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Maths, Addmaths and Statistics: Post your doubts here!

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In question no 5 (b)Why we will multiply the probablities of Sarah and Terry and not add them .And why answer is 1/36 not
2/3.
 

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b(ii) They both threw a six:

this question has two events: 1) sarah throws a six and 2) terry throws a six

the question asks you the probability that both the events 1 AND 2 happen at the same time. thus, we will multiply.

if the question asked you to find the probability that either sarah threw a six OR terry threw a six, we would have added the probabilities.
 
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thank you for your help.. i have 2 more questions .. http://papers.xtremepapers.com/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_w10_qp_11.pdf
same paper.. question no.25 part A.. i am really really bad at speed time graph.. :/ and question no 26 part B.
25 (a), kind of a trick question (but it's very easy).
The Acceleration or Retardation = Gradient of Speed-Time Graph.
Since the Retardation is constant (Straight Line), Just take the gradient of that line,
Acceleration = 0-4/25-20 = -4/5 = -0.8 m/s^2
 
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thank you for your help.. i have 2 more questions .. http://papers.xtremepapers.com/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_w10_qp_11.pdf
same paper.. question no.25 part A.. i am really really bad at speed time graph.. :/ and question no 26 part B.
For Question 26 (b),
If you've shown the triangles properly then you would know that,
BC = base of triangle ABC, BD = base of triangle BCD.
Taking ratios for similar triangles,
BC/BD = AB/BC
6/4 = AB/6
36 = 4 AB
AB = 9 cm
AB = AD + DB
9 = AD + 4
AD = 5 cm.
 
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(a)Let the runs in first 3 matches be x, y, z respectively.
Let z be lowest score and x be highest score.
Mean = 9, Median = 2nd Term = 8 (Therefore y = 8),
x+8+z/3 = 9
x+8+z = 27
x+z = 19
From given information,
(z+7)+z = 19
2z = 12
z = 6
x=13

x=13, z=6, y=8.

(b) Mean = 11
Let fourth match runs = w
27 + w/4 = 11
27+w = 44
w = 17.

Hope that helps! :)
 
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b(ii) They both threw a six:

this question has two events: 1) sarah throws a six and 2) terry throws a six

the question asks you the probability that both the events 1 AND 2 happen at the same time. thus, we will multiply.

if the question asked you to find the probability that either sarah threw a six OR terry threw a six, we would have added the probabilities.
Thanx, understood the question
 
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Written as products of their prime factors, N = 2^p × 5^q × 7^r and 500 = 2^2 × 5^3 .
The highest common factor of N and 500 is 2^2 × 5^2 .
The lowest common multiple of N and 500 is 2^3 × 5^3 × 7 .
Find p, q and r.
 
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Since HCF of N and 500 has 5^2 in it, N must have 5^2 so q=2.

Since LCM of N and 500 has 2^3 in it, N must have 2^3 because 500 only has 2^2, so p=3.

Since LCM of N and 500 has 7 in it, N must also have a 7 as 500 does not have a 7, so r=1.
 
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If I leave out Relative velocity, how difficult does getting an A* become. Relative Velocity is freaking me out.
 
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If I leave out Relative velocity, how difficult does getting an A* become. Relative Velocity is freaking me out.
If relative velocity does come, it can range around 6-10 marks and depending on the ease of the paper, you're A* could well and truly be still possible.
However, you shouldn't just give up. Here's a playlist, search for the relative velocity videos from here and watch them. They'll help.
http://www.youtube.com/playlist?list=PL503679CCDE65357E
 
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