put it this way: the probability is dependent on the first card you pull out so there fore it doesn't matter what you pull out... you get a club and then u have 39 non clubs to choose from from the 51 cards that are left
does that clear it up??
Tracing/ Drawing the diagram also helps the markers to see wot u were doing as an extra to explain your working...
thats y they want it ( but don't need ) if u've made the rest of it clear
got a question -- in Q7b) i) where √3cosx = sinx would it be reasonable to go:
√3 = tanx
therefore x = π/3
is that correct?
actually.. don't worry.. it is right, and the other solution is 4π/3 ... all good, just 4got the full ?ion for a second
Probability Q9
b) (i) was 1 mark - answer = 39/51
b) (ii) 2 marks - answer = 39.26.13/51.50.49
The Caine Family
(ii) at 3rd bday was 2 marks
(iii) 18th - 1 mark (answer about 51000)
yes, thats right: √3 sq . 5π/3 . 1/2 = 5π/2
thats my only error so far other than Q10b ii (hopefully they give me 1 mark for it to compensate for this silly one)
good point - just lost a mark :(
The major sector would be the lightly shaded part which would be the majority of the large circle...
ie A = (√3.√3. 5pi/3 ) / 2
and total area is area of quad in (iii) and major sector from (iv) and then lightly shaded sector of small circle
no it wasn't million
N=Ae^(kt)
Where N is the estimate for the number of mobile phones in use (in millions) and t in years...
so when t=1, N=1600 not 1,600,000,000
therefore A is 948 not 948 million
t=7.2 because distance of +ve velocity = 8.5 units
therefore with -ve velocity, u have area of triange = 2.5 units then 6 units at 5units / second
does anybody else agree with that??
Question 5b) (iv) was ln 2
Question 7a) (iii) k was 15/4
Question 7a) (iv) P (1/2, 4.25) ??
Question 7b) (ii) total shaded region was 4 units sq
Question 8a) (ii) November 2010
Question 8b) (iv) p=12 q=18
Question 9b) (i) 39/51
Question 9b) (ii) (39*26*13)/(51*50*49)
Question 10a) (iii)...