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Mathematics: Post your doubts here!

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STATS (Can someone help?)
The random variable X has the distribution B (20,p)
Given that p=0.7 and P(X≥x) >0.1, find the greatest possible value of x.
Is there a straight forward method rather than the long method of trying?

The answer is 17 just in case.
Thanks :)
Which year is this from?
 
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Which year is this from?
I'm sorry I forgot. I believe its somewhere between 2005 and 2008.
It is not straight forward you literally have to calculate the probability and you can't use normal distribution concept but actually they kind off gave you a hint, the part before it they asked to calculate the probability when X=16 and we got it greater than 0.1, and we know when you are powering a fraction the greater the power the lower the value so you could expect to start ahead with 17 )
 
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Help needed in 12 part a please solve it.
TIA
 

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Does anyone have Notes about rotation of complex number point when it is mulitiplied or divide by other complex number?
 
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Someone plz post the solution to the (ii) part
 

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Actually its an olevels question but i wasnt getting answer in o level thread so i decided to ask it from a level thread members
i hope no one minds it :::))))))))
help me with part b and part c of this question :
upload_2019-4-12_1-14-23-png.64570

upload_2019-4-12_1-14-57-png.64571
Okay, first of all, keep in mind the fact that they are similar
Use this concept:
(CY / CO) = (BY / BA )
They said the ratio of CB: BY is 1:2, therefore, you can say that CB magnitude is 1/3 of CY )
I kept getting AB as 2c instead of 4c
I took 0607 last year and you take another code, check if you have any extra rules I probably didn't take
Or you can take OAB and OBC since they're similar too
you can say (AC /OB)=(3a /AB) which you'll also get 2c
or even if you use (3a= kAB) and (6c=kBC) you'' get the same value.
Please check with what I'm saying and tackle this., the thing is its no isosceles or equilateral so we cant assume anything
 
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Okay
Actually its an olevels question but i wasnt getting answer in o level thread so i decided to ask it from a level thread members
i hope no one minds it :::))))))))
help me with part b and part c of this question :
upload_2019-4-12_1-14-23-png.64570

upload_2019-4-12_1-14-57-png.64571
so lets look at this
It might seem confusing if youre stressed because thats what happened to me
They clearly said ratio or CB:BY is 1:2 therefore we can say that BY is 2/3 of CY
so you can say
(CY / CO)= (BY / BA)
Since BY is 2/3 of CY then
(CY / CO)= (2/3CY / BA)
Reconstruct the formula to get (CY/ 2/3 CY)= (OC / BA)
So 3/2 =(-6c = BA)
BA- -4c
and since they want AB then reverse the direction so it becomes AB=4c
When you find AB you can find OB since AB= OB=OA
And you can find the rest using the same concept
DO NOT ASSUME ANYTHING THIS IS A SCALENE TRIANGLE NO SIDES ARE EQUAL SO NO VECTORS ARE EQUAL
 
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Measurements of wind speed on a certain island were taken over a period of one year. A box-and-
whisker plot of the data is obtained where the LQ= 39 and UP=63. It is suggested that wind speed can be modelled approximately by a normal distribution with mean μ and standard deviation σ.
(i) Estimate the value of μ.
(ii) Estimate the value of σ.


The value of μ=(63+39)/2=51, but can someone help me how to find the σ?
 
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Measurements of wind speed on a certain island were taken over a period of one year. A box-and-
whisker plot of the data is obtained where the LQ= 39 and UP=63. It is suggested that wind speed can be modelled approximately by a normal distribution with mean μ and standard deviation σ.
(i) Estimate the value of μ.
(ii) Estimate the value of σ.


The value of μ=(63+39)/2=51, but can someone help me how to find the σ?
Feeling lazy so won't solve it, just guide. You make any two equations and solve them simultaneously for example:
P(39<X<63)=0.5 - since the probability of values lying between LQ and UQ is 0.5 so solve this and get answer for standard deviation(using mean from (i))
You may also use the equations: p(X<39)=0.25 and P(X>63)=0.25 and solve simultaneously.
 
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Someone plz post the solution to the (ii) part
|z-10i|=2|z-4i|
(z-10i)(z+10i)=4(z-4i)(z+41) - using the concept from first equation in (i)
(z-10i)(z-10i)*=4(z-4i)(z-4i)* - now use concept from 2nd equation in (ii)
(z-10i)(z*+10i)=4(z-4i)(z*+4i)
zz*+10iz-10iz*+100=4(zz*+4iz-4iz*+16)
3zz*+6iz-6iz*-36=0
3(zz*-2iz*+2iz-12)=0 - first part proven
|z-2i|=(z-2i)(z*+2i)=zz*-2iz*+2iz+4
Now compare this with the earlier equation we have proven. If we subtract 16 from this new equation, both the equations would be equal right? So...
zz*-2iz*+2iz+4=4^2 - notice i out 4^2 not 16 because we were squaring both sides of the equation as the question had told us to do earlier.
Hence radius=4 making the equation |z-2i|=4
 
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|z-2i|=(z-2i)(z*+2i)=zz*-2iz*+2iz+4
Now compare this with the earlier equation we have proven. If we subtract 16 from this new equation, both the equations would be equal right? So...
zz*-2iz*+2iz+4=4^2 - notice i out 4^2 not 16 because we were squaring both sides of the equation as the question had told us to do earlier.
Hence radius=4 making the equation |z-2i|=4
Can you please explain this part again? I don't get the concept of 4^2 ?
 
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Can you please explain this part again? I don't get the concept of 4^2 ?
Ok so we are trying to find out the radius of the (circle) complex number |z-4i| right? So how our earlier proof showed us that we need to square both the sides and gather the terms to show that one side of complex number is equal to the other. Basically, the proof zz*-2iz*+2iz-12=0 shows us that every complex number equation must boil down to this if both sides of equation are equal.
Let radius of the complex number=X so |z-4i|=X and now square both sides
zz*-2iz*+2iz+4-X^2=0
If you compare this to the equation earlier, both equations are almost identical with the difference being that the constant term is -12 in the other equation and in this equation it's 4-X^2.
We need to make the constant terms equal so that the equations are identical.
So -12=4-X^2... X will turn out to be 4 meaning that the radius is 4.
 
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Can someone help me with Part (ii)
View attachment 64584
Oct/Nov/2016/61
For a blue pen to be in the left pocket once operation T takes place, a red pen must've been chosen from the left pocket followed by red pen from right pocket (1st) OR a blue pen chosen from left pocket and a blue pen chosen from right pocket(2nd).
For 1st scenario: (3/4)*(4/5)=0.6
2nd: (1/4)*(2/5)=0.1
0.6+0.1=0.7
 
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You must find the combinations when only the batsman is being included in the squad(1st), only the bowler is being included in the squad(2nd) and when none of them are being included(3rd).
1st: 1C1(for the batsman having a confirmed place)*6C4*4C4(bowler has been excluded here)*2C1*2C1=60
2nd: 6C5*1C1(bowler having confirmed place)*4C3*2C1*2C1=96
3rd: 6C5(batsman excluded)*4C4(bowlers excluded)*2C1*2C1=24
24+96+60=180
 
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