can someone solve q2 http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_32.pdf
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did anyone help you?i had some ambiguity in below quest:
1) P(Z>a)=0.8888 , a is negativ,
1-fi(-a)=o.8888
1-[1-fi(a) ]=o.8888
the other solution of quest i found:
2)p(z<t)=0.25, t is negativ
fi(-t)=0.25
1-fi(t)=0.25..
can anyone tell me this difference and concept how to solve such type of quest plzzz
its 2 + 2^(1/2)?Alright... ii) part please ?
THANX MAN you Are A Big Help In This Crucial Tym INSHALLAH ALLAH will Reward U(3-λ)i + (-2+2λ)j + (1+λ)k = (4 +µa)i + (4 +µb)j + (2-µ)k
comparing coefficients of k
(1+λ) = (2-µ)
λ = 1 - µ
comparing coefficients of i
(3-λ) = (4 +µa)
3 -1 + µ = 4 + µa
µ = 2/(1-a)
comparing coefficients of j
(-2+2λ) = (4 +µb)
-2 + 2 - 2µ = 4 + µb
0 = 4 + µ(2+b)
0 = 4 + (4+2b)/(1-a)
2a-b = 4
its 2 + 2^(1/2)?
thanx for the prayer..THANX MAN you Are A Big Help In This Crucial Tym INSHALLAH ALLAH will Reward U
wait ..
Alright... ii) part please ?
Thank you so much.These might help, but do as many past paper questions on these topics as you can.
And most importantly pray to Allah, insha'Allah you will get it.
Sorrry for the delay .. am not good at paint ..
It's ok. Chill !
This really helped! Got it now! Thanks a lot
no need to say thanx again and again or I wil have to say "anytime" again and again ..
Esme Can you plz help me with q.4 of this paper.: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_32.pdf
u dont have to keep sayin anytime everytime...just like the post dude liking it is like sayin u welcomeno need to say thanx again and again or I wil have to say "anytime" again and again ..
really ?? .. ohkay..u dont have to keep sayin anytime everytime...just like the post dude liking it is like sayin u welcome
can someone solve q2 http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_32.pdfas far as i can comprehend you would be having difficult in integrating
cos(2x)/sin(2x) dx
u see differential of sin(2x) is 2cos(2x) ...
to get 2 cos(2x) in numerator just multiply the integral with 2/2
u shall get 1/2 S 2 cos(2x)/sin(2x) ( S denotes integral sign)
now u can integrate using integration by recognition.. it will become
1/2 ln (sin(2x))
i got it thanksits logarithm question
substitute 5^x by a
yu shall get
a/5 = a - 5 (if yu dont get this step .. tell me)
a = 5a - 25
a =25/4
now
5^x = 25 /4 (apply logs to both sides)
x log (5) = log (25/4)
x = 1.14
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