Represent the solutions of the following inequalities graphically.

Solving Linear Inequalities in One Unknown The techniques involved are similar to those in solving linear equations. Equation Inequality 1 3 2 = - x 1 3 2 > - x 3 1 2 + = - x 3 1 2 + > - x Add 3 to both sides. 4 2 = x 4 2 > x 2 4 = x 2 4 > x Divide both sides by 2. 2 = x 2 > x

If n is an odd number, If n is an even number, The mode of a set of data is the datum with the highest frequency.

Medians of Ungrouped Data When a set of data is arranged in ascending / descending order, the datum in the middle is the median of the data. For example, The heights of these 5 girls are in ascending order. 140 cm 138 cm The middle datum is 135 cm. Therefore, it is the median. 135 cm 130 cm 133 cm

Medians of Grouped Data Step 2. Find the value on the horizontal axis corresponding to the cumulative frequency of (i.e. n  50%). This value is the required median. Cumulative frequency Heights of a group of students 10 20 Height (cm) 30 144.5 149.5 154.5 159.5 164.5 139.5 169.5 40 ® = 20) 2 40 ( median

The following table summarizes the characteristics of the three averages: Arithmetic mean Median Mode 1. All the data are involved in the calculation 2. Affected by data which are extremely large or extremely small 3. Must have on value only 4. Data have to be arranged in order before calculation 5. Must be one of the data 1. All the data are involved in the calculation    2. Affected by data which are extremely large or extremely small    3. Must have one value only    4. Data have to be arranged in order before calculation    5. Must be one of the data   

Let’s consider the data set 1, 2, 4, 6, 6: Mean Median Mode Original data set 1, 2, 4, 6, 6 3.8 4 6 New data set (add 5 to each datum in the original set) 8.8 9 11 6, 7, 9, 11, 11 (3.8 + 5) (4 + 5) (6 + 5) New data set (multiply each datum in the original set by 2) 7.6 8 12 2, 4, 8, 12, 12 (3.8  2) (4  2) (6  2)

Consider another data set 1, 2, 4, 5, 6, 6: Mean Median Mode Original data set 1, 2, 4, 5, 6, 6 4 4.5 6 New data set (insert a datum ‘0’ in the original set) 3.167 3 6 0, 1, 2, 4, 5, 6, 6 (< 4) (< 4.5) (= 6) New data set (delete a datum ‘2’ in the original set) 4.4 5 6 1, 4, 5, 6, 6 (> 4) (> 4.5) (= 6) New data set (delete a datum ‘5’ in the original set) 3.8 4 6 1, 2, 4, 6, 6 (< 4) (< 4.5) (= 6)

Test Examination Marks 25 65 Weight 2 8 10 8 65 2 25  + = ∴ The weighted mean mark of Harry 57 = In general,