18.91
19.75
25.63
38.23
Correct answer is C
\(y = 19.33 + 0.42x\)
\(\text{The value of y when x = 15} = 19.33 + (0.42 \times 15)\)
= \(19.33 + 6.30\)
= 25.63
\(\frac{1}{5}\)
\(\frac{2}{5}\)
\(\frac{3}{5}\)
\(\frac{4}{5}\)
Correct answer is B
No explanation has been provided for this answer.
| Age(in years) | 1 - 5 | 6 - 10 | 11 - 15 |
| Frequency | 3 | 5 | 2 |
Calculate the standard deviation of the distribution.
1.10
2.36
3.50
7.50
Correct answer is C
|
Age (years) |
Freq (f) |
Mid-value (x) |
fx | \(d = (x - \bar{x})\) | \(d^{2}\) | \(fd^{2}\) |
| 1 - 5 | 3 | 3 | 9 | - 4.5 | 20.25 | 60.75 |
| 6 - 10 | 5 | 8 | 40 | 0.5 | 0.25 | 1.25 |
| 11 - 15 | 2 | 13 | 26 | 5.5 | 30.25 | 60.5 |
| \(\sum =\) | 10 | 75 | 122.5 |
\(Mean (\bar {x}) = \frac{\sum fx}{\sum f} = \frac{75}{10} = 7.5\)
\(SD = \sqrt{\frac{\sum fd^{2}}{\sum f}}\)
= \(\sqrt{\frac{122.5}{10}}\)
= \(\sqrt{12.25} = 3.50\)
1.5 s
3.0 s
4.0 s
6.0 s
Correct answer is B
\(F = ma \)
\(10 = 2.5a \implies a = 4 ms^{-2}\)
Since it is a retarding movement, then \(a = -4 ms^{-2}\).
\(v = u + at; v = 0 ms^{-1}, u = 12 ms^{-1}\)
\(0 = 12 + (-4t) \implies 0 = 12 - 4t\)
\(4t = 12 \implies t = 3 s\)
1 cm
\(\sqrt{2}\) cm
2 cm
4 cm
Correct answer is C
The area of sector in radians = \(\frac{r^{2} \theta}{2}\)
\(3 cm^{2} = \frac{1.5 \times r^{2}}{2}\)
\(r^{2} = \frac{3 \times 2}{1.5} = 4\)
\(r = 2 cm\)