- Messages
- 8,393
- Reaction score
- 9,461
- Points
- 573
Re: O'level Maths, Addmaths and Statistics help available he
Thank you very much. Can you also help me with
N05/P2/Q9. ?
Thank you very much. Can you also help me with
N05/P2/Q9. ?
-
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)
Maths, Addmaths and Statistics: Post your doubts here!
- Thread starter OakMoon!
- Start date
Re: O'level Maths, Addmaths and Statistics help available he
(i) class bounds c.f
14.5-19.5 5
19.5-24.5 13
24.5-34.5 26
34.5-44.5 46
44.5-54.5 86
54.5-64.5 100
median = 100/2 = 50th term which lies in 44.5-54.5 class (check d c.f table)
median age = l + h/f(N/2 - C)
=44.5 + 10/40(50-46)
=44.5 + 10/40(4)
=44.5 +1 = 45.5yrs
(ii)Q1=100/4=25th term which lies in class 24.5-34.5
Q1 = l + h/f(N/4 - C)
=24.5 = 10/13(25-13)
=33.73yrs approx
Q3 =100 x 3/4 = 75th term which lies in class 44.5-54.5
Q3 = l + h/f(3N/4 - C)
=44.5 +10/40(75-46)
=51.75yrs approx
IQR = 51.75 - 33.73 = 18.02
SalmanPakRocks said:
(i) class bounds c.f
14.5-19.5 5
19.5-24.5 13
24.5-34.5 26
34.5-44.5 46
44.5-54.5 86
54.5-64.5 100
median = 100/2 = 50th term which lies in 44.5-54.5 class (check d c.f table)
median age = l + h/f(N/2 - C)
=44.5 + 10/40(50-46)
=44.5 + 10/40(4)
=44.5 +1 = 45.5yrs
(ii)Q1=100/4=25th term which lies in class 24.5-34.5
Q1 = l + h/f(N/4 - C)
=24.5 = 10/13(25-13)
=33.73yrs approx
Q3 =100 x 3/4 = 75th term which lies in class 44.5-54.5
Q3 = l + h/f(3N/4 - C)
=44.5 +10/40(75-46)
=51.75yrs approx
IQR = 51.75 - 33.73 = 18.02
Re: O'level Maths, Addmaths and Statistics help available he
(iii)a. x(midpoint) d(i.e x-a) fd
17 I -13 I -65
22 I -8 I -64
29.5 I -0.5 I -6.5
39.5 I 9.5 I 190
49.5 I 19.5 I 780
59.5 I 29.5 I 413
total = 1247.5
mean = a + fd/f
=30 + 1247.5/100
=42.5 approx
(b) apply d original formula.all u need 2 do is put 'd' instead ov 'x'.put in al d values n ul get d answer
i hope this helped=)
(iii)a. x(midpoint) d(i.e x-a) fd
17 I -13 I -65
22 I -8 I -64
29.5 I -0.5 I -6.5
39.5 I 9.5 I 190
49.5 I 19.5 I 780
59.5 I 29.5 I 413
total = 1247.5
mean = a + fd/f
=30 + 1247.5/100
=42.5 approx
(b) apply d original formula.all u need 2 do is put 'd' instead ov 'x'.put in al d values n ul get d answer
i hope this helped=)
- Messages
- 8,393
- Reaction score
- 9,461
- Points
- 573
Re: O'level Maths, Addmaths and Statistics help available he
thank you very much. It helped alot !!aleezay said:salmanpakrocks
- Messages
- 29
- Reaction score
- 0
- Points
- 0
Re: O'level Maths, Addmaths and Statistics help available he
CAN ANYONE SOLVE Q.9 AND Q.10 OF REVIEW QUESTIONS OF CHAPTER 12 OF MATHEMATICS D3 OXFORD
CAN ANYONE SOLVE Q.9 AND Q.10 OF REVIEW QUESTIONS OF CHAPTER 12 OF MATHEMATICS D3 OXFORD
- Messages
- 8,393
- Reaction score
- 9,461
- Points
- 573
Re: O'level Maths, Addmaths and Statistics help available he
What does this means ? Source N08/P1/Q7
Using a scale of 2 cm to represent 10 kg, starting at −10 kg, on the horizontal axis, and a
scale of 2 cm to represent 2 people per 5 kg on the vertical axis, draw on graph paper a
histogram to illustrate these data.
Should we use Frequency Density in the vertical axis or not ? I m confused here.
What does this means ? Source N08/P1/Q7
Using a scale of 2 cm to represent 10 kg, starting at −10 kg, on the horizontal axis, and a
scale of 2 cm to represent 2 people per 5 kg on the vertical axis, draw on graph paper a
histogram to illustrate these data.
