Mathematics: Post your doubts here!

[PhyZac]

That is one question which I still haven't understood.. I myself need help in that question

Tkp , Dug , PhyZac you guys got any idea how this one would be solved?

This is very tricky question. Took me time to understand, anyway,


Alice123
you are right about (20+12<x<20-12)

now you have to imagine the graph in your head, and you have to come into a conclusion that it is symmetrical.

Now imagine the z of 20+12 that is 32 is x
So the z of 20-12 that is 8 will be a negative number -x

Both of them are x but one is negative, because they are the same distance from the mean

Let the probability of x is p
so the probability of -x is 1-p

we know that difference of the two probability is 0.94

so p - (1-p) = 0.94
p - 1 + p = 0.94
2p = 0.95 + 1 = 1.94
p = 0.97

Now is p is 0.97 z is from the table, 1.882 (or 1.881)

now continue normally

1.882 = 32-20 / sd
sd = 12/1.882
=6.38

 

[syed1995]

This is very tricky question. Took me time to understand, anyway,


Alice123
you are right about (20+12<x<20-12)

now you have to imagine the graph in your head, and you have to come into a conclusion that it is symmetrical.

Now imagine the z of 20+12 that is 32 is x
So the z of 20-12 that is 8 will be a negative number -x

Both of them are x but one is negative, because they are the same distance from the mean

Let the probability of x is p
so the probability of -x is 1-p

we know that difference of the two probability is 0.94

so p - (1-p) = 0.94
p - 1 + p = 0.94
2p = 0.95 + 1 = 1.94
p = 0.97

Now is p is 0.97 z is from the table, 1.882 (or 1.881)

now continue normally

1.882 = 32-20 / sd
sd = 12/1.882
=6.38

Damn! that's real tricky.. I would never have reached that conclusion ever!

 

[Dudu]

Help on this one please?

On a certain road 20% of the vehicles are trucks, 16% are buses and the remainder are cars.
(i) A random sample of 11 vehicles is taken. Find the probability that fewer than 3 are buses. [3]
(ii) A random sample of 125 vehicles is now taken. Using a suitable approximation, find the
probability that more than 73 are cars.

 

[leosco1995]

Prove the identity: cosec 2Ɵ - cot 2Ɵ ≡ tan Ɵ

 

[Iffat]

oct/nov 08
2. 1+sinx/cosx = cosx/1+sinx take common denominator
= [(1+sinx)^2 + (cos^2)x]/cosx(1+sinx)
= (1 + 2sinx + (sin^2)x + (cos^2)x)/cosx(1+sinx) here an identity needs to be used : (cos^2)x +(sin^2)x = 1, so (sin^2)x = 1 - (cos^2)x
= (1 + 2sinx + 1 - (cos^2)x + (cos^2)x )/ cosx(1+sinx) , the '(cos^2)x' cancels out
= 2 + 2sinx/cosx(1+sinx)
= [2(1 + sinx)]/cosx(1+sinx) , the '1+sinx' cancels out
= 2/cosx hence equal to right hand side
hope you understood

thanx...wat bout the oda 3qs?

 

[syed1995]

Help on this one please?

On a certain road 20% of the vehicles are trucks, 16% are buses and the remainder are cars.
(i) A random sample of 11 vehicles is taken. Find the probability that fewer than 3 are buses. [3]
(ii) A random sample of 125 vehicles is now taken. Using a suitable approximation, find the
probability that more than 73 are cars.

Probability of buses = 0.16
Probability of trucks = 0.2
Probability of cars = 0.64

for i) we will use binomial distribution..

n=11
buses = 0.16 = p
not buses= 0.84 = q

Fewer than 3 are buses means .. there can be either 0,1 or 2 buses.

0.84^11 + 11C1*(0.16)*(0.84)^10 + 11C2*(16)^2*(0.84)^9

= 0.7479
= 0.748 (3sf)

for part ii) we use normal approximation to binomial..

n=125
Cars = 0.64 = p
Not Cars = 0.36 = q

first we check if both np >5 and nq > 5 or not..

np = 125*0.64 = 80 and nq = 125*0.36 = 45 so normal can be used...

np = 80 = mean or µ
npq = 28.8 = variance (SD^2)

=>P(X>73)
= P(z>73.5-80/√28.8) // REMEMBER since it's Normal approximation to binomial .. we need to do continuity correction.. i always forget that.. (z>73 means + 0.5 z<73 means - 0.5)


=P(z>-6.5/√28.8)
=P(z>-1.211)
=P(z<1.211)
=Phi(1.211)
=0.8869 + 2*10^-4 (from the table)
=0.8871
=0.887 (3sf) Answer

 

[Tkp]

That is one question which I still haven't understood.. I myself need help in that question

Tkp , Dug , PhyZac you guys got any idea how this one would be solved?

dnt have s1

 

[syed1995]

dnt have s1

you goin for m1,m2 .. or just have p1,m1 this year?

