First thing first, many congrats on your Hifz :)
Now to query, I can only give basic suggestion here. The most difficult part is starting to study, once you get a hang of what is actually required it becomes easy to manage and divide your time then.
I would recommend starting with your...
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As long as your candidate number is accurate, you are generally good to go but it would be advisable to contact the relevant board representative in your country just to be on the extra safe side for example your school, british council etc
Generally they are not harder than May/June but people do get less A's than May/June due to the fact that less people appear and therefore grade thresholds help scale your grade less. But its not impossible to score higher, just focus, believe in yourself and you'll ace it surely! (y)
Ideally you should elaborate a little more on that but as long as all the points above are included one way or another, you will get full marks.
These are all the significant details that the examiner expects in your answer but you can (and should) answer in your own words.
Most likely you will be awarded 2 or 3 marks for the correct plot even though you did frequency density.
This falls in the same category as "error carry forward".
Don't worry and focus on the rest of the exams as you'll still get 2-3 marks :)
Good luck for the remainder of your papers!
1. I like to take the step-by-step approach here. We need to find probability that there are exactly 2 of 8 days \(\text{(}^8C_{2}\text{)}\) on which martin is playing computer games when his friend phones. So there are 2 days when he calls in his gaming time and 6 days when he might call...
There are two unknowns here, starting speed \(u\) and acceleration \(a\). Using \(s=ut+\frac{1}{2}at^2\), we’re given that in 12 seconds from X to Y:
\(\qquad 40=12u+\frac{1}{2}12^2a=12u+72a\)
and in 18 seconds from X to Z:
\(\qquad 80=18u+\frac{1}{2}18^2a=18u+162a\)
Multiplying the first...
a) Since you have points P and Q, their equation would be in the form \(y=mx+c\) where
\(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
\(m=\frac{5-3}{9-2}=\frac{2}{7}\)
Substituting back into \(y=mx+c\) and substituting x and y coordinates from either P or Q should give \(c=\frac{17}{7}\).
Thus equation...
The formula I used earlier is derived from this under special conditions for elastic collisions.
Since velocity is a vector quantity, the signs need to be changed with directions.
The pulse sent by transmitter is reflected and returns to the transmitter so it takes it \(6.3\mu s\times2\) to...
Since the collision is perfectly elastic, we can use
\(v_{1}i+v_{1}f=v_{2}i+v_{2}f\) (You might have a different variant of this formula depending on your board)
Substituting in our values, we get
\(6+2=0+v_{2}f\)
And therefore,
\(v_{2}f=8\)
Hope this helps!
From a velocity-time graph, the acceleration at time t is given by the gradient at time t. This gives the instantaneous acceleration at that time.
In this graph, the gradient and hence the acceleration is changing.
Note that the gradient at time t = 3s is obtained by drawing a tangent at that...