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The distribution curve is symmetrical about its mean which is 20.THANK YOU SO MUCH!!
could you please explain why P(X < 8) = P(X > 32) = 0.03?
THANK YOU SOOO MUCH XDThe distribution curve is symmetrical about its mean which is 20.
Total area under any normal distribution curve = 1
In the above sketch, if the yellow area = 0.94,
P(X < µ-2σ) = P(X > µ+2σ) = 0.3
thanks for the helpFor Q.10.iii., it's the usage of the iteration formula simply. The root lies between 1.2 and 1.3, so let's assume it is 1.25. Now
Tn =T1= 1.25
Use the formula given to calculate further T(n+1) values
T2 = 1.2703
T3 = 1.2745
T4 = 1.2745
T5 = 1.2756
T6 = 1.2756
At this point, you see that the iteration has converged successfully. Hence, the value of the root is 1.276 (correct to 3 decimal places).
(If you're unsure about the concept then refer to this http://www.examsolutions.net/maths-...erical-methods/roots/iteration/tutorial-1.php)
For Q.10.iv., you just use the value found for t, that is, 1.276 and solve the equation. But before that, simplify the equation, especially the part in tan 2x.
I got it... thanks
you dint get velanes explanation?
that was for paper 31 and this is for paper 32, awwww, come on
okay i will try doing it
okay sooo ..........
okay sooo ..........
part 1 ) you have to show that after a time 't' the distance between the particles is 2.6t
so you use the formula s=ut+1/2at^2
then d= (1.3t + 1/2at^2) - (-1.3t + 1/2at^2)
solve this and you get d=2.6t m
we take speeds as negative as they are moving in opposite directions
part 2 ) now i attached a picture for part 2 so you could get a better understanding
the general formula you use to find acceleration in these questions is a=g sin alpha where g=10ms^-2
and sin alpha = O/H which you can see from the diagram O=1.6m and H=2.6t and t given in the question is 2.5 so H=2.6 x 2.5=6.5 m
now put in the values in the equation a=g sin alpha
you get a=10 x 1.6/6.5 =2.46 ms^-2
part 3) this part of the question asks us to find how much of distance did P move when Q was zero
so for Q >>>> when v=0m/s >>>>> 0=-1.3 + 2.46t then you solve this and get time as 0.528 s
distance moved by P at time of 0.528 s so d=1.3 x 0.528 + 1/2 2.46 x 0.528^2 = 1.03 m
i hope you got it
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