 

[Tkp]

you goin for m1,m2 .. or just have p1,m1 this year?

p1,m1

 

[Dug]

Prove the identity: cosec 2Ɵ - cot 2Ɵ ≡ tan Ɵ

LHS = 1/sin2Θ - cos2Θ/sin2Θ
= (1 - cos2Θ)/sin2Θ
= [1 - (1 - 2sin²Θ)]/2sinΘcosΘ
= tanΘ

 

[leosco1995]

LHS = 1/sin2Θ - cos2Θ/sin2Θ
= (1 - cos2Θ)/sin2Θ
= [1 - (1 - 2sin²Θ)]/2sinΘcosΘ
= tanΘ

Got it, thanks. I need to practice identities a bit more.

 

[Kumkum]

thanx...wat bout the oda 3qs?

sorry forgot bout the others
here:
may/june 09
4ii)
From 4i) u got ur values for a,b and c.
a=6 b=2 and c=3, right?
so u jst put these values into this equation ur given and solve for x
====> y = 6sin(2x) + 3
so when y=0,
6sin(2x) +3 = 0
sin(2x) = -1/2
since you can't solve for negative values u can say:
let sin2x = 0.5
==> 2x = pi/6 (angles are in radians)
what i do frm here i let 2x = alpha (for alpha i'll use @)
so @ = pi/6
now you find the angles for sin where it is negative that will be in the 3rd and 4th quadrant
so,
in 3rd quadrant:
@ = pi + pi/6 = 7pi/6
in 4th quadrant:
@ = 2pi - pi/6 = 11pi/6 <-----this angle u can eliminate it coz u can see it's going to be the largest angle
now u can find x:
2x = 7pi/6
so, x = 7pi/12
hence the smallest is 7pi/12
i know its kinda long but thats how i do it, hope u understood

 

[Kumkum]

thanx...wat bout the oda 3qs?

oct/nov 09
9iii)
so frm 9ii) u had to find the gradients of AD and CD,
AD = 8/h and CD = -8/h-12
from the rectangle u shud see that gradient of AD and CD are perpendicular so u use the concept of 'm1m2 = -1', to equate the gradients nd find the x-coordinates.
so,
m1m2 = -1
(8/h)*(-8/h-12) = -1
-64/h^2 - 12h = 1
-64 = -h^2 + 12h
h^2 - 12h - 64 = 0
h^2 -16h + 4h - 64 = 0
h(h - 16) + 4(h -16) = 0
(h + 4)(h - 16) = 0
h = -4 or 16
so frm here u can straight away see that x-coordinate of B = -4 while for D = 16
hope u got it

 

[wajiman]

I have almost tried this several times but then I give up..Help me

sry i did not take maths

 

[Iffat]

sorry forgot bout the others
here:
may/june 09
4ii)
From 4i) u got ur values for a,b and c.
a=6 b=2 and c=3, right?
so u jst put these values into this equation ur given and solve for x
====> y = 6sin(2x) + 3
so when y=0,
6sin(2x) +3 = 0
sin(2x) = -1/2
since you can't solve for negative values u can say:
let sin2x = 0.5
==> 2x = pi/6 (angles are in radians)
what i do frm here i let 2x = alpha (for alpha i'll use @)
so @ = pi/6
now you find the angles for sin where it is negative that will be in the 3rd and 4th quadrant
so,
in 3rd quadrant:
@ = pi + pi/6 = 7pi/6
in 4th quadrant:
@ = 2pi - pi/6 = 11pi/6 <-----this angle u can eliminate it coz u can see it's going to be the largest angle
now u can find x:
2x = 7pi/6
so, x = 7pi/12
hence the smallest is 7pi/12
i know its kinda long but thats how i do it, hope u understood

wooow that is 1 looong ans..bt i understood ..thanx

 

[Iffat]

oct/nov 09
9iii)
so frm 9ii) u had to find the gradients of AD and CD,
AD = 8/h and CD = -8/h-12
from the rectangle u shud see that gradient of AD and CD are perpendicular so u use the concept of 'm1m2 = -1', to equate the gradients nd find the x-coordinates.
so,
m1m2 = -1
(8/h)*(-8/h-12) = -1
-64/h^2 - 12h = 1
-64 = -h^2 + 12h
h^2 - 12h - 64 = 0
h^2 -16h + 4h - 64 = 0
h(h - 16) + 4(h -16) = 0
(h + 4)(h - 16) = 0
h = -4 or 16
so frm here u can straight away see that x-coordinate of B = -4 while for D = 16
hope u got it

thanx..

 

[Iffat]

can sum 1 plz help wif d foll qs:
2009 oct-nov p12:
q10(i)(c)
2010 may-jun p11:
q1
q5(i)
2010 may-jun p12:
q1(i)
q11(iii)(iv)(v)

 

[Fatima18]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w05_qp_6.pdf
Number 2 pls.

 

[Fatima18]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s03_qp_6.pdf
Number 6 iii )
Please guys.???

 

[Rutzaba]

can sum 1 plz help wif d foll qs:
2009 oct-nov p12:
q10(i)(c)
2010 may-jun p11:
q1
q5(i)
2010 may-jun p12:
q1(i)
q11(iii)(iv)(v)

y dun u ppl buy solved solutions... they are very very help ful and they help u thru with p1 p3 s1 s2 m1 ...
there the explanations are way more lengthy