First consider P(draw)= 1/5. As regardless of who goes first there will be 1/5 chance the second person gets the same number.
That leaves 4/5 left- now since we don't know who goes first or any other information they have equal chance of winning. Therefore, 1/2 * 4/5= 2/5.
The committees setting the papers are independent I believe. Might mean the difficulty was purely coincidental. But to be honest for a decent student the 2U paper wasn't that challenging. Not sure about 4U obviously
For theta I found x to be root 200/3 or something like that. Then I considered that the curved surface area of cone is equal to the area of the cut out sector they gave us. So I just found the max value of the the cone and then used the area of sector to find theta by equating the areas. I got...
Look at raw marks.org never in recent history has requirement b6 been higher than 88. I'm thinking our paper was much harder than 2014 and the last couple of years
Thinking I got around 96,97 what would that align to?
For shoe multi it was 1/7 right?
Also for the max min was I like the only person who equated the curved surface area of the cone and the sector area of the circle to find theta lol there must have been an easier way
Displacement is represented by the area under the velocity curve. So we just calculate the area of the triangle we see i.e. 1/2 * 4 * 8= 16. But the particle starts at 2 so we have 16+2=18. Thus D
Blit probably knows better than me but I would interpret that as requiring two distinct headings from Operations strategies; i.e. you would do global strategies and the four subpoints and then you would pick another like quality management and cover the three subpoints under that as well.