Should we use Frequency Density in the vertical axis or not ? I m confused here.
- Messages
- 29
- Reaction score
- 0
- Points
- 0
Re: O'level Maths, Addmaths and Statistics help available he
@salman bro can u please solve these questions
@salman bro can u please solve these questions
- Messages
- 8,393
- Reaction score
- 9,461
- Points
- 573
Re: O'level Maths, Addmaths and Statistics help available he
I would have solved, but the problem is I gave maths this May/June and after the exam I gave away my books to my junior. So currently I don't have a book.
I would have solved, but the problem is I gave maths this May/June and after the exam I gave away my books to my junior. So currently I don't have a book.
- Messages
- 29
- Reaction score
- 0
- Points
- 0
Re: O'level Maths, Addmaths and Statistics help available he
Q.10( http://www.youtube.com/watch?v=T9zonvYVa-8)
Q.9(http://www.youtube.com/watch?v=M3WViKMt3K4)
here you go buddy see the questions and then answer me now please
Q.10( http://www.youtube.com/watch?v=T9zonvYVa-8)
Q.9(http://www.youtube.com/watch?v=M3WViKMt3K4)
here you go buddy see the questions and then answer me now please
- Messages
- 29
- Reaction score
- 0
- Points
- 0
Re: O'level Maths, Addmaths and Statistics help available he
what did u get in maths ??
what did u get in maths ??
- Messages
- 2
- Reaction score
- 0
- Points
- 0
Re: O'level Maths, Addmaths and Statistics help available he
guys where shud i post my question??
guys where shud i post my question??
- Messages
- 2
- Reaction score
- 0
- Points
- 0
Re: O'level Maths, Addmaths and Statistics help available he
Someone plz tell me the advantages/disadvantages of dual bar charts and percentage bar charts
Someone plz tell me the advantages/disadvantages of dual bar charts and percentage bar charts
- Messages
- 1,882
- Reaction score
- 1,331
- Points
- 173
Re: O'level Maths, Addmaths and Statistics help available he
In the end ov it, 3r = 48, so r = 16 cm
For the second part,
Area of shaded region = Area of triangle OQP - Area of segment formed in smaller circle by QP
So QP = ?
Using cosine rule,
QP^2 = 16^2 + 16^2 - 2(16)(16).cos120
Solving this,
QP = 27.71 cm
Then,
QP=OQ=OP=27.71 cm (Its an equilateral triangle because the angles formed are the same ) !
Area of triangle OQP = 1/2 x 27.71 x 27.1 x sin60 = 332.554 cm^2
Area of segment = area of sector in circle - area of triangle in it
= 120/360 x pie x 16^2 MINUS 1/2 x 16 x 16 x sin120 = 157.23 cm^2
Now, shaded area 332.55 - 157.23 = 175 cm^2 !!
Q10 first part goes like This !!captainhyderabad said:
In the end ov it, 3r = 48, so r = 16 cm
For the second part,
Area of shaded region = Area of triangle OQP - Area of segment formed in smaller circle by QP
So QP = ?
Using cosine rule,
QP^2 = 16^2 + 16^2 - 2(16)(16).cos120
Solving this,
QP = 27.71 cm
Then,
QP=OQ=OP=27.71 cm (Its an equilateral triangle because the angles formed are the same ) !
Area of triangle OQP = 1/2 x 27.71 x 27.1 x sin60 = 332.554 cm^2
Area of segment = area of sector in circle - area of triangle in it
= 120/360 x pie x 16^2 MINUS 1/2 x 16 x 16 x sin120 = 157.23 cm^2
Now, shaded area 332.55 - 157.23 = 175 cm^2 !!
- Messages
- 29
- Reaction score
- 0
- Points
- 0
Re: O'level Maths, Addmaths and Statistics help available he
buddy can u please answer Q9 also please
buddy can u please answer Q9 also please
- Messages
- 1,882
- Reaction score
- 1,331
- Points
- 173
- Messages
- 8,393
- Reaction score
- 9,461
- Points
- 573
Re: O'level Maths, Addmaths and Statistics help available he
Thanks Ghalya for doing this, Actually Captainhyderabad I have my papers 10 days from now, so have very little time.
Yes i got A* in maths luckily.
Yes Ghalya is 100% correct.
Thanks Ghalya for doing this, Actually Captainhyderabad I have my papers 10 days from now, so have very little time.
Yes i got A* in maths luckily.
Yes Ghalya is 100% correct.