Mathematics: Post your doubts here!

[19islandprincess96]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_62.pdf

How do we do
1) Question 2?
2) 5 iii?
30 7 ii?

 

[littlecloud11]

Can someone please show me how to solve question 6 of this paper?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_61.pdf

6i) P(<6) = 63/600 = .105
p(>6) - 1-1.05 =.895
z value for .895 = 1.253
for the P(>6)-

6-μ /σ = -1.253
μ - 1.253 σ = 6 ------1

P(>12) = 155/600 =.2583
P(<12) = 1-.2583 =.7417
z value for .7417 = .648
for P(<12) -

12- μ/σ = .648
rearrange:
μ + .648 σ =12 -----2

solve eq 1 and 2 simultaneously:

μ - 1.253 σ = 6
-μ - .648 σ =-12
-1.901 σ = -6
σ = 3.15

therefore μ = 9.95

ii) When the distribution of a variable Y, lies within +/-σ the Z value is +/-1
So the range for this is Z>1 and Z<-1
P (Z<-1 and Z>1)
For Z>1 = 1 -Ф1
For Z<-1 = Ф(-1) = 1 -Ф1
So,
=1 -Ф1 + Ф(-1)
=1 -Ф1 +1 -Ф1
=2 - 2* .8413
P=.3174
number of feathers =.317* 1000 = 3174

 

[Scafalon40]

Okay sorry for late reply, but i was solving it in order to understand.

Well it is because see lets take the example (z +2)
so one root is
z = -2

now if we take p(z^2)
then
(z^2 + 2)
z^2 = -2
z = √-2
z = +/- i √2

basically the square root of the answers! get it?

Yeah I think so, thanks
So, if we let y=z^2, we get the same equation as in part (i)
and the same solutions i.e y=-2 etc
but y=z^2
so z^2=-2
and z=sqrt(-2)
so z=+/- sqrt(2)
Thanks

 

[~`Heba`~ :)]

since work done here is the work done against gravity+ gain in potential energy + gain or loss in kinetic energy
to calculate gain or loss of kinetic energy we need final and initial velocity
now we are available with initial velocity = 6 ms^-1 =V1
power at bottom of the hill=P1
Force at the bottom of the hill =F1
now at top f the hill velcity =?
Power=P2=5*P1
Force=F2=3*F1
recall the formula P=Fv
take the ratio of both
(P1/5*P1)=(F1/3*F1)*(6/V2)
now we have
(5*6/3)=10=V2
now
work done against gravity+ gain in potential energy =945000
kinetic energy change = 0.5*1250*(10^2-6^2)=625*64=40000
therefore
945000+40000=985000=work done

Alternative method... similar but I think it's easier

F2=3F1
P2=5P1

F2V2=5F1V1
Substitute F2=3F1 and V1=6
3F1V2=5F1(6)
3F1V2=30F1
F1V2=10F1
Cancel out F1
V2=10

I think you already got the height which is 50, using trigonometry, we'll use that to get P.E.

Work done by driving force <What we want> = Change in K.E. + Change in P.E. + Work done against resistance to motion
W = 1/2m(v^2-u^2) + mg(h2-h1) + Work done against resistance to motion

Work done against resistance to motion = Fd, and because F of the resistance to motion = 800, and d = 400, 800x400 =320000

W= 1/2(1250)(10^2-6^2) + 1250(10)(50) + 320000
W = 985000 J

Thanks a lot!!
and please if you could do the graph part in #7 it would be great if you explain it to me

 

[Amaryllis]

Well from one you got
2(x-2)^2 + 3

now you have to know the possible values of y.

The x values are all real numbers ( as stated in the question)

Now what i did was, make the red bit of the equation zero , to do this, you have to make x as 2, now 2(2-2)^2 + 3 will equal 3, then i try a negative number, i choose -1, 21 came, so this bigger than 3, i tried a positive number as 1, and 5 came, so in all cases, a number bigger than three comes, so
y => 3, [i mean y is more than or equal to 3]

Check this, might help you, it has everything you need for statistic.
http://www.examsolutions.net/maths-revision/syllabuses/CIE/period-1/S1/module.php

awesome ! thank you alot .. jazak allah 5air

 

[PANDA-]

Thanks a lot!!
and please if you could do the graph part in #7 it would be great if you explain it to me

Lol.... i don't think it's possible to show a graph here
Unless we draw it scan etc...

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_62.pdf

How do we do
1) Question 2?
2) 5 iii?
30 7 ii?

That is Question part 1 , the second part needs just the formula!

