| Age in years | 10 - 14 | 15 - 19 | 20 - 24 | 25 - 29 | 30 - 34 |
| Frequency | 6 | 8 | 14 | 10 | 12 |
Find the mean of the distribution.
23.4
23.6
24.3
24.6
Correct answer is B
| Age in years |
Classmark (x) |
Frequency (f) |
fx |
| 10 - 14 | 12 | 6 | 72 |
| 15 - 19 | 17 | 8 | 136 |
| 20 - 24 | 22 | 14 | 308 |
| 25 - 29 | 27 | 10 | 270 |
| 30 - 34 | 32 | 12 | 384 |
| Total | 50 | 1170 |
\(Mean = \frac{\sum fx}{\sum f}\)
= \(\frac{1170}{50} = 23.4\)
| Age in years | 10 - 14 | 15 - 19 | 20 - 24 | 25 - 29 | 30 - 34 |
| Frequency | 6 | 8 | 14 | 10 | 12 |
In which group is the upper quartile?
15 - 19
20 - 24
25 - 29
30 - 34
Correct answer is C
The upper quartile occurs at the \(\frac{3N}{4}\) position.
= \(\frac{3 \times 50}{4} = 37.5\)
This occurs in the class 25 - 29.
| Age in years | 10 - 14 | 15 - 19 | 20 - 24 | 25 - 29 | 30 - 34 |
| Frequency | 6 | 8 | 14 | 10 | 12 |
What is the class mark of the median class?
17
22
27
32
Correct answer is B
\(\text{Total frequency} = 50\)
Median occurs at \(\frac{50}{2} = \text{25th position}\)
= Class 20 - 24 with classmark \(\frac{20 + 24}{2} = 22\)
\(8\pi cm^{2}s^{-1}\)
\(16\pi cm^{2}s^{-1}\)
\(24\pi cm^{2}s^{-1}\)
\(48\pi cm^{2}s^{-1}\)
Correct answer is D
Surface area of sphere, \( A = 4\pi r^{2}\)
\(\frac{\mathrm d A}{\mathrm d r} = 8\pi r\)
The rate of change of radius with time \(\frac{\mathrm d r}{\mathrm d t} = 3cm s^{-1}\)
\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)
= \(8\pi \times 2cm \times 3cm s^{-1} = 48\pi cm^{2}s^{-1}\)
\(6\)
\(\frac{11}{6}\)
\(\frac{11}{4}\)
\(\frac{5}{3}\)
Correct answer is B
\(h : x \to 2 - \frac{1}{2x - 3}\)
\(h(x) = \frac{2(2x - 3) - 1}{2x - 3} = \frac{4x - 7}{2x - 3}\)
Let x = h(y)
\(x = \frac{4y - 7}{2y - 3}\)
\(x(2y - 3) = 4y - 7 \implies 2xy - 4y = 3x - 7\)
\(y = \frac{3x - 7}{2x - 4}\)
\(h^{-1}(x) = \frac{3x - 7}{2x - 4}\)
\(\therefore h^{-1}(\frac{1}{2}) = \frac{3(\frac{1}{2}) - 7}{2(\frac{1}{2}) - 4}\)
= \(\frac{\frac{-11}{2}}{-3} = \frac{11}{6}\)