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Mathematics: Post your doubts here!

Dug

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Loss occurs when X<0.
P(X<0) = P(Z < 0 - 6.4/5.2)
= P(Z < - 1.231)
= 1 - Φ(1.231)
= 1 - 0.8909
= 0.1091

This is the probability of loss on any random given day.

Now finding the probability that loss occurs on exactly 1 of the next 4 days:
n = 4
r = 1
p = 0.1091
q = 0.8909

Probability = [4C1] (0.1091) (0.8909)³ = 0.309
 
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Loss occurs when X<0.
P(X<0) = P(Z < 0 - 6.4/5.2)
= P(Z < - 1.231)
= 1 - Φ(1.231)
= 1 - 0.8909
= 0.1091

This is the probability of loss on any random given day.

Now finding the probability that loss occurs on exactly 1 of the next 4 days:
n = 4
r = 1
p = 0.1091
q = 0.8909

Probability = [4C1] (0.1091) (0.8909)³ = 0.309
Thank you!! :)
 
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in first one
after finding a put it in the equation and find its derivative,put the derivative as 0 so that u find the stationary values of x,now do double derivative and put these values of x and see if points are a minimum if they are then put in the following step if one is then put that and forget the other :p,put the values in f(x) and see if u get negative answer which u shudnt as f(x) stays positive !
this proves that the graph never went beyond x axis and was always +,u use stationary values to find the values of x for minimum points because these values have f(x) at the lowest value so if they dont give u minus nothing will :D :D
U will get a cubic equation so u will need to solve it :p :p (hahaha thats why i didnt solve it its really boring ) :p

For the other one try it urself it really is straightforward,u need to use those different forms of expressing complex no. to get w then its quite easy ! Ill INSHAALLAH solve it tomorrow or soon as for now im really out of time sorry :( :(
thank you :D
i think i kinda sorta get the first part :D
but question 5 maybe straight forward for you ....... but not for me :p
so i would like it if you solved it :D
and about the boring part.......when is maths ever not boring ?? :S
 
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part 4 (ii)

We know by part (i) that the factors are (x^2 +2x +2) (x^2 -4x +4)
to factorize (x^2 -4x +4)
this can be done as (x-2) ^2
then we d factorize (x^2 +2x +2)
we wud get ((x+1) ^2) +1)
now if we multiply ((x+1)^2) +1)) (x-2) ^2 all the negative terms wud be squared to give positive results thus
((x+1)^2) +1)) (x-2) ^2 > 0
thank you very very much :D
 
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in first one
after finding a put it in the equation and find its derivative,put the derivative as 0 so that u find the stationary values of x,now do double derivative and put these values of x and see if points are a minimum if they are then put in the following step if one is then put that and forget the other :p,put the values in f(x) and see if u get negative answer which u shudnt as f(x) stays positive !
this proves that the graph never went beyond x axis and was always +,u use stationary values to find the values of x for minimum points because these values have f(x) at the lowest value so if they dont give u minus nothing will :D :D
U will get a cubic equation so u will need to solve it :p :p (hahaha thats why i didnt solve it its really boring ) :p

For the other one try it urself it really is straightforward,u need to use those different forms of expressing complex no. to get w then its quite easy ! Ill INSHAALLAH solve it tomorrow or soon as for now im really out of time sorry :( :(
ummm...its okay you don't have to do question 5 .....i got it now :D
after thinking really really hard of course :p
 
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f(x)=g(x)
So
k-x=9/(x+2)
(k-x)(x+2)=9
kx+2k-x^2-2x=9
-x^2+(k-2)x+2k-9=0
or
x^2+(2-k)x+9-2k=0

For Two Equal and real roots > b^2-4ac=0

(k-2)^2-4(-1)(2k-9)=0
k^2-4k+4+8k-36=0
k^2+4k-32=0

k = 4 or -8

now solve substituting values of k

for k=4
x^2+(2-k)x+9-2k=0
x^2+(2-4)x+9-2(4)=0
x^2-2x+1=0

root = 1

for k=-8
x^2+(2-k)x+9-2k=0
x^2+(2-(-8))x+9-2(-8)=0
x^2+10x+25=0

root = -5

so the values of k are 4 or -8. and the roots for these values are 1 and -5.

PS. Well i might have made some sign mix-ups while copying the solution from my copy ..
That was rele well explained! Thanx :D
 
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Q5. The 2 points r (3,6) and (0.75,3) and the distance is 3.75 units.. right?
u have 2 eqns (y2=12x) and (3y=4x+6)
Write 2nd eqn as (x=(3y-6)/4) and substitute in 1st equation, u will get 6 and 3 as values of y... then substitute in any of the equations to get x.. now we have 2 pts.. find the distance b/w them..

Q10 v ,, here u have to equate the g(x) to 10.
U shd take (square root of x as any other variable,, let it be z
then solve the equation as a quadratic--> z2-3z-10
u will get 5 and -2.. so equate each of them to (square root of x) so answers will be 5 squared and (-2)squared to get 25 and 4.
 
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can any one help me with June 2005 Q7 iii
June 2009 Q10 ii - iii
pure 1 .
Q7.. when u sketch the curve, from 0 to 90 it curves down then bends upwards.. this bending shd mean that the curve doesnt have an inverse but if u stop at 90 degrees, it can have an inverse... so answer is 90.(am sure u didnt get what i meant but try,i think it is considered a fact)
Q10.. ii) minimum point has x=3 as the line of symmetry.. and since lower limit is zero, u multiply 3 by 2 (they said has a line of symmetry) if they would have said "has an inverse" the answer would have been 3.
iii) The minimum point has y coordinate as -5, so f(x) >-5 , then when u substitute x=0, u get y = 13 as upper limit.
(Math is difficult to explain when i dont have the person in front of me LOL :LOL: )
 
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Q7.. when u sketch the curve, from 0 to 90 it curves down then bends upwards.. this bending shd mean that the curve doesnt have an inverse but if u stop at 90 degrees, it can have an inverse... so answer is 90.(am sure u didnt get what i meant but try,i think it is considered a fact)
Q10.. ii) minimum point has x=3 as the line of symmetry.. and since lower limit is zero, u multiply 3 by 2 (they said has a line of symmetry) if they would have said "has an inverse" the answer would have been 3.
iii) The minimum point has y coordinate as -5, so f(x) >-5 , then when u substitute x=0, u get y = 13 as upper limit.
(Math is difficult to explain when i dont have the person in front of me LOL :LOL: )
Ya rahmaaa thank youu :) .. hahaha bas msa shr7ty 7elw (y) :D
 
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Pleasee Mechanics1 June2005 no.4 .

4.i) In this one we use the formula F=ma instead of F, i will use T, since T(tension) is the force working here.
So for Particle A , (there is no friction, only weight and tension) 2-T=0.2a
For particle B , (there is friction and tension) T - uR= 0.3a << in this equation i used Friction=uR(u is the coefficient of friction)
Now we have 2 equations 2-T=0.2a
T-uR=0.3a
Cancel out the Ts. And since its not moving(limiting equilibrium) acceleration = O
We end up with 2-u3=o , u= 2/3 (3 is the normal contact force=R)

4.ii)Since there is upward force acting upon it, we need to find the new normal contact force. So, R = downward force - upward = 3 - 1.8 =1.2
Now also find out the friction =uR = 2/3 x 1.2 = 0.8
We make the equation for new force X.(its going to right away from pulley) X = Frictional Force + Tension
We don't have the tension we find it out by the help of particle A.(tension is same on the both sides)
For particle A, only weight is working. so Weight - Tension = 0 (cause there is no acceleration)
2 - T = 0 , T= 2
Now we can find out X, X= Frictional Force + Tension
= 0.8 + 2 = 2.8
Hope it helped :)
 
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