Mathematics: Post your doubts here!

[Bushrafareed]

five particular boys and five particular girls are selected and placed in mixed pairs for tennis. Find the total number of different mixed pairs which can be made using these 10 children? Can anyone solve this? . Ans is 120

 

[hm12]

nov 2011 p61 q.5 part (i)

it says letters are within 12g of mean i.e 20 g so you basically have two values 1) 20+12=32 (2)20-12=8 now you have to find
P(8<X<32)

i have done the tricky part hope you can do the rest

 

[AdeelRox]

five particular boys and five particular girls are selected and placed in mixed pairs for tennis. Find the total number of different mixed pairs which can be made using these 10 children? Can anyone solve this? . Ans is 120

its simple.
with the 1st girl then can 5 be five boys.
with the second one there can be 4 boys.
with the 3rd girl there will be 3 remaining boys.
with the 4th girl there can be 2 remining boys.
and with the last girl there will be the last boy remaining.
so, its gona be 5 x 4 x 3 x 2 x 1=120.
i hope this will help yeh dude

 

[AdeelRox]

it says letters are within 12g of mean i.e 20 g so you basically have two values 1) 20+12=32 (2)20-12=8 now you have to find
P(8<X<32)

i have done the tricky part hope you can do the rest

dude i also did that part this aint triky i found it dificult that what to do next

 

[Bushrafareed]

can you tell me what's the logic behind it? I didn't get what the question is actually asking

its simple.
with the 1st girl then can 5 be five boys.
with the second one there can be 4 boys.
with the 3rd girl there will be 3 remaining boys.
with the 4th girl there can be 2 remining boys.
and with the last girl there will be the last boy remaining.
so, its gona be 5 x 4 x 3 x 2 x 1=120.
i hope this will help yeh dude

 

[hm12]

dude i also did that part this aint triky i found it dificult that what to do next

this method is confusing leave it . try this
0.94 is the probability between the two values if we subtract it from 1 and divide by two i.e 1-0.94/2=0.03 we get the probability of one small area so P(X<32)=1-0.03=0.97

 

[Bushrafareed]

oh i got it . Thanks for the help, i wasted around 30 min solving this question..

its simple.
with the 1st girl then can 5 be five boys.
with the second one there can be 4 boys.
with the 3rd girl there will be 3 remaining boys.
with the 4th girl there can be 2 remining boys.
and with the last girl there will be the last boy remaining.
so, its gona be 5 x 4 x 3 x 2 x 1=120.
i hope this will help yeh dude

 

[hm12]

i cant explain more than this hope you get it

 

[AdeelRox]

can you tell me what's the logic behind it? I didn't get what the question is actually asking

dude its very easy.its like fill in the blanks.
suppose in the first blank u have one of five letters to fill.
in the next 1 u will have one of 4.
in the next one u will have one of 3 to fill
then in the next blank u will have one of 2 left.
then in the last blank u will have only 1 option to fill since u will have used all the other options in the other blanks
so ur options r gona be 5 then 4 then 3 then 2 then 1.
so it will be like 5 x 4 x 3 x 2 x 1=120
i hope u get it now

 

[AdeelRox]

this method is confusing leave it . try this
0.94 is the probability between the two values if we subtract it from 1 and divide by two i.e 1-0.94/2=0.03 we get the probability of one small area so P(X<32)=1-0.03=0.97

i got it all except for just 1 minute thing. dude why did u divided "1-0.94" by 2. i mean both the small sectors mite not be equal,its not given that they r equal

 

[AdeelRox]

oh got it

 

[driggerbilal]

suggestions anybody ??

Its better to make it in horizontal ..

 

[Khan_971]

yeah i tried that but couldnt get the answer..could you plz do it?

P(k-125/4.2 <Z< 128-125/4.2) =.7465
Fi(3/4.2) - fi(k-125/4.2) = .7465
fi(k-125/4.2) = .0158
since its lower than .5,
1-fi(125-K/4.2) = .0158
.9842 = fi(125-K/4.2)
Using normal table:
2.15 = 125-k/4.2
9.03 = 125-K
K = 115.97 = 116

 

[angelicsuccubus]

Is it just me or is M2 actually easier than P3?

its just you

 

[Mustehssun Iqbal]

Assalamu alaikum,
Q.3 part 1 pls...
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_61.pdf
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_61.pdf

 

[Esme]

Its better to make it in horizontal ..

ok thanks.

 

[Esme]

its better you draw it horizontally as done in the marking scheme because in class and in books i have always seen it horizontally

alright thank you !

 

[Diamona151294]

um last minute problems.....:/

Um just part ii of Q3 And part iii of Q7 thanku sooooooooooooooo much!!!

 

[Mustehssun Iqbal]

um last minute problems.....:/
View attachment 10645View attachment 10646
Um just part ii of Q3 And part iii of Q7 thanku sooooooooooooooo much!!!

same questions here...

 

[JulyMei]

um last minute problems.....:/
View attachment 10645View attachment 10646
Um just part ii of Q3 And part iii of Q7 thanku sooooooooooooooo much!!!

first attachment is Q3

second attachment is Q7

hope it helps
gonna have my math paper 6 later.. wish me luck!