"Both cyclists travelled the same distance for 16 seconds" - this means that the area under both graphs is the same.
Alonso area = 1/2 (16 + 10) * 10 = 130
Boris area = 1/2 * 16 * v = 8v
So 8v = 130.
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
"Both cyclists travelled the same distance for 16 seconds" - this means that the area under both graphs is the same.
Which question?Sum1 please du dis....
23/O/N/12
Draw a line from A to C then look at triangle ABC:
Draw a line from A to C then look at triangle ABC:
This is an isosceles triangle (since AB = BC). Also, we know from (a) that angle ABC = 90. So can you use this information to find the other two angles in the triangle?
Once you know angle BAC, you can subtract this from 168 to find the bearing.
If you're still stuck, let me know and I'll give you a full solution.
easy now here you go
Ok that would be 168-45=123 . Tks .
Damn it i wasted my time making that Loleasy now here you go
no i don't think its any thing like that its just 0.5 x total frequency...so in this case it would b 160 x 0.5=80guys for cumulative frequency is its odd set of data to find the median using quartiles are we suppose to do n+1/2 or just n/2 and its odd ex 160 and we do 160/2 we get 80 are we suppose to do the position of 80 and the next number thus 81 or just 80?? HELP PLZ
what if it was 161 as in the frequency.. then how do we find the median and they give us a cf curve..no i don't think its any thing like that its just 0.5 x total frequency...so in this case it would b 160 x 0.5=80
0.5 x 161 that is wat i would do!!!!what if it was 161 as in the frequency.. then how do we find the median and they give us a cf curve..
Finding the median from a CF curve is an estimation so it doesn't really matter if you use (n+1)/2 or n/2.what if it was 161 as in the frequency.. then how do we find the median and they give us a cf curve..
OK SO we never do n+1 rite... for grouped data... THANK U0.5 x 161 that is wat i would do!!!!
and i think that's how its done
yes nvr for cf.....ur velcum any time just remember me in ur prayersOK SO we never do n+1 rite... for grouped data... THANK U
its pretty easy its just the perpendicular bisector of AC and BD...Can someone please post the solution for 0580/22/M/J/09
Question 16
PLEASE PLEASE PLEASEEEEE
What does it mean by "...which are equidistant from the lines AC and BD"its pretty easy its just the perpendicular bisector of AC and BD...
Oh, and thanks so much for the speedy reply!its pretty easy its just the perpendicular bisector of AC and BD...
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now