up to get total possibilities. Divide each value above by this Total possibilities to get its probability. Hope this is it.
X..........1...2....3....4....
P(X)....

oh I got it.. i took time until I realised that 3C3 is actually 1(for boy).. I appreciate ur help. thanks

 

[Khan_971]

Why can't we use 2^12/7!5!.....Sorry~~

BTW 20C7 IS 12!/7!5!

 

[Khan_971]

yes thankyou =]

But where was this from? u just had to multiply what i gave with V right?

 

[Wanzi21]

How about this?? is it possible to construct tree diagram?? I think it'll be big :S

 

[Khan_971]

someone please explain how to calculate the mean and standard deviation using a histogram? i cant understand the ms..:'(

Didnt I clear this?

 

[angelicsuccubus]

But where was this from? u just had to multiply what i gave with V right?

I donno .. I think Im still confused =[ but it helped that you tried.

And its from A2 mechanics (which Ive never studied, long story)... my paper is tomorrow and there's no one here who does A2 mech so......

 

[mominzahid]

Didnt I clear this?

Yupp.. you didd... i was using the same method but the answer was wrong somehow. :/
lemme try it agn and get back to u.

 

[reina81]

The 94% probability is the area in the MIDDLE, which means there is a "tail" on either side, of 3% probability. When you look for z value you need to count from the left side, or negative infinity. So here the probability is 0.94 + 0.03 = 0.97
View attachment 10051

how should i know when to use this method and when to simply use the probability given?