300144
Find the mean of the data 10, 15, 17, 19, 20 and 21:
1 11
2 16
3 18
4 21
Explanation:
18 Data observations are: 10, 15, 17, 19, 20 and 21 \(\text{mean} = \frac{\text{Sum}}{\text{Number}}\) \(\text{mean} = \frac {10+15+17+19+20+21}{6}\) \(\text{mean} = \frac{102}{6} = 17\) since, the number of observations are even, the median will be the mean of the two middle observations: \(\text{median} = \frac{{3}^{\text{rd}} + {4}^{\text{th}}}{2}\) \(\text{median} = \frac{{17} +{19}}{2}\) \(\text{median} = {18}\)
DATA HANDLING
300145
The mean of x, x + 3, x + 6, x + 9 and x + 12 is:
1 x + 6
2 x + 3
3 x + 9
4 x + 12
Explanation:
x + 6 By definition, \(\text{Average} =\frac{ \text{x}+(\text{x+3})+({\text{x+6}})+({\text{x+9}})+({\text{x+12}})}{5}\) \(\frac{{5}+{30}}{5} = {\text{x+6}}\)
DATA HANDLING
300146
If 6, p, 12, 8 and 9 mean of the data is 9 then p = ?
1 7
2 8
3 9
4 10
Explanation:
10 Arithmetic mean, A \( = \frac{\text{S}}{\text{N}} = \frac{6+{\text{p}}+12+8+9}{5} = {9}\) \(\frac{{35+}{\text{p}}}{5} = {9}\) 35 + p = 45 p = 45 - 35 = 10
DATA HANDLING
300147
If the mean of x and \(\frac{1}{\text{x}}\) is M, then the mean of x\(^{\1}\) and \(\frac{1}{\text{x}^{2}}\) is:
300144
Find the mean of the data 10, 15, 17, 19, 20 and 21:
1 11
2 16
3 18
4 21
Explanation:
18 Data observations are: 10, 15, 17, 19, 20 and 21 \(\text{mean} = \frac{\text{Sum}}{\text{Number}}\) \(\text{mean} = \frac {10+15+17+19+20+21}{6}\) \(\text{mean} = \frac{102}{6} = 17\) since, the number of observations are even, the median will be the mean of the two middle observations: \(\text{median} = \frac{{3}^{\text{rd}} + {4}^{\text{th}}}{2}\) \(\text{median} = \frac{{17} +{19}}{2}\) \(\text{median} = {18}\)
DATA HANDLING
300145
The mean of x, x + 3, x + 6, x + 9 and x + 12 is:
1 x + 6
2 x + 3
3 x + 9
4 x + 12
Explanation:
x + 6 By definition, \(\text{Average} =\frac{ \text{x}+(\text{x+3})+({\text{x+6}})+({\text{x+9}})+({\text{x+12}})}{5}\) \(\frac{{5}+{30}}{5} = {\text{x+6}}\)
DATA HANDLING
300146
If 6, p, 12, 8 and 9 mean of the data is 9 then p = ?
1 7
2 8
3 9
4 10
Explanation:
10 Arithmetic mean, A \( = \frac{\text{S}}{\text{N}} = \frac{6+{\text{p}}+12+8+9}{5} = {9}\) \(\frac{{35+}{\text{p}}}{5} = {9}\) 35 + p = 45 p = 45 - 35 = 10
DATA HANDLING
300147
If the mean of x and \(\frac{1}{\text{x}}\) is M, then the mean of x\(^{\1}\) and \(\frac{1}{\text{x}^{2}}\) is:
300144
Find the mean of the data 10, 15, 17, 19, 20 and 21:
1 11
2 16
3 18
4 21
Explanation:
18 Data observations are: 10, 15, 17, 19, 20 and 21 \(\text{mean} = \frac{\text{Sum}}{\text{Number}}\) \(\text{mean} = \frac {10+15+17+19+20+21}{6}\) \(\text{mean} = \frac{102}{6} = 17\) since, the number of observations are even, the median will be the mean of the two middle observations: \(\text{median} = \frac{{3}^{\text{rd}} + {4}^{\text{th}}}{2}\) \(\text{median} = \frac{{17} +{19}}{2}\) \(\text{median} = {18}\)
DATA HANDLING
300145
The mean of x, x + 3, x + 6, x + 9 and x + 12 is:
1 x + 6
2 x + 3
3 x + 9
4 x + 12
Explanation:
x + 6 By definition, \(\text{Average} =\frac{ \text{x}+(\text{x+3})+({\text{x+6}})+({\text{x+9}})+({\text{x+12}})}{5}\) \(\frac{{5}+{30}}{5} = {\text{x+6}}\)
DATA HANDLING
300146
If 6, p, 12, 8 and 9 mean of the data is 9 then p = ?
1 7
2 8
3 9
4 10
Explanation:
10 Arithmetic mean, A \( = \frac{\text{S}}{\text{N}} = \frac{6+{\text{p}}+12+8+9}{5} = {9}\) \(\frac{{35+}{\text{p}}}{5} = {9}\) 35 + p = 45 p = 45 - 35 = 10
DATA HANDLING
300147
If the mean of x and \(\frac{1}{\text{x}}\) is M, then the mean of x\(^{\1}\) and \(\frac{1}{\text{x}^{2}}\) is:
300144
Find the mean of the data 10, 15, 17, 19, 20 and 21:
1 11
2 16
3 18
4 21
Explanation:
18 Data observations are: 10, 15, 17, 19, 20 and 21 \(\text{mean} = \frac{\text{Sum}}{\text{Number}}\) \(\text{mean} = \frac {10+15+17+19+20+21}{6}\) \(\text{mean} = \frac{102}{6} = 17\) since, the number of observations are even, the median will be the mean of the two middle observations: \(\text{median} = \frac{{3}^{\text{rd}} + {4}^{\text{th}}}{2}\) \(\text{median} = \frac{{17} +{19}}{2}\) \(\text{median} = {18}\)
DATA HANDLING
300145
The mean of x, x + 3, x + 6, x + 9 and x + 12 is:
1 x + 6
2 x + 3
3 x + 9
4 x + 12
Explanation:
x + 6 By definition, \(\text{Average} =\frac{ \text{x}+(\text{x+3})+({\text{x+6}})+({\text{x+9}})+({\text{x+12}})}{5}\) \(\frac{{5}+{30}}{5} = {\text{x+6}}\)
DATA HANDLING
300146
If 6, p, 12, 8 and 9 mean of the data is 9 then p = ?
1 7
2 8
3 9
4 10
Explanation:
10 Arithmetic mean, A \( = \frac{\text{S}}{\text{N}} = \frac{6+{\text{p}}+12+8+9}{5} = {9}\) \(\frac{{35+}{\text{p}}}{5} = {9}\) 35 + p = 45 p = 45 - 35 = 10
DATA HANDLING
300147
If the mean of x and \(\frac{1}{\text{x}}\) is M, then the mean of x\(^{\1}\) and \(\frac{1}{\text{x}^{2}}\) is:
