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Mathematics: Post your doubts here!

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aleezay

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Kumkum said:
can someone pls help me with question 5 and 6bii) nd 6c
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_63.pdf
thnx in advance :)
Click to expand...
Q. 5(b)(ii) The first digit has to be a 2 if we are to get a no. between 20,000 and 30,000. the no of ways the remaining 5 digits can be used= 5P4= 120
(c) The probability of placing a white tile ( or any other colored tile) first is 1/3. The second tile must not be white so P(not white)=1 - 1/3=2/3. The same goes for all the remaining 6 tiles i.e the probability of placing a tile would be 2/3 since it has to be different from the previous tile. Now multiply your result by 3 (since you can place any of white,black or grey tiles first).
I hope this helps :)
 
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aleezay

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According to some friends of mine, using ES series calculators will straightaway result in a deduction of 10 marks. Is this true?? :O
 
Kumkum

Kumkum

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aleezay said:
Q5.(i) Apply continuity correction. If u are rounding off a number to 84, it must be in the range 83.5<X<84.5. Now calculate the probability of the result ( by converting the data to normal distribution by subtracting mean from each value and dividing by standard deviation)

(ii) Simply calculate P(X>87)- no need to apply continuity correction since there is no mention of rounding off numbers. U will get 0.3282
now convert the data to binomial distribution: X - B(5,0.3282) and calculate the result for P(X=0) + P(X=1) (at most 1 out of 5 trials- I'm assuming you'd be knowing how to evaluate this expression)

(iii) Convert data to normal distribution
Fi{(k-82)/under root 126} - Fi(5/under root 126) = 0.3
The rest is all simple algebra.
Please let me know if you still have a confusion. :)
Click to expand...
aleezay said:
Q. 5(b)(ii) The first digit has to be a 2 if we are to get a no. between 20,000 and 30,000. the no of ways the remaining 5 digits can be used= 5P4= 120
(c) The probability of placing a white tile ( or any other colored tile) first is 1/3. The second tile must not be white so P(not white)=1 - 1/3=2/3. The same goes for all the remaining 6 tiles i.e the probability of placing a tile would be 2/3 since it has to be different from the previous tile. Now multiply your result by 3 (since you can place any of white,black or grey tiles first).
I hope this helps :)
Click to expand...
thank u sooo much!! it did help :) (y)
 
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aleezay

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Minato112

Minato112

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princess787 said:
can anyone help me in 9709_w12_qp_31 question 3 and 10 ii) ! please! :)
Click to expand...
3) sin(θ + 45◦) = 2 cos(θ − 30◦)
sin θ. cos 45 + cos θ. sin 45 = 2 [ cos θ.cos 30 + sin θ.sin 30]

cos 45 = (|2/2), sin 45 = (|2/2), cos 30 = (|3/2) and sin 30 = (1/2)

(|2/2) sin θ + (|2/2) cos θ = 2 [ (|3/2) cos θ + (1/2) sin θ ]
(|2/2) sin θ + (|2/2) cos θ = |3 cos θ + sin θ
(|2/2) sin θ - sin θ = |3 cos θ - (|2/2) cos θ
((|2/2) - 1) sin θ = (|3 - (|2/2)) cos θ

Dividing everywhere by cos θ,

tan θ = [ (|3 - (|2/2))/((|2/2) - 1) ]

θ = 180 - *tan inverse* - [ (|3 - (|2/2))/((|2/2) - 1) ] = 105.9

As for the question 10 (ii), if noone answers it, I'll do it.
Also Note that "|" is used for root
 
P

PhyZac

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Alice123 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_62.pdf
no 3ii n 5iii . thnx in advance Dug PhyZac syed1995
Click to expand...
For question 5iii


there are 2 restrictions.

Either both question 1 and 2 of part A or None

If we consider 1st restriction.
Then we take question 1 and 2, and now 4 questions are left (we need total of 6)
No. of questions are 11 take the first 2 we get 9 left to choose 4 from.
therefore
9C4 = 126

Then if we consider 2nd restriction
Then we have to choose 6 from 9 questions ( since we arent allowed with question 1 and 2)
so 9C6 = 84

126 + 84 = 210
 
Alice123

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PhyZac said:
For question 5iii


there are 2 restrictions.

Either both question 1 and 2 of part A or None

If we consider 1st restriction.
Then we take question 1 and 2, and now 4 questions are left (we need total of 6)
No. of questions are 11 take the first 2 we get 9 left to choose 4 from.
therefore
9C4 = 126

Then if we consider 2nd restriction
Then we have to choose 6 from 9 questions ( since we arent allowed with question 1 and 2)
so 9C6 = 84

126 + 84 = 210
Click to expand...
very clearly explained. thanks a lot
 
whiz-kidd

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can anybody help me with this question for mechanics (M2)

O/N 2009 QP 52
Question 2 and 5

thanks!
 
Alice123

Alice123

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PhyZac said:
For question 5iii


there are 2 restrictions.

Either both question 1 and 2 of part A or None

If we consider 1st restriction.
Then we take question 1 and 2, and now 4 questions are left (we need total of 6)
No. of questions are 11 take the first 2 we get 9 left to choose 4 from.
therefore
9C4 = 126

Then if we consider 2nd restriction
Then we have to choose 6 from 9 questions ( since we arent allowed with question 1 and 2)
so 9C6 = 84


126 + 84 = 210
Click to expand...


 
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PhyZac

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Alice123 said:
Click to expand...
b (i) For getting a even number u need the number to end with even number (ofc) well in that number we have 2 4 6 8 as even,
now we consider each separtely
with 2 as end
we have 1 44 687
we rearrange 6! / 2! (we permutate 6 and we need to divide with 2 cuz there are 2 4s)
with 6 and 8 as end it is same thing as with 2
so we get 6!/2! and 6!/2!
with 4 as end we wont have 2 4s, so it will be only 6!
now add
6! + (6!/2!)*3 = 1800

i will try to solve rest soon In Sha Allah

b (ii) to get a number between 20 000 and 30 000 we have to start with number 2 and only, (we cant have a number smaller than 30000 starting with 3 ryt!)
so basically we have 2 and 4 blank
2_ _ _ _
so we permutate 4 from the left 5 numbers
we get 5P4 = 120!

(c) wat a tricky one!
anyway,
see, now since all have same prob, u get that each prob is 1/3

so we start any color and then take any of the other two in next shot
I am not sure if u got my point,
so we 1 * (2/3 )^7

first try is one, becuz all is possible..and power is 7 since 8 - 1 = 7
 
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