Mathematics: Post your doubts here!

[The Sarcastic Retard]

Part (ii) help pleaseView attachment 59965

AZ = 5cos(0.6)
AQ = 5
ZQ = AQ - AZ = 5 - 5c0s(0.6) = 0.8733
ZP = ZQ = 0.8733
so PQ = 0.8733 + 0.8733 = 1.75

 

[mohmed ahmed soliman]

At least one damaged tape means (P>=1) which is the same as (1-P (0)).
1-[ nC0*(0.2)^0*(0.8)^n ] >=0.85
1-(0.8)^n >=0.85
Therfore solve to get n>=8.5 so the smallest value will be n=9 (we need an integer value)

(P>=1) which is the same as (1-P (0)) so then why u didnt do any calculation based on this rule

 

[i_try9621]

Doubt in iv) . It's from http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_43.pdf

 

[Frizzy03]

Hi guys!May someone please assist me in answering part i of question 9 for 9709/31/M/J/10.Explanation(s) and method would be helpful.Thanks in advance for your assistance

 

[farhan141]

Thanks in advance . Please show all the steps if u can

 

[Bhaijan]

math p3 year 05 may/june
my question is part ii)
should i solve by bringing cosec to right or the other way

 

[The Sarcastic Retard]

I just expl

Hi guys!May someone please assist me in answering part i of question 9 for 9709/31/M/J/10.Explanation(s) and method would be helpful.Thanks in advance for your assistance

I just explained it yesterday check recent posts.

 

[The Sarcastic Retard]

math p3 year 05 may/june
my question is part ii)
should i solve by bringing cosec to right or the other way

View attachment 59981

To do this.
cosec(x) - 1/2x - 1 = f(x)
Let f(x) = 0
Then put f(0.5) and f(1) then one wwill be < 0 and one will be > 0 so it means root lies b/w this points.

 

[The Sarcastic Retard]

Thanks in advance . Please show all the steps if u can

Find AB.
AB = 2rcos@
Area of sector ABC = 4@r^2cos^2@
Area of triangle BOA = r^2cos@sin@
Area of sector OBA = 1/2r^2(pi - 2@)
Aera of segment = area of sector OBA - Area of triangle BOA = r^2(1/2(pi - 2@) - cos@sin@)
Area of 2 segements = 2 *r^2(1/2(pi - 2@) - cos@sin@)
Total shaded region = Aera of sector ABC + area of 2 segments = 2r^2(pi/2 - @ - cos@sin@) + 4@r^2cos^2@
Now manipulate it to get r^2(pi - sin2@ + 4@cos^2@ - 2@)
Now we know Total area of shaded region = 1/2 * pir^2
So r^2(pi - sin2@ + 4@cos^2@ - 2@) = 1/2 * pir^2
So manipulating this will give us cos2@ = 2sin2@ - pi/4@

P.S. I am in hurry.. If u have any doubts in this ask me. Just to make sure in triangle OBA there are 2@ drawn in diagram below the third angle will be pi - 2@ so maybe if this can arise as ur doubt, this is the solution to it. (area of segment OBA))

 

[iSean97]

S1 Help 9709/61/O/N/13/Q4

Spoiler

 

[qwertypoiu]

S1 Help 9709/61/O/N/13/Q4

Spoiler

Do you need help for all the parts or a particular one? And wow this is the first time I saw a spoiler button here!

 

[iSean97]

Do you need help for all the parts or a particular one? And wow this is the first time I saw a spoiler button here!

Nevermind just solved an hour ago. Miscalculated IQR.

Erm how to use the info to plot this?
I know Mean is the position of the Peak. But how to determine the height of the peak? How to use the Variance here???
W13/O/N/61/Q1

Spoiler

 

[Saad the Paki]

Nevermind just solved an hour ago. Miscalculated IQR.

Erm how to use the info to plot this?
I know Mean is the position of the Peak. But how to determine the height of the peak? How to use the Variance here???
W13/O/N/61/Q1

Spoiler

View attachment 59986

I dont know how to do this but seriously though how do you use that spoiler button?

 

[Saad the Paki]

(P>=1) which is the same as (1-P (0)) so then why u didnt do any calculation based on this rule

I did do the calculations.
If u see, P (0) = nC0*(0.2)^0*(0.8)^n = (0.8)^n [ PS " nC0 " is equal to 1 and " 0.2^0 " is also equal to 1 so you're just left with (0.8)^n ]
1-P (0) = 1-(0.8)^n

 

[mohmed ahmed soliman]

help in II

 

[mohmed ahmed soliman]

help in II

 

[mohmed ahmed soliman]

help in B

 

[Saad the Paki]

help in IIView attachment 59987 View attachment 59988

At least 2 observations means P (2) + P (3) + P (4)
OR 1 - [ P (0) +P (1) ]
The probability is given in part (i) so just plug in the values in the formula

 

[mohmed ahmed soliman]

help in 3 iii

 

[mohmed ahmed soliman]

in I why o.256^4 why not 0.256*4
i cant do anything in II