Mathematics: Post your doubts here!

[yousef]

Don't have M1. I don't know how it will be solved but.. If you know the speed as well as the height of the hill .. Potential energy changes into kinetic..

mgh = 1/2*mv^2
2gh=v^2
v=√(2gh)

Don't Have M1 so I don't really know how it will be solved...

thanks .. but thats not the way )) :: can u plz tell me how to do ..mj 2010 varient 22 physics ... q 4

 

[Rutzaba]

Q2

To solve for the three unknowns, we need to construct three equations using the data provided.

ΣP(X=x) = 1
p + q + r + 0.4 = 1 --- (i)

E(X) = 2.3
-3p + 2r +1.6 = 2.3 --- (ii)

Var(X) = 3.01
9p + 4r + 6.4 - 2.3² = 3.01 --- (iii)

I will leave the arithmetic to you.

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Q5 iii

P(87 < X < k) = 0.3

P[(k - 82)/√126] - P[(87 - 82)/√126] = 0.3

Let t = (k - 82)/√126

Φ(t) - Φ(0.445) = 0.3

Φ(t) = 0.9718

Φ(t) = Φ(1.978)

t = 1.978

(k - 82)/√126 = 1.978

k = 104

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I don't understand what you wrote next...

sum one else solved...
now tell me in june 2012 ppr 13 question 3.... is b wrong in the answer sheet...
just solve the quadratic eq mentioned in the second last line of question 3

 

[Tkp]

sum one else solved...
now tell me in june 2012 ppr 13 question 3.... is b wrong in the answer sheet...
just solve the quadratic eq mentioned in the second last line of question 3

nope why b will be wrong

 

[Rutzaba]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_ms_13.pdf

 

[Minato112]

Im srry for this post everyone but I posted a long answer here and it just wont show it... badrobot14 is this another problem?

I thought I would upload a word file which contain all these answers, but it says uploading failed! I have sme answers pending...

 

[Rutzaba]

nope why b will be wrong

chek fr ur self... 15x^2 -24x +4
if u solve this... how will u get -4.25

 

[Tkp]

chek fr ur self... 15x^2 -24x +4
if u solve this... how will u get -4.25

ok.check out the value of a first which is 0.5 then put the value of a
15*.25-24*.5+4

 

[Rutzaba]

ok.check out the value of a first which is 0.5 then put the value of a
15*.25-24*.5+4

y??? the eq i got was also this... y wud we use 24.5 instead?

 

[Rutzaba]

ok.check out the value of a first which is 0.5 then put the value of a
15*.25-24*.5+4

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

 

[Rutzaba]

ok.check out the value of a first which is 0.5 then put the value of a
15*.25-24*.5+4

thankyouuuuuuuuuuu

 

[badrobot14]

Im srry for this post everyone but I posted a long answer here and it just wont show it... badrobot14 is this another problem?



I thought I would upload a word file which contain all these answers, but it says uploading failed! I have sme answers pending...

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.....

 

[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