Mathematics: Post your doubts here!

[hush puppy]

HI! does anybody know the approx. grade threshold for 9709 AS level pure mathematics 2013?

 

[Scafalon40]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_61.pdf
Why are the midpoints in Q6 part (iv) 50.5, 125.5 etc instead of 50, 125...

Anyone? PhyZacDug

 

[Wajih ul hassan]

ummm okay
here's part 1 )
dy/d(t)=3asin^2t(cost)
d(x)/d(t)=-3a sint cos^2t
d/d(x)=3a sin^2t cost/-3a sint cos^2t
dy/d(x)=-tant

now for part 2 (the tricky bit )
.gradient=-tant , x=acos^3t, y=asin^3t
y-asin^3t= -tant(x-acos^3T)
y-asin^3t=-(sint/cost)x+(sint/cost)acos^3t
ycost-asin^3tcost=-xsint+asintcos^3t
xsint +ycost=asintcost(cos^2t+sin^2t)
xsint+ycost= asintcost

and part 3
for y-intercept put y=0,
xsint+o= asintcost
x=acost
for y-intercept put x=0
0+ycost=asintcost
y=asint
coordinates of x are (acost,0)
coordinate of y are (0, asint)
lenght=((acost)^2+(asint)^2)^1/2
=(a^2cost^2+a^2sint^2)^1/2
=(a^2(cos^2t+sin^2t))^1/2
=(a^2)^1/2
=a
you get it ?

but what about c in part 2
and why is it y-asin^3t= -tant(x-acos^3T)
shouldn't it be asin^3t= -tant(acos^3T)
because its y=tant+c

 

[applepie1996]

but what about c in part 2
and why is it y-asin^3t= -tant(x-acos^3T)
shouldn't it be asin^3t= -tant(acos^3T)
because its y=tant+c

umm i substituted my values in (y-y1)=m(x-x1) where y1 and x1 are the co-ordinates
if you want it in the y=mx + c format i guess you could do it
if you don't get it .........then i will try it with y=mx+c and do it for you

 

[Wajih ul hassan]

umm i substituted my values in (y-y1)=m(x-x1) where y1 and x1 are the co-ordinates
if you want it in the y=mx + c format i guess you could do it
if you don't get it .........then i will try it with y=mx+c and do it for you

thank you can u please ues that

 

[PhyZac]

PhyZac Can you please help me with November 2001 paper 6 Question 7.I'd be grateful if you gave a detailed explanation.Thanks!!
Here is the question:
A bag contains 7 orange balls and 3 blue balls.4 balls are selected at random from the bag,without replacement.Let X denote the number of blue balls selected.
i.Show that P(X=0)=1/6 and P(X=1)=1/2
ii.Construct a table to show the probability distribution of X.

(i)
So P(X=0) means when no blue is taken,
it will be
7/1o x 6/9 x 5/8 x 4/7 = 1/6
first 7 orange from 10 total, then 6 from 9 total (because one is taken) etc..

P(X=1)
(7x6x5x3) / (10x9x8x7) x 4 = 1/2
we will take 3 from orange so first 7 then 6 then 5 left, for blue we take once so only 3
down is total number, that is, 10 then 9 , 8 ,7
x4 because blue can be taken first second third or fourth

(ii)
for 0 we know 1/6
for 1 we know 1/2
for 2 it will be 4C2 x (3x2x7x6) / (10x9x8x7) = 3/10 [i take 4 combine 2 to get how many ways can we get 2 blues]
for 3 it will be 4C3 x (3x2x1x7) / (10x9x8x7) = 1/30

X=x..... || 0 || 1 || 2 |||| 3 |
P(X=x) |1/6|1/2|3/10| 1/30|

 

[PhyZac]

Anyone? PhyZacDug

I had same doubt, but apparently the reason is, it is less than or equal to the specific value. When it is only less than then only 50 not 50.5
i guess! not sure.

 

[applepie1996]

thank you can u please ues that

here
y= mx + c
a sin^3 t=-tan t (acos^3 t) + C
C = a sin^3 t + a tan t cos^3 t
y=-tan t x + a sin^3 t + a tan t cos^3 t
y + tan t x = a sin^3 t + a (sint/cost) cos^3 t
y + (sint/cost)x = a sin^3 t + a sint cos^2 t
y cost + sint x = cost (a sin^3 t + a sint cos^2 t)
y cost + sint x = a cost sint (sin^2 t + cos^2 t )
y cost + sint x = a cost sint

 

[sma786]

Stats people, i need the answer to this weird question, (Probability Distribution)
Will anyone reply? the question is so long, if anyone can reply so i will type it out, please ^_^

 

[daredevil]

Stats people, i need the answer to this weird question, (Probability Distribution)
Will anyone reply? the question is so long, if anyone can reply so i will type it out, please ^_^

wats the question??

 

[sma786]

whats the question??

two fair dice are thrown simultaneously. The random variable H is the highest common factor of the two scores. Tabulate the probability distribution of H, combining together all the possible ways of obtaining the same value

 

[daredevil]

two fair dice are thrown simultaneously. The random variable H is the highest common factor of the two scores. Tabulate the probability distribution of H, combining together all the possible ways of obtaining the same value

look i cant very well make the table here but u must have done these questions somewhere where u have a table.... its mostly in the dice questions.. u have numbers for one die in the top row and for the other in the first column and u give the results in all the 36 boxes that come from the 6 rows and 6 columns .

 

[daredevil]

two fair dice are thrown simultaneously. The random variable H is the highest common factor of the two scores. Tabulate the probability distribution of H, combining together all the possible ways of obtaining the same value

look at the table... fill in the solution boxes urself for 1 and 1 the HCF will be 1 so u write one, for 4 and 2 it will be 2 and so on... if u don't get it then get back to me but i think this is pretty much how u're supposed to do it. and tag me in any stats question uu hav becaus it'll help me prepare too

 

[Dug]

Anyone? PhyZacDug

Class Intervals | X
1 - 100 | 50.5
101 - 150 | 125.5
151 - 200 | 175.5
...

 

[Scafalon40]

I had same doubt, but apparently the reason is, it is less than or equal to the specific value. When it is only less than then only 50 not 50.5
i guess! not sure.

Class Intervals | X
1 - 100 | 50.5
101 - 150 | 125.5
151 - 200 | 175.5
...

Interesting. Dug's method make sense. Thanks for sharing, Dug! Any thoughts, PhyZac?

 

[Scafalon40]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_63.pdf
ms http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_63.pdf
Have a look at Q3 part iii. Could you tell me the class boundries for this one Dug
I think its something like 19.5-34.5 etc. No?

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_63.pdf
ms http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_63.pdf
Have a look at Q3 part iii. Could you tell me the class boundries for this one Dug

i thikn it will be like for the first one: 0=< x < 10

include the first value and not the last one.... there can be no decimals i think bcz there are marks

 

[Dug]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_63.pdf
ms http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_ms_63.pdf
Have a look at Q3 part iii. Could you tell me the class boundries for this one Dug
I think its something like 19.5-20.5 etc. No?

0 - 9
10 - 19
20 - 34

 

[Scafalon40]

0 - 9
10 - 19
20 - 34

Yeah exactly. So the boundries are something like:
-0.5-9.5
9.5-19.5
19.5-34.5
Correct or not?

 

[Scafalon40]

i thikn it will be like for the first one: 0=< x < 10

include the first value and not the last one.... there can be no decimals i think bcz there are marks

But that means the class width is 10, and there is no way you can get a midpoint of 4.5, as it says in the markscheme.