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Mathematics: Post your doubts here!

Rizwan Javed

Rizwan Javed

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Konstantino Nikolas said:
Yeah yeah, no problem!
Click to expand...
let distance moved by P be s. and the distance moved by Q be s'.
s - s' = 5

use the equation : ut +0.5at^2

I have rounded off 0.5 * 9.81 = 4.9 to 5 for ease.

17(t+2) - 5(t+2)^2 - (7t - 5t^2) = 5
t = 0.9s

We get the other time if we substract this way, but this time using: ' t-2 ' for q
s - s' = 5

17(t) - 5(t)^2 - 7(t - 2) + 5(t - 2)^2 = 5
t = 2.9s
 
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P

Pixie Twinkles

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Rizwan Javed said:
let distance moved by P be s. and the distance moved by Q be s'.
s - s' = 5

use the equation : ut +0.5at^2

I have rounded off 0.5 * 9.81 = 4.9 to 5 for ease.

17(t+2) - 5(t+2)^0.5 - (7t - 5t^2) = 5
t = 0.9s

We get the other time if we substract this way, but this time using: ' t-2 ' for p
s' - s = 5

7t - 5t^2 - 17(t+2) + 5(t+2)^0.5
t = 2.9s
Click to expand...
Could you please help me out on this question :
Why a runner's acceleration cannot exceed 10μms^-2?. where μ is the coefficient of friction between her shoes and the track. The highest speed that a runner can keep during 800 m race is 8 ms^-1. Show that the fastest time she can hope to achieve by standing start is (100+ 2/5μ )
 
Konstantino Nikolas

Konstantino Nikolas

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Rizwan Javed said:
let distance moved by P be s. and the distance moved by Q be s'.
s - s' = 5

use the equation : ut +0.5at^2

I have rounded off 0.5 * 9.81 = 4.9 to 5 for ease.

17(t+2) - 5(t+2)^0.5 - (7t - 5t^2) = 5
t = 0.9s

We get the other time if we substract this way, but this time using: ' t-2 ' for p
s' - s = 5

7t - 5t^2 - 17(t+2) + 5(t+2)^0.5
t = 2.9s
Click to expand...

Isn't it supposed to be either : tp = t + 2 or tq = t - 2 ?
Why should we take t+2 AND t-2 for p itself?? o_O
 
nehaoscar

nehaoscar

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upload_2016-2-28_19-40-45.png
upload_2016-2-28_19-41-13.png
The ms just says 1 ... can someone please show me how it is 1 ... ??
I tried using the reverse normal distribution method for Z but I'm not getting it.... :/
 
Copy Cat

Copy Cat

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nehaoscar said:
View attachment 59524
View attachment 59525
The ms just says 1 ... can someone please show me how it is 1 ... ??
I tried using the reverse normal distribution method for Z but I'm not getting it.... :/
Click to expand...
Draw up a probability distribution table.
Then just try up different values for eg:- a=2 (add up the probabilities from -2 to 4 and check whether it gives out the answers).
You only have two possibilities for "a" (1 & 2).
 
nehaoscar

nehaoscar

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Copy Cat said:
Draw up a probability distribution table.
Then just try up different values for eg:- a=2 (add up the probabilities from -2 to 4 and check whether it gives out the answers).
You only have two possibilities for "a" (1 & 2).
Click to expand...
Why doesn't it work by doing
P(X<2a) - P(x<-a) = 17/35
and then standardizing to Z and working in reverse to find a??
And also why are there only two possibilities for "a" (1 & 2)??
 
Rizwan Javed

Rizwan Javed

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nehaoscar said:
Why doesn't it work by doing
P(X<2a) - P(x<-a) = 17/35
and then standardizing to Z and working in reverse to find a??
And also why are there only two possibilities for "a" (1 & 2)??
Click to expand...
X is not a continuous random variable, thus doesnot follow the Normal Distribution.

Now, look that X lies in between two values, one is negative, and the other is +ive. That means this range must contain X=0.
Now solve:
1/10 + x(9/70) = 17/35
x = 3 (this shows that X can take only 4 values, with zero inclusive)

Now, see the set of values for X. You'll notice that a can't be equal to -2 because this would mean that X has larger than 4 values. So the only value satisfying this is a=1.

You don't need to do all this stuff as this just a one mark question. This is all just for explanation. :)
 
nehaoscar

nehaoscar

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Rizwan Javed said:
X is not a continuous random variable, thus doesnot follow the Normal Distribution.

Now, look that X lies in between two values, one is negative, and the other is +ive. That means this range must contain X=0.
Now solve:
1/10 + x(9/70) = 17/35
x = 3 (this shows that X can take only 4 values, with zero inclusive)

Now, see the set of values for X. You'll notice that a can't be equal to -2 because this would mean that X has larger than 4 values. So the only value satisfying this is a=1.

You don't need to do all this stuff as this just a one mark question. This is all just for explanation. :)
Click to expand...
Thankyouu soooo much for such a clear explanation! :D
 
Konstantino Nikolas

Konstantino Nikolas

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Is there a mark for solving simultaneous equations? I mean, in the ms they give a method mark for "solving simultaneously". Does that mean if we skip the step and use the calc to do it directly, we will lose a mark?
 
nehaoscar

nehaoscar

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upload_2016-3-2_13-50-39.png
upload_2016-3-2_13-51-2.png
Ok so i know how to solve it but the thing is here they have classed them as 0-9 then 10-19 then 20-34 and so on so the midpoints will be 4.5 and 14.5 and 27 etc...
Why can't you split the data like 0≤X<10 and 10≤X<20 and 20≤X<35 and so on??
 
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