Mathematics: Post your doubts here!

[smartangel]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_61.pdf
Q6 iv..:S also in part i..why"d we plot upper class boundaries..i means the sign says greater than or equal to..so shudnt we plot that value like 100, 150 and so on?

 

[mominzahid]

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

 

[Khan_971]

for aii the MS says

3 digit odd 500+ = 4 ways
3 digit odd 600+ = 3 × 2
= 6 ways
4 digit odd 1000+ = 4 ways
4 digit odd 3000+ = 4 ways
4 digit odd 5000+ = 4 ways
4 digit odd 6000+ = 6 ways
OR 4 digit odd, last digit in 3 ways,
2
nd
to last in 3 ways, 2
nd
in 2 ways
first in 1 way = 18
Total = 28 ways
can u explain this?

500+ means 513, 531, 561, 563
do the same for the rest.
Need I say more?

 

[mominzahid]

500+ means 513, 531, 561, 563
do the same for the rest.
Need I say more?

Ahh got ittt... Thanksss..
i posted a histogram question above please explain that aswell? ur a lifesaver.

 

[Khan_971]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_63.pdf
Quesion no. 6 part 3 plz how did they do this narking scheme main tricky answer hai plz clarify me soon tommorrow is my papers.

Im gonna try my best. Okay so Brown can only take 1 of the 3 seats right? 3C1 = 3
Lin will only get behind a student. So take two fingers and place them on two side-by-side seats. assume brown take the front seat at the right corner. Count all positions Lin can sit in: 10
Only 1 of 5 students will be sitting ahead of Lin: therefore 5C1 = 5
3 people are cleard as well as 3 seats. 11 seats and 9 people remaining. therefore 11P9.
therefore, 3*5*10*11P9/ (total arrangement in i)

 

[Khan_971]

Why 20C7??? Still don't get it.....The marking scheme uses 12!/7!5!......

WOOPS. My bad. thanks for checking. since there are 7 same heads and 5 same tails. Remember if letter/objects are repeated u divide the total possible arrangments by n!, where n is the number of times it is repeated.
so 12 coins, 12!
7 heads, so divide by 7!. 5 tails so divide by 5!

 

[Wanzi21]

first part is direct 4!. 2nd part u gotta write down all possible combinations. I already showed the third part

thanksss ^^

 

[Khan_971]

Ahh got ittt... Thanksss..
i posted a histogram question above please explain that aswell? ur a lifesaver.

Ok well u know we simply cant leave spaces in a histogram. so thats why u have to make a connection between the intervals. In O levels u can recall adding .5 to the upper boundary and subtracting .5 from the lower boundary. Meaning first interval will be .5-20.5. 2nd will be 20.5-30.5. and so on. u can get class width by subtracting Upper boundary - lower boundary.

Freq density is frequency/ class width. find it for all the intervals.

For mean mark for group data. u need to find the midpoint values of each interval. Midpoint= (Upper boundary + lower boundary)/2. Let these midpoints be x
then find the corresponding fx. Then there's the simple formula Mean = Sum of fx/ 234 students

Variance = sum of [f * x^2] - (Mean)^2
Standard deviation is square root of variance

 

[Khan_971]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_61.pdf
Q5 b..no idea what to do in this one!!

the N (x,y) means x is the Mean (that miuw symbol) and y is the Variance, (sigma squared). havent u done normal distribution?
Subtract mean from 33+a and 33-a, divide by standard deviation??

 

[sunnyexo]

WOOPS. My bad. thanks for checking. since there are 7 same heads and 5 same tails. Remember if letter/objects are repeated u divide the total possible arrangments by n!, where n is the number of times it is repeated.
so 12 coins, 12!
7 heads, so divide by 7!. 5 tails so divide by 5!

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

 

[angelicsuccubus]

guys help me out here, if I have (3 - 90(v+30)) (dv/dx) + 1 =0 ... and a=v(dv/dx)

Im having a mind block... how am I supposed to find a here?

 

[Khan_971]

I dont get it..

can u please explain it in more simpler terms?
i dont get wht u assumed x1 and x2 to be and what 2fi(z)-1=0.34 means?

He meant ke look. if it is (-y<Z<y). Where y is any number
Then the prob. will be fi - (1-fi) [since fi(-y) will be 1-fi
open the brackets, 1 becomes negative, the fi's add up
2fi-1 is the final thing.

 

[Khan_971]

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

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

 

[Khan_971]

guys help me out here, if I have (3 - 90(v+30)) (dv/dx) + 1 =0 ... and a=v(dv/dx)

Im having a mind block... how am I supposed to find a here?

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

 

[Khan_971]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_62.pdf
Q6!! part ii

Draw a tree diagram. 1-P (0 boys) is your answer

 

[Khan_971]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_62.pdf
Q1.im getting 7.95..are any of you getting the same answer as marking scheme which is 7.68..idk where im going wrong in this one!
also Q4 ii of the same paper! how do we do it????

Have u tried taking 3.sf??

 

[Wanzi21]

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

 

[smartangel]

the N (x,y) means x is the Mean (that miuw symbol) and y is the Variance, (sigma squared). havent u done normal distribution?
Subtract mean from 33+a and 33-a, divide by standard deviation??

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..

 

[Khan_971]

help

a column has 8 places. 2 holes to get in. 8C2. Im not sure about the toher 2 parts. Mind telling me which paper is this from?

 

[hassam]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_63.pdf
q2 Khan_971