can anyone explain the 6 b ii question in this paper http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w08_qp_04.pdf
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no need .. u'r welcomeThanks
its okie..i am glad if i helpedThanks alot! Really helped me
But, why do we add 2/6?-----------------------
we have 2 cards with no. 1
the probability of it is 2/6 the we solve bii) the first not 1 is 4/6 and the second is 1 is 2/5 ((without replacement)) so the probability that one card is 1 from 2 cards is 4/6 x 2/5
then he wants that 1 card chosed is 1 before the third try ---so we have the probability that 1 from 2 cards has the no.1 . add this to the probability of getting 1 which is 2/6-------->2/6+ 4/6 x 2/5 = 3/5
cuz we have 2 card and the probability that e multiplied are for 1 card the first not 1 and the second is 1But, why do we add 2/6?
the height for 22-24 is straight forward just 26/10= 2.6can anyone explain the 6 b ii question in this paper http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w08_qp_04.pdf
can u tell me how did u find i first?can anyone please help me with Q10 ii) on http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w10_qp_42.pdf and the steps on working it out please.
Appreciate it
Simultaneous Equationscan u tell me how did u find i first?
with the nth term formula right?Simultaneous Equations
i didnt use any nth term formula just derived a simultaneous equation from the difference between the terms in c i) but wasnt able to do so in c ii)with the nth term formula right?
its quite simple:can anyone please help me with Q10 ii) on http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w10_qp_42.pdf and the steps on working it out please.
Appreciate it
thank u sooo much!!! =Dthe height for 22-24 is straight forward just 26/10= 2.6
(the 10 comes from 115/11.5)
then for 10-20... the gap between 10 and 20 = 10
so we divide 10 by 2 which was the gap between 20 and 22
then multiply the ans=5 by 10
that = 50
thus we divide 35/50 = 0.7
the last one is similar...
the gap is 6 this time
so 6/2 = 3
then 10*3= 30
therefore 24/30 = 0.8
hope it helps!!!
its quite simple:
1) make 3 eqns. (since there are 3 variables)
2) they are:
a) -33+x=y
b) x+y=z
c) y+z=18
3) then replace y with -33 + x
i.e. x + -33 + x= z
and -33 + x + z =18
4) reorder and make it into a simultaneous eqn.
i.e. 2x-z=33 and x +z = 51
5) solve the eqn.
6) u will get the value of x, = 28
7) then put the value of x (28) in the first eqn. (a)
8) u will get y as -5
9) then put the value of y (-5) in the last eqn. (c)
10) u will get z as 23
so there u have it
hope it helps and best of luck!!!
LOL u seem to be going over the top XDi make the same steps with the same numbers then the answer is completely different from the mark scheme
what's wrooong ???????// is it my calculator????
LOOOL i don't mean this Q i mean with other questions using sine and cosine and things like thatLOL u seem to be going over the top XD
just do this:
1) leave ur calculator aside (its unnecessary for the question)
2) create the 2 eqns.
a) 2 + d=e
b) d + e =10
3) reorder eqn. (a)
i.e. d-e = -2
4) solve the simultaneous eqn.
5) 2d = 8 is what u should get
6) so d = 8/2 which is 4
7) put the value of d in eqn. (b)
8) therefore 4-e=-2
9) so e should be 4 + 2 = 6
10) answers: d = 4, e = 6
now u can calm down
good luck!!
if u hv the paper printed out it would be easier for u.Thanks always smile.
I have some other doubt. How to answer question 6 b) iii:
http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w09_qp_4.pdf
oh ok....my badLOOOL i don't mean this Q i mean with other questions using sine and cosine and things like that
IT IS DRIVING ME CRAAAAAAAAAAAAZY
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