Date | 14/11/2011 (Monday) |
Time | 8.15 a.m. – 9.25 a.m. |
Class | 6AS |
Subject | Mathematics T |
Topic | Continuous Probability Distribution |
Learning Area | Normal Distribution |
Learning Outcome | Students should be able to determine the standard deviation of a normal distribution. |
Activity |
- Discussing example 28 (textbook, page 399)
- Working out questions 3 and 4 (textbook, page 400)
|
Teaching Aid | STPM Mathematics T (Penerbitan Pelangi, Paper 2) |
Reflection | Students have been able to determine the standard deviation of a normal distribution. |
Date | 14/11/2011 (Monday) |
Time | 11.30 a.m. – 12.40 p.m. |
Class | 6RS |
Subject | Mathematics T |
Topic | Matrices |
Learning Area | Determinant of a 2X2 Matrix |
Learning Outcome | Students should be able to determine the determinant of a 2X2 matrix. |
Activities |
- Discussing Example 4.6 (textbook, page 158)
- Working out questions 1a and 1b (textbook, page 161)
|
Teaching Aids | Ace Ahead STPM Text Mathematics T (Oxford Fajar, Volume 1) |
Reflection | Students have been able to determine the determinant of a 2X2 matrix. |
Date | 15/11/2011 (Tuesday) |
Time | 8.15 a.m. – 9.25 a.m. |
Class | 6RS |
Subject | Mathematics T |
Topic | Matrices |
Learning Area | Determinant of a 3X3 Matrix |
Learning Outcome | Students should be able to determine the determinant of a 3X3 matrix. |
Activities |
- Discussing Example 4.9 (textbook, page 161)
- Working out question 4a (textbook, page 162)
|
Teaching Aids | Ace Ahead STPM Text Mathematics T (Oxford Fajar, Volume 1) |
Reflection | Students have been able to the determinant of a 3X3 matrix. |
Date | 15/11/2011 (Tuesday) |
Time | 9.45 a.m. – 10.55 a.m. |
Class | 6AS |
Subject | Mathematics T |
Topic | Continuous Probability Distribution |
Learning Area | Normal Distribution |
Learning Outcome | Students should be able to determine the standard deviation of a normal distribution containing two normal random variables. |
Activity |
- Discussing example 29 (textbook, page 400)
- Working out question 1 (textbook, page 404)
|
Teaching Aid | STPM Mathematics T (Penerbitan Pelangi, Paper 2) |
Reflection | Students have been able to determine the standard deviation of a normal distribution containing two normal random variables. |
Date | 16/11/2011 (Wednesday) |
Time | 7.05 a.m. – 8.15 a.m. |
Class | 6RS |
Subject | Mathematics T |
Topic | Differentiation |
Learning Area | Rates of change |
Learning Outcome | Students should be able to determine the small change of a variable in a function. |
Activities |
- Discussing Example 7.36 (textbook, page 325)
- Working out question 7 (2010 STPM Trial Paper 1)
|
Teaching Aids | Ace Ahead STPM Text Mathematics T (Oxford Fajar, Volume 1) |
Reflection | Students have been able to determine the small change of a variable in a function. |
Date | 16/11/2011 (Wednesday) |
Time | 8.50 a.m. – 9.25 a.m. |
Class | 6AS |
Subject | Mathematics T |
Topic | Continuous Probability Distribution |
Learning Area | Normal Distribution |
Learning Outcome | Students should be able to determine the mean of a binomial distribution using the method of normal distribution approximation. |
Activity |
- Discussing example 35 (textbook, page 407)
- Working out questions 1 and 2 (textbook, page 408)
|
Teaching Aid | STPM Mathematics T (Penerbitan Pelangi, Paper 2) |
Reflection | Students have been able to determine the mean of a binomial distribution using the method of normal distribution approximation. |
17/11/2011 (Thursday) : Majlis Restu Ilmu
18/11/2011 (Friday) : No teaching and learning (students have been given leave)
2 comments:
hey,
here i will post regarding quickest solve the math problems with vedic mathematics for solve Fractions in 8th Class...if you are in 8th class then you should read this......
multiply 2 numbers close to 100,1000,10000,etc
Step 1) First, we multiply the offsets 2 and 3
We get 6.
Since our base is 100, which has 2 zeros, the product of offsets must also have 2 digits. Hence we write 6 as 06 and write it down as last 2 digits of our answer.
98 -2
97 3
-------------------------
06
Step 2) Now for the previous digits, just add any 2 numbers in the above figure crosswise i.e. either (98-3) or (97-2)
We get 95. This is written before 06 as follows: -
98 -2
97 3
---------------------------
95 / 06
That gives us 9506. That s our answer.
Hence 98 x 97 = 9506
Multiply 888 by 998: -
888 -112
998 -002
---------------------------
886 / 224
Here, the base is 1000 i.e. offsets are differences from 1000
-112 x 002 = 224 ( Simple enough )
And 888 002 = 886
Hence, 888 x 998 = 886224
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