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39/51 is for part (i) and ur answer is for (ii)

Question 8a) N = Ae^(kt) when t=1, N=1600 1600=Ae^k e^k = 1600/A ----------(1) when t=2, N=2600 2600=Ae^(2k) from (1) : 2600 = A (1600/A)^2 2600A = 1600^2 A = 1600^2 / 2600 A = 984.62 (2dp) 1600 = 984.62 * e^k ln (1600/984.62) = k . ln e k = 0.49 (2dp)

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??

not sure, i simplified it but don't know if we needed to. POAD: we were aloud to take it out of exam room at the end unless you left early...

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

I just did lots of past papers... been doing them since the end of term 2 when we finished the core. doing 3 unit helps as well

yeah thanks.. 5π/2 not 5π/3 thanks for the solutions too.

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

incorrect solution: 4c) iv) Find area of major sector ACB correct answer should be 1/2 . √3^2. 5π/3 = 5π/3 units sq

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

just the 1600

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??

hey mc88, in mobile fone growth... A = 948.62 (2dp) k = 0.49 (2dp)

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)...