I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.com and I will add you to the mailing list.
Test 2 Results
You are identified by the last five digits of your student number. The first column is your Test 1 result while the second is your Test 2 result. The third is your continuous assessment mark out of 30. The fourth is the percentage you must have on the final to pass and the last is the percentage you must have on the final to get 70%. I will have these test papers with me over the remaining classes. One person never put their name on their test and there will be a – where your score should be. The zeroes, unfortunately are genuine zeros. One man is still on for the ton.
S/N | T1 | T2 | CA/30 | Pass | 70% |
51004 | 100 | 100 | 30 | 14 | 57 |
68681 | 97 | 97 | 29.1 | 16 | 58 |
45948 | 98 | 92 | 28.5 | 16 | 59 |
49905 | 98 | 82 | 27 | 19 | 61 |
44985 | 94 | 86 | 27 | 19 | 61 |
93761 | 90 | 82 | 25.8 | 20 | 63 |
49846 | 82 | 85 | 25.05 | 21 | 64 |
93763 | 76 | 90 | 24.9 | 22 | 64 |
73974 | 76 | 86 | 24.3 | 22 | 65 |
72110 | 76 | 78 | 23.1 | 24 | 67 |
70110 | 82 | 68 | 22.5 | 25 | 68 |
74033 | 71 | 64 | 20.25 | 28 | 71 |
61502 | 76 | 54 | 19.5 | 29 | 72 |
72743 | 61 | 69 | 19.5 | 29 | 72 |
74238 | 76 | 51 | 19.05 | 30 | 73 |
61904 | 47 | 72 | 17.85 | 32 | 75 |
74049 | 65 | 44 | 16.35 | 34 | 77 |
74564 | 16 | 92 | 16.2 | 34 | 77 |
73685 | 76 | 28 | 15.6 | 35 | 78 |
59158 | 55 | 49 | 15.6 | 35 | 78 |
43371 | 95 | – | 14.25 | 37 | 80 |
61788 | 50 | 41 | 13.65 | 38 | 81 |
60831 | 50 | 41 | 13.65 | 38 | 81 |
70213 | 44 | 44 | 13.2 | 38 | 81 |
72983 | 86 | – | 12.9 | 39 | 82 |
71112 | 40 | 44 | 12.6 | 39 | 82 |
22204 | 81 | – | 12.15 | 40 | 83 |
59616 | 42 | 33 | 11.25 | 41 | 84 |
66644 | 39 | 36 | 11.25 | 41 | 84 |
46581 | 45 | 26 | 10.65 | 42 | 85 |
67367 | 39 | 32 | 10.65 | 42 | 85 |
69699 | 13 | 55 | 10.2 | 43 | 85 |
56925 | 21 | 37 | 8.7 | 45 | 88 |
77537 | 47 | 10 | 8.55 | 45 | 88 |
70398 | 55 | – | 8.25 | 45 | 88 |
73289 | 52 | 3 | 8.25 | 45 | 88 |
61939 | 50 | 3 | 7.95 | 46 | 89 |
56695 | 47 | 0 | 7.05 | 47 | 90 |
58461 | 24 | 23 | 7.05 | 47 | 90 |
??? | – | 42 | 6.3 | 48 | 91 |
61941 | 18 | 21 | 5.85 | 49 | 92 |
72127 | 23 | 10 | 4.95 | 50 | 93 |
74242 | 29 | 0 | 4.35 | 51 | 94 |
60829 | 21 | 0 | 3.15 | 53 | 96 |
76010 | – | – | 0 | 57 | 100 |
17751 | – | – | 0 | 57 | 100 |
44439 | – | – | 0 | 57 | 100 |
46134 | – | – | 0 | 57 | 100 |
15818 | – | – | 0 | 57 | 100 |
Week 12
Review tutorials: questions — Winter 2012 — one-to-one help
Week 13
Review tutorials in the lecture slots: questions — Winter 2012 — one-to-one help
Nothing in the tutorial slot Wednesday 13:30 – 15:30
Math.Stack Exchange
If you find yourself stuck and for some reason feel unable to ask me the question you could do worse than go to the excellent site math.stackexchange.com. If you are nice and polite, and show due deference to these principles you will find that your questions are answered promptly. For example this question on factorising quadratics: a very simple task which, incredibly, some of us are still unable to do.
4 comments
Comments feed for this article
May 9, 2013 at 11:15 am
Student 38
For the 2012 winter exam paper what method would you use for question Q1B(i)?
Implement this method to estimate the values of x and y correct to one decimal places
May 9, 2013 at 11:17 am
J.P. McCarthy
You could either use partial pivoting, the Jacobi Method or the Gauss-Siedel Method.
Regards,
J.P.
August 2, 2013 at 12:45 pm
Student 44
Hi JP
I was wondering could you help me with 2 questions.
1. Laplace Winter 2012 Q. 4 (b)
After completing the square
What is happen here were did the
come from?
2. Laplace Winter 2012 Q. 4 (c)
What is going on from the part
Regards.
August 2, 2013 at 12:54 pm
J.P. McCarthy
1. We are trying to send
back to the
-domain. We know from the tables that
Therefore we need the three on top to match the three on the bottom to make it look like this. We can do this by taking out the two (e.g.
).
After this it needs a first shift theorem.
2. The equation there holds for ALL
for unique values of
. If it holds for ALL values of
then in particular it holds for
.
See what happens when you substitute these values into both sides of the equation… bearing in mind that you are trying to find the unique values of
.
Regards,
J.P.
P.S. I might not be able to respond to further questions until Tuesday but will try!