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Not a problem. Not so dumb after all to try putting this into what I askedooooops sorry for the other picture I gave you was for Nov 2012, really sorry
You're welcomeOhh thank you
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Not a problem. Not so dumb after all to try putting this into what I askedooooops sorry for the other picture I gave you was for Nov 2012, really sorry
You're welcomeOhh thank you
You were doing it right, but your answer's incomplete yet.Any one can solve this for me?
You were doing it right, but your answer's incomplete yet.
After what you did, use equations from y and z simultaneously to find s and t.
-1+s = 2+t 4-s = 1+t
-1 +s = 2+3-s <-- t = 3-s
s= 3
this gives t = 3-3 = 0. This is different from what the previous two equations gave, hence the lines do not intersect.
(Had the been intersecting, you'd get s and t from both comparisons same.)
the fastest way is to use lami's rule to get the angle opposite to P1 then get the angle AP1X using subtraction from 360 degrees, then lami again to get the value of W.
Are you sure? Because mark scheme said all three component should not be satisfied. =/
Question 5) no 2 , can you explain the work energy principle here? I dont understand why the work done against friction is negative, and why the ke loss is greater than the pe gain? How do I figure out when the loss is greater than the gain to be precise.
ok guys heres the deal
6 ii 5 students sit anywhere by the window... 5! =120
3 business men sit anywhere= 3!= 6
now for the married couples... there are 2 couples =2 two seats each =2 and 3 seats remaining for each couple =3p2
3p2 *2*2= 24
5!=120
3!=6
24*6*120 = 17280
very well explained.... thankyou sooooo much littlecloud11Okay, this is a tricky one.
View attachment 25529
See the picture. The red seats highlight the ones that Mrs. Lin can occupy. She has to sit directly behind a student, so if she occupies those seats then it's possible for a student to be placed in front of her. The reason I kept one seat in the second row uncolored is because one of the seats in the front row is occupied my Mrs. Brown. A student can only take two seats in the front row.
So for Mrs. Lin the possible combinations are 10P1
Mrs. Brown has to sit in one of the front seats so 3P1
She can sit behind any of the 5 students, so it's 5P1, the reason you don't count the possible seats for the student is because it doesn't matter where they are placed as long as one of them is in front of Mrs. Lin
And finally after Mrs. Lin, a student and Mrs. brown has been placed there are 9 remaining passengers and 11 remaining seats. There is no restrictions. So it's 11P9
Total no. of ways = 10P1 * 3P1* 5P1 * 11P9
If all 12 passangers are seated randomly = 14P12
so probability = 10P1 * 3P1* 5P1 * 11P9/ 14P12 =.0687
Hope you get this!
Okay, this is a tricky one.
View attachment 25529
See the picture. The red seats highlight the ones that Mrs. Lin can occupy. She has to sit directly behind a student, so if she occupies those seats then it's possible for a student to be placed in front of her. The reason I kept one seat in the second row uncolored is because one of the seats in the front row is occupied my Mrs. Brown. A student can only take two seats in the front row.
So for Mrs. Lin the possible combinations are 10P1
She can sit behind any of the 5 students, so it's 5P1, the reason you don't count the possible seats for the student is because it doesn't matter where they are placed as long as one of them is in front of Mrs. Lin
Mrs. Brown has to sit in one of the front seats so 3P1
And finally after Mrs. Lin, a student and Mrs. brown has been placed there are 9 remaining passengers and 11 remaining seats. There is no restrictions. So it's 11P9
Total no. of ways = 10P1 * 3P1* 5P1 * 11P9
If all 12 passangers are seated randomly = 14P12
so probability = 10P1 * 3P1* 5P1 * 11P9/ 14P12 =.0687
Hope you get this!
ok guys heres the deal
6 ii 5 students sit anywhere by the window... 5! =120
3 business men sit anywhere= 3!= 6
now for the married couples... there are 2 couples =2 two seats each =2 and 3 seats remaining for each couple =3p2
3p2 *2*2= 24
5!=120
3!=6
24*6*120 = 17280
Some help with Part ii as well.. Since I don't understand why the mark scheme is ignoring the other window seats... and just doing a 3P3*4P4*5P5 ..
Would really appreciate it ...
Thanks
Thanks for solving all our problems....View attachment 25598
The students cannot take the window seats in the first row because the 3 business people have to be seated there. Each married couple must be seated in a different row, but they have to be on the same side of the isle. So one of the couples have to take two window seats so that both couples are in different rows. So you see, only 5 window seats remain, and the students have to take these 5 seats in any order. So it's 5! and not 9P5.
Thanks for solving all our problems....
bother? you aren't bothering meSorry to bother you but can you please solve it for me ? I didn't get you..
View attachment 25598
The students cannot take the window seats in the first row because the 3 business people have to be seated there. Each married couple must be seated in a different row, but they have to be on the same side of the isle. So one of the couples have to take two window seats so that both couples are in different rows. So you see, only 5 window seats remain, and the students have to take these 5 seats in any order. So it's 5! and not 9P5.
well this was a fun questionhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_4.pdf
@Obsidian Fl1ght
@Pie-man
its question 6(iii) ?
JazakAllah
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