Mathematics: Post your doubts here!

[Tkp]

s

ohhhhhhhhhhh u mean instead of solving we will simply substitute the value of a into it? oh

o now u got it rght

 

[salvatore]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_13.pdf
Please help me with question no. 9 (i & iii) and 10 (iii).

I'll really appreciate if anyone could help me solve the above questions (Hope they aren't too many )
Thanks

 

[Rutzaba]

yes, uploading seems to b giving problems... still.... sorry..



try uploading it to a file sharing site or maybe ur google drive n share the link in the meantime.....

bhai... also copy paste nhi horha... kuch bhi nhi -_- bari mushkil se ye link hua tha kal

 

[Dudu]

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

Good man!
Sorry on the late acknowledgement, really appreciate it mate

 

[airyhat]

Q5

a. ii)

5 _ _ _

No. of arrangements = 6³ = 216

b. ii)

n(B) = 6

n(G) = 8

If the cousins are in, 3 boys are already selected. The number of ways to select the remaining team is 11C2.

If the cousins are out, we need to select 5 students for the team which is simply 11C5.

No. of ways = 11C2 + 11C5 = 517

Can u explain why is it 6^3?
And if three people are already chosen won't it be 8C2+8C5? Why are we still taking 11 when we know 3 of the boys are cousins

 

[hassanbmj]

Please help with my question too people

 

[Fatima18]

Fatima18

S03 qp 6
Q6 iii

P(G) = 17/42

They are asking for the probability of choosing a house where a parent lives, such that a grandparent is already chosen.

P(H|G) = P(H1|G) + P(H2|G)

P(H1|G) = P(H1∩G)/P(G) = [(1/3)(2/7)]/(17/42) = 4/17

P(H2|G) = P(H2∩G)/P(G) = [(1/3)(3/7)]/(17/42) = 6/17

P(H|G) = 10/17

---------------

w07 qp 6
Q7

ii)

P(RB) = P(W And R) + P(R And R)

= (1/6)(7/10) + (5/6)(8/10)

= 47/60

iii)

P(RA|RB) = P(RA∩RB)/P(RB)

= (2/3)/(47/60)

Thanx alot!!

 

[Fatima18]

i)

P(A) = P(B) = P(C) = 1/3

P(2T) = P(A and 2T) + P(B and 2T) + P(C and 2T)

= (1/3)(6/10)(5/9) + (1/3)(5/8)(4/7) + (1/3)(3/10)(2/9)
= 53/210

ii)

P(A|2T) = P(A and 2T)/P(2T)

= [(1/3)(6/10)(5/9)]/(53/210)

= 70/159

Thanx..xD

 

[asexamskillme111]

lny - ln70 = e^-3t - 1

Can someone rearrange this to make y the subject? If you would show me steps that would be helpful

 

[Rutzaba]

lny - ln70 = e^-3t - 1

Can someone rearrange this to make y the subject? If you would show me steps that would be helpful

by ln proprty if the are being substracted it mean they are eing divided

so take ln common... ln(y/70) =e^-3t - 1

then wen we will remove the ln on the other side the whole thing wud become the power of e

y/70 =e^ (e^-3t - 1)
y=70 e^ (e^-3t - 1)

 

[Minato112]

Please help with my question too people

Can U plz repost it?

 

[peter lucantoni]

I NEED SOME HELP! In the question 6(ii) ,where is the 1.282 come from? thank you so much!



please !

 

[hassanbmj]

Mechanics problem:
Q: A particle is moving along a straight line with constant acceleration. In an interval of T sec ot moves D meters; in the next interval of 3T sec it moves 9D meters. How far does it move in a further interval of T seconds ?
Answer 5D.
Your help will be much appreciated.​

 

[Rutzaba]

Mechanics problem:​

Q: A particle is moving along a straight line with constant acceleration. In an interval of T sec ot moves D meters; in the next interval of 3T sec it moves 9D meters. How far does it move in a further interval of T seconds ?​

Answer 5D. ​

Your help will be much appreciated.​

dude i got a perfect eye sight...wat i aint got is... mechanics xD

 

[syed1995]

dude i got a perfect eye sight...wat i aint got is... mechanics xD

same xD

 

[Dug]

Mechanics problem:​

Q: A particle is moving along a straight line with constant acceleration. In an interval of T sec ot moves D meters; in the next interval of 3T sec it moves 9D meters. How far does it move in a further interval of T seconds ?​

Answer 5D. ​

Your help will be much appreciated.​

s = ut + ½ at²

At t = T, s = D
D = uT + ½ aT²
uT = D - ½ aT² --- i

At t = T + 3T, s = D + 9D
10D = 4uT + 8aT² --- ii

Put (i) in (ii)
10D = 4D - 2aT² + 8aT²
aT² = D --- iii

Put this back in (i)
uT = D - ½ D
uT = ½D --- iv

At t = 5T,
s = 5uT + (25/2) aT² --- v

Plug (iii) and (iv) into (v)
s = (5/2)D + (25/2)D
s = 15D

Distance covered in the final interval = Total distance - Distance covered in the first 4T seconds
= 15D - 10D
= 5D

 

[Sandhya Mahat]

Help me solving this guys: Solve the equation:
Mod(x-1) + Mod(x+1) -1 = Mod(x)

 

[princess787]

can anyone help me in 9709_w12_qp_31 question 3! please! and if you have solved it please upload the whole paper! i solved the papers... bt d MS is not clear! blank spaces everywhere! loads of doubt!! neeed to understand the method! if nt the whole paper! please help me in question 3! thanx! :*

 

[yubakkk]

pLz help this question. its from book.

For a biased cubical dice the probability of any particular score between 1 and 6 (inclusive) being obtained is inversely proportional to that score. Find the probability of scoring a 1.

 

[Scafalon40]

Q5

a. ii)

5 _ _ _

No. of arrangements = 6³ = 216

b. ii)

n(B) = 6

n(G) = 8

If the cousins are in, 3 boys are already selected. The number of ways to select the remaining team is 11C2.

If the cousins are out, we need to select 5 students for the team which is simply 11C5.

No. of ways = 11C2 + 11C5 = 517

wait could you elaborate please? why did you take 6^3?