Mathematics: Post your doubts here!

[Yousif Mukkhtar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_61.pdf
please help me with(question 6 ii )
and for question 7 (a ii ) 7! x 7! (is for arranging the friends and the partners ) but we are multiplying by 2 because either the friends could be first or the partners ? am i right ?? i want to confirm .

Q6ii) Firs you need to find the probability of lengths more than one standard deviation from mean. To get this idea, imagine the STANDARD NORMAL DISTRIBUTION curve in which 0 is the mean. When it says more than one standard deviation from mean, it means more than +1 and -1 from the mean 0. See the attached image for better clarification.

When you get this idea then:
P(more than 1 standard deviation)= P(Z greater than 1) or P(Z less than -1)
=P(Z greater than 1) x 2 (as both sides are symmetrical)
=(1- fi(1))x2
=(1-0.8413)x2
=0.3174

Therefore number of feathers= 0.3174 x1000= 317 feathers

 

[Yousif Mukkhtar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_61.pdf
please help me with(question 6 ii )
and for question 7 (a ii ) 7! x 7! (is for arranging the friends and the partners ) but we are multiplying by 2 because either the friends could be first or the partners ? am i right ?? i want to confirm .

For q7) Yes you have got the idea.

 

[Yousif Mukkhtar]

Can anyone explain why they gave Z= 1.882 and p=0.97? Q5) I)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf

Can someone explain this?

 

[Zenia ZZ]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_61.pdf
( Q7 - c part ) I don't get it at all . brain not working ! please help me
Jiyad Ahsan

 

[Zenia ZZ]

part 3, sorry it took so much time
oh and i forgot to mention ( though its obvious)
the denominators cancel out a part of the numerators so only the 8P8 's are left

Honestly you are so helpful . Thank a tonne for the help you are providing me May you pass with bright colours in all your exams ! Good Luck !

 

[Anika Raisa]

Can anyone explain why they gave Z= 1.882 and p=0.97? Q5) I)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf

part (i)
12P8
=19958400

part (ii)
so frances and mary do not sit together,
just find out the number of ways in which they DO sit together and subtract from the total
if they sit together then 11P7 x 2P2 - - - - (11P7 because since they sit together you count them as one unit, 2P2 because they can arrange themselves in their group, sit left or right)
11P7 x 2P2
=3326400
this is when they do sit together, so subtracting from total ( part (i))
19958400 - 3326400
=16632000

part (iii)
i'll just upload the working for it in a bit
"Jiyad Ahsan, post: 564620, member: 31829"]part 3, sorry it took so much time
oh and i forgot to mention ( though its obvious)
the denominators cancel out a part of the numerators so only the 8P8 's are left

All credit to JiyadAhsan ....

 

[19islandprincess96]

Somebody help me with 3b, please.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_6.pdf

 

[Yousif Mukkhtar]

All credit to JiyadAhsan ....

Thanks. Q1 Is it okay to write the probabilities in simplified form? For eg P(0)= 7/24 instead of 210/720?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_61.pdf

 

[syed1995]

Thanks. Q1 Is it okay to write the probabilities in simplified form? For eg P(0)= 7/34 instead of 210/720?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_61.pdf

YOU NEED TO WRITE in standard form!

 

[Yousif Mukkhtar]

YOU NEED TO WRITE in standard form![/quote)

So 7/24 is accepted even though it is not written in the marking scheme?

 

[syed1995]

So 7/24 is accepted even though it is not written in the marking scheme?

Yes.

 

[syed1995]

So 7/24 is accepted even though it is not written in the marking scheme?

Yes.

 

[Rutzaba]

is this right?
jub < hota hai to direct table read krte hain

​

jub ≤ hota hai tou 1 se minus krte hain

​

jub > hota hai tou 1 se minus krte hain

​

jub ≥ hota hai tou direct lete hain

​

agar jo hum nikal rhe hain wo negative ho

​

tou 1 mai se minus krskte hain

 

[19islandprincess96]

is this right?
jub < hota hai to direct table read krte hain

​

jub ≤ hota hai tou 1 se minus krte hain

​

jub > hota hai tou 1 se minus krte hain

​

jub ≥ hota hai tou direct lete hain

​

agar jo hum nikal rhe hain wo negative ho

​

tou 1 mai se minus krskte hain

P(z<a) = direct table
P(z>a) = 1- table value
P(z<-a) = 1 - table value
P(z>-a) = direct table value
P(a<z<b) = value of (b) from table - value of (a) from table
P(-a<z<b) = [value of (b) from table + value of (a) from table] -1
P(-a<z<-b) = Value of (a) from table - Value of (b) from table

 

[19islandprincess96]

Somebody help me with 3b, please.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_6.pdf

Please?

 

[syed1995]

So 7/24 is accepted even though it is not written in the marking scheme?

Yes.

 

[aniketjain]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_62.pdf
qoestion no. 1 ..... how do u calculate s.d (standard deviation )

 

[syed1995]

can someone please solve 4 i http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_62.pdf

Didn't you give the papers a long time ago ?

u=4s
P(X>5)=0.15
(z>(5-4s/s))=0.15
1-Phi(5-4s/s) = 0.15
Phi(5-4s/s)=0.85
5-4s/s=invPhi(0.85)

now derive the value of s.. and mean = 4*s

 

[Rutzaba]

Didn't you give the papers a long time ago ?

u=4s
P(X>5)=0.15
(z>(5-4s/s))=0.15
1-Phi(5-4s/s) = 0.15
Phi(5-4s/s)=0.85
5-4s/s=invPhi(0.85)

now derive the value of s.. and mean = 4*s

yeah i was helping sum one and i got it and i del the post xD

 

[Rutzaba]

Didn't you give the papers a long time ago ?

u=4s
P(X>5)=0.15
(z>(5-4s/s))=0.15
1-Phi(5-4s/s) = 0.15
Phi(5-4s/s)=0.85
5-4s/s=invPhi(0.85)

now derive the value of s.. and mean = 4*s

listen can you do 5 part three of the same year?please