Edexcel AS Maths Paper 2- Stats and Mechanics *UNOFFICIAL MARK SCHEME *

For question 7 the total time was 264 not 248

because (1/2x80x24)+(24xt)+(48x24x1/2)=4800

So 960+24t+576=4800

24t + 1536 = 4800

24t = 3264

t = 136s

Total time: 136+80+48=264s

For question 7 the total time was 264 not 248

because (1/2x80x24)+(24xt)+(48x24x1/2)=4800

So 960+24t+576=4800

24t + 1536 = 4800

24t = 3264

t = 136s

Total time: 136+80+48=264s

but the velocity was never negative so how is the
displacement not the same as the distance?
genuinely confused

The displacement is never negative. If you plotted the graph of x against t, you'd see that the velocity is negative from t=1/2 to t=1, which means that at that point, the particle returns to its original position. So the distance travelled is 2 times the distance travelled from t=0 to t=1/2. The displacement from t=0 to t=1 is 0 but the distance is not that.
Does that help a bit?

how did people work out the mean
famm i think my way i did was wrongg

is it not the sum of the frequency /n ?

2a I think was 6%?
I think 3 and 4 were the other way round? (not that it matters)
I wrote down that 3ai was 0.0599 and 3aii was 0.381 so that's probably your question 4.
5b I think was 0.005889898 so to 3sf that's 0.00589

I thought I got 264 for the distance in 7 but I might be misremembering.

Thanks for this!
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yeah man I'm sure i did that, but i just dont remember getting something like a 10.1/2 answer for it
have no idea yikes

I think it was 31 / 10 = 3.1
Then the deviation i used sqrt of Sum on f^2/n - (sum of f/n)^2