Tuesday, November 15, 2011

Week 46 (14 - 18/11/2011)

Date14/11/2011 (Monday)
Time 8.15 a.m. – 9.25 a.m.
Class6AS
Subject Mathematics T
Topic Continuous Probability Distribution
Learning AreaNormal Distribution
Learning OutcomeStudents should be able to determine the standard deviation of a normal distribution.
Activity
  1. Discussing example 28 (textbook, page 399)
  2. Working out questions 3 and 4 (textbook, page 400)
Teaching AidSTPM Mathematics T (Penerbitan Pelangi, Paper 2)
ReflectionStudents have been able to determine the standard deviation of a normal distribution.


Date14/11/2011 (Monday)
Time11.30 a.m. – 12.40 p.m.
Class6RS
SubjectMathematics T
TopicMatrices
Learning AreaDeterminant of a 2X2 Matrix
Learning OutcomeStudents should be able to determine the determinant of a 2X2 matrix.
Activities
  1. Discussing Example 4.6 (textbook, page 158)
  2. Working out questions 1a and 1b (textbook, page 161)
Teaching AidsAce Ahead STPM Text Mathematics T (Oxford Fajar, Volume 1)
ReflectionStudents have been able to determine the determinant of a 2X2 matrix.


Date15/11/2011 (Tuesday)
Time8.15 a.m. – 9.25 a.m.
Class6RS
SubjectMathematics T
TopicMatrices
Learning AreaDeterminant of a 3X3 Matrix
Learning OutcomeStudents should be able to determine the determinant of a 3X3 matrix.
Activities
  1. Discussing Example 4.9 (textbook, page 161)
  2. Working out question 4a (textbook, page 162)
Teaching AidsAce Ahead STPM Text Mathematics T (Oxford Fajar, Volume 1)
ReflectionStudents have been able to the determinant of a 3X3 matrix.


Date15/11/2011 (Tuesday)
Time 9.45 a.m. – 10.55 a.m.
Class6AS
Subject Mathematics T
Topic Continuous Probability Distribution
Learning AreaNormal Distribution
Learning OutcomeStudents should be able to determine the standard deviation of a normal distribution containing two normal random variables.
Activity
  1. Discussing example 29 (textbook, page 400)
  2. Working out question 1 (textbook, page 404)
Teaching AidSTPM Mathematics T (Penerbitan Pelangi, Paper 2)
ReflectionStudents have been able to determine the standard deviation of a normal distribution containing two normal random variables.


Date16/11/2011 (Wednesday)
Time7.05 a.m. – 8.15 a.m.
Class6RS
SubjectMathematics T
TopicDifferentiation
Learning AreaRates of change
Learning OutcomeStudents should be able to determine the small change of a variable in a function.
Activities
  1. Discussing Example 7.36 (textbook, page 325)
  2. Working out question 7 (2010 STPM Trial Paper 1)
Teaching AidsAce Ahead STPM Text Mathematics T (Oxford Fajar, Volume 1)
ReflectionStudents have been able to determine the small change of a variable in a function.


Date16/11/2011 (Wednesday)
Time 8.50 a.m. – 9.25 a.m.
Class6AS
Subject Mathematics T
Topic Continuous Probability Distribution
Learning AreaNormal Distribution
Learning OutcomeStudents should be able to determine the mean of a binomial distribution using the method of normal distribution approximation.
Activity
  1. Discussing example 35 (textbook, page 407)
  2. Working out questions 1 and 2 (textbook, page 408)
Teaching AidSTPM Mathematics T (Penerbitan Pelangi, Paper 2)
ReflectionStudents 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:

Anonymous said...
This comment has been removed by the author.
Anonymous said...

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