• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

XPFMember

XPRS Moderator
Messages
4,555
Reaction score
13,290
Points
523
last 2 parts of this questions please

how'd you solve the first one? 14C12/2?

Assalamoalaikum wr wb!
Alright, his question is a bit tricky, but anyways..InshaAllah, you’ll find it easy ...

So first part’s simple, it’s 14P12
Second part...they say 3 business people in the first row...so they may exchange seats...so 3! arrangements for the 3 business people...
Let’s move on..2 seats on the same side of aisle reserved for the two couples. We’re left with 5 window seats, so students will have 5! Arrangements...as for the couples..first will have to chose out of 3, 3C1 and then they may exchange seats within themselves...so the arrangements for first couple is 3C1 x 2!
Similarly work out for the second couple, from 2 of them...they gotta chose one..2C1 x 2!
Multiply all these....[3!] x [5!] x [3C1 x 2!] x [2C1 x 2!] = 17, 280

Part three..
Mrs Brown in the first row..so 3C1
‘Mrs Lin sits directly behind a student’....count, 10 such possible combinations are there...ignore one of the first seats as Mrs Brown has to sit there ;) and then it may be one of the 5 students...so we have 10 x 5C1
q.jpg


Now there are 11 seats left, and 9 people...so there seating arrangements will be 11P9
Multiply all these... [3C1] x [10 x 5C1] x [11P9]
Don’t forget, they asked to find the probability...total combinations were 14P12...Divide this result by 14P12...
[3C1] x [10 x 5C1] x [11P9] / 14P12 = 0.0687

Hope this helped...
Remember me in your prayers...
 
Messages
1,476
Reaction score
1,893
Points
173
AOA every one..........
can any body help me out with 9709/04/O/N/07.............???
Question 2........>
sorry to write it.....:oops::confused:
Aoa,
Im taking upwards direction positive and downward direction negative
Distance travelled by particle A = 10 - distance travelled by particle B
u1t + 0.5 a1 (t)^2 = 10 - [u2t + 0.5 a2 (t)^2]

[ u1 = 12.5 , a1 = - g , u2 = 0 , a2 = -g ]
Substituting these values

12.5t - 0.5g (t)^2 = 10 - [ 0.5 g (t)^2]
12.5t - 0.5g (t)^2 + [ 0.5 g (t)^2] = 10
12.5t = 10
t = 10 / 12.5 = 0.8

height above O,
h = 12.5t - 0.5g (t)^2
h = 12.5(0.8) - 0.5(9.81) (0.8)^2
h = 10 - 3.14
h = 6.8 m
:)
 
Messages
1,476
Reaction score
1,893
Points
173
Can anyone help me out with the second part of question no 2. (i.e the magnitude of the resultant force)
QP : OCT/NOV 2011 PAPER 41
MS : OCT/NOV 2011 PAPER 41
Aoa,
There is tension in the string in both parts, i.e there is a tension T opposite to weight of particle A and there is also a tension T opposite to weight of particle B
That is why resultant force on the pulley due to string = 2T = 2(4.55) = 9.1 N
:)
 
Messages
1,476
Reaction score
1,893
Points
173

Attachments

  • awe.png
    awe.png
    118.8 KB · Views: 14
Messages
483
Reaction score
199
Points
53
I haven't seen the question, but it must be asking for the exact answer. . .
Um...I posted the link to the paper
Do yourself a favour and look at the question
I don't think it says anything about an exact answer...
 
Messages
772
Reaction score
149
Points
38
Um...I posted the link to the paper
Do yourself a favour and look at the question
I don't think it says anything about an exact answer...
Alright, in these sort of cases, if you get an exact answer(usually up to 2 decimal places), mark schemes prefer to write it in its original form. You won't loose a mark for writing it up to 3 significant figures. The examiner report doesn't say anything about deducting a mark.
 
Messages
1,476
Reaction score
1,893
Points
173
The e.r does nt say that marks of students who rounded off the answer were deducted. so i think they must have been credited
 
Messages
483
Reaction score
199
Points
53
Alright, in these sort of cases, if you get an exact answer(usually up to 2 decimal places), mark schemes prefer to write it in its original form. You won't loose a mark for writing it up to 3 significant figures. The examiner report doesn't say anything about deducting a mark.
The e.r does nt say that marks of students who rounded off the answer were deducted. so i think they must have been credited
Guys its just that the examiner report does not mention incidence where candidates round correctly, only incidences where they do incorrect rounding
Though that 2 dp thing has got me thinking
 
Messages
43
Reaction score
11
Points
18
t
Aoa,
Im taking upwards direction positive and downward direction negative
Distance travelled by particle A = 10 - distance travelled by particle B
u1t + 0.5 a1 (t)^2 = 10 - [u2t + 0.5 a2 (t)^2]

[ u1 = 12.5 , a1 = - g , u2 = 0 , a2 = -g ]
Substituting these values

12.5t - 0.5g (t)^2 = 10 - [ 0.5 g (t)^2]
12.5t - 0.5g (t)^2 + [ 0.5 g (t)^2] = 10
12.5t = 10
t = 10 / 12.5 = 0.8

height above O,
h = 12.5t - 0.5g (t)^2
h = 12.5(0.8) - 0.5(9.81) (0.8)^2
h = 10 - 3.14
h = 6.8 m
:)
thnx bro...........
 
Messages
21
Reaction score
6
Points
13
I need help in W11,p61, Q5 . Please explain the interpretation of the sentence underlined.w11.p61.JPG

Also w11/p62 Q5w11.p62.JPGMay Allah bless you all. JazakAllah
 
Messages
971
Reaction score
532
Points
103
I need help in W11,p61, Q5 . Please explain the interpretation of the sentence underlined.View attachment 7001

Also w11/p62 Q5View attachment 7002May Allah bless you all. JazakAllah
The underlined phrase means the weight of 94% of the letters have a standard deviation of 12g. So let's say the mean is 20g. 94% of the letters have a weight of something between 18g and 42g, i.e. 12g off from the mean value.

As for the second question..

(i) Let the probability of the spinner landing on the blue side be 'x'. Since the probability of the spinner landing on the red and green sides are three and four times P(blue) respectively, their probabilities are '3x' and '4x'. Probabilities always add up to 1, so..

x + 3x + 4x = 1
8x = 1
x = 1/8

Hence the probability of the spinner landing on the blue side is 1/8.

(ii) You want a red AND a green AND a blue, so you multiply their probabilities:

1/8 * 4/8 * 3/8 = 12/512.

Also, since there are 3! different ways for this to happen (you could have RBG, BGR, GRB, etc.), you multiply the result by 3!

12/512 * 3! = 9/64.

(iii) Use normal approximation for this.

n = 136
p (success of landing on blue) = 1/8
q (failure of landing on blue) = 7/8

mean = 136 x 1/8 = 17
standard deviation = √(17 * 7/8) = √14.875

Also note that you will use continunity correction, so your value becomes 19.5, not 20. Using formula,

z = (19.5 - 17) / √14.875

Once you figure out the value of z, use the table to figure out the probability.

I hope this helped. :)
 
Messages
389
Reaction score
202
Points
53
Please someone help me this problems ;

1. Find the maximum value of 1/√cosx + sinx + 5 , and the value of x where the maximum value occurs.

2. Solve the equation 10sin^2 (1/2x) - 5sinx = 4 , giving the values of x between 0 and 360 to the nearest 0.1

Thank You in anticipation & may ALLAH bless you :)
 
Top