And for that question, Y value is 2 X values minus eachother.
[Two independent values of X are chosen at random. The random variable Y takes the value 0 if the
two values of X are the same. Otherwise the value of Y is the larger value of X minus the smaller
value of X.] this is from the paper....
anyway so the possible values of Y are
0 [when (2,2) or (4,4) or (6,6) are chosen]
2 [when (4,2) or (6,4) or (2,4) or (4,6) are chosen]
4 [when (6,2) or (2,6) are chosen]

So now we know the values, we have to find the probabilities
for 0, (0.5 x0.5) + (0.4 x 0.4) + (0.1 x 0.1) = 0.42 [P.S, i sub the probabilities of the numbers above from table.]
for 2, (0.4 x 0.5) + (0.1 x 0.4) + (0.5 x 0.4) + (0.4 x 0.1) = 0.48
for 4, (0.1 x 0.5) + (0.5 x 0.1) = 0.1

 

[Mayedah]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_11.pdf
question 7 part (ii) please anyone solve this !

 

[A star]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf
question 7 part (ii) please anyone solve this !

from solving first part you are able to get to sinx=0.5 . now if you compare both equations the only diff is n so we can say that
30/10=n hence n= 3 now for greatest solution find all solutions up to 3 x 360. then divide the highest with 3 you will get the answer

 

[Sadiqa Ali]

can anyone please provide me the worked solutions for m/j 2011 p3 questions??

 

[Esme]

This is from M/J 08 p4


For ii) can you explain why the speed at C is equal to the speed at A. Is it because both the points are at the same level? Even then, can someone provide the logic behind it? Maybe from a practical point of view?

 

[Tabi Sheikh]

This is from M/J 08 p4
View attachment 24331

For ii) can you explain why the speed at C is equal to the speed at A. Is it because both the points are at the same level? Even then, can someone provide the logic behind it? Maybe from a practical point of view?

at same point the potential energy gained or loss is equal for a body and here we assume that change in potenial energy is equal to change in kinetic energy

 

[PANDA-]

This is from M/J 08 p4
View attachment 24331

For ii) can you explain why the speed at C is equal to the speed at A. Is it because both the points are at the same level? Even then, can someone provide the logic behind it? Maybe from a practical point of view?

Since it is smooth, there is no loss of energy because there is no friction, so potential energy is directly converted to kinetic energy. Gain in potential energy = Loss in kinetic energy. So as it goes down from A to B, it loses P.E. and gains K.E., when it goes up to C, it gains the lost P.E., and loses the gained K.E., velocity returns to as it was in A.

 

[~`Heba`~ :)]

Lol.... i don't think it's possible to show a graph here
Unless we draw it scan etc...

oh okay

 

[ahmed abdulla]

ANY HELP >>> THIs question really conFUSES and u can call it hard question....
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_11.pdf q 7 (ii) .... ? with clear expaination plzz

 

[~`Heba`~ :)]

can anyone help me with #7 (ii)

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_11.pdf
how do i get the largest value?

 

[ahmed abdulla]

can anyone help me with #7 (ii)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf
how do i get the largest value?

same doubt

 

[PANDA-]

ANY HELP >>> THIs question really conFUSES and u can call it hard question....
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf q 7 (ii) .... ? with clear expaination plzz

can anyone help me with #7 (ii)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf
how do i get the largest value?

2cos^2(nθ) = 2-2sin^2(nθ)

2-2sin^2(nθ) = 3sin(nθ)

nθ = 30 or nθ = 150

Smallest value = 10
θ = 10

n(10) = 30
n=3

Now for the largest value, well I don't know, because logically, it would be 50 degrees... because

nθ = 150
3θ = 150
θ = 50 <--- But this isn't the correct answer according to the ms... PhyZac Could you help?

 

[PhyZac]

ANY HELP >>> THIs question really conFUSES and u can call it hard question....
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf q 7 (ii) .... ? with clear expaination plzz

can anyone help me with #7 (ii)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf
how do i get the largest value?

PANDA-
Okay, see.
From first part u get 150 as well as 30

Now they said smallest value now is 10 meaning
sin (n(10))= 1/2
10n = 30
n = 3

Okay now to find the largest value...first change the range
from 0 < q < 360
by multiplying with 3 (since now 3q)
to
0< q<1080

that is a three rounds in the quadrants [hope u get this point]


I drew a sketch, every round is different colour. So the largest value is on the third round, on the blue point, from the sketch u find, 2 rounds plus 150
(2 rounds, 360 x 2 = 720
3q = 720 + 150 = 870
q = 290

 

[PANDA-]

PANDA-
Okay, see.
From first part u get 150 as well as 30

Now they said smallest value now is 10 meaning
sin (n(10))= 1/2
10n = 30
n = 3

Okay now to find the largest value...first change the range
from 0 < q < 360
by multiplying with 3 (since now 3q)
to
0< q<1080

that is a three rounds in the quadrants [hope u get this point]


I drew a sketch, every round is different colour. So the largest value is on the third round, on the blue point, from the sketch u find, 2 rounds plus 150
(2 rounds, 360 x 2 = 720
3q = 720 + 150 = 870
q = 290

I see. Thanks for the explanation.