| Marks | 5 - 7 | 8 - 10 | 11 - 13 | 14 - 16 | 17 - 19 | 20 - 22 |
| Frequency | 4 | 7 | 26 | 41 | 14 | 8 |
The table above shows the marks obtained by 100 pupils in a test. Find the probability that a student picked at random scored at least 14 marks.
0.22
0.41
0.49
0.63
Correct answer is D
No explanation has been provided for this answer.
| Marks | 5 - 7 | 8 - 10 | 11 - 13 | 14 - 16 | 17 - 19 | 20 - 22 |
| Frequency | 4 | 7 | 26 | 41 | 14 | 8 |
The table above shows the marks obtained by 100 pupils in a test. Find the upper class boundary of the class containing the third quartile.
13.0
16.0
16.5
22.5
Correct answer is C
The third quartile can be found in the \((\frac{3N}{4})^{th}\) position
= \((\frac{3 \times 100}{4})^{th}\) position
This is found in the class 14 - 16.
The upper class boundary = 16.5
-11
-9
-3
4
Correct answer is C
Using the remainder theorem, the remainder when a polynomial \(ax^{2} + bx + c\) is divided by \((x - a)\) is equal to \(f(a)\).
\(2x^{3} + 3x^{2} + qx - 1\) divided by \((x + 2)\), the remainder = \(f(-2)\)
\(\implies f(-2) = f(1)\)
\(f(-2) = 2(-2^{3}) + 3(-2^{2}) + q(-2) - 1 = -16 + 12 - 2q - 1 = -5 - 2q\)
\(f(1) = 2(1^{3}) + 3(1^{2}) + q(1) - 1 = 2 + 3 + q - 1 = 4 + q\)
\(4 + q = -5 -2q \implies 4 + 5 = -2q - q = -3q\)
\(q = -3\)
Find the value of p for which \(x^{2} - x + p\) becomes a perfect square.
\(-\frac{1}{2}\)
\(\frac{1}{4}\)
\(\frac{1}{2}\)
\(1\)
Correct answer is B
The equation \(ax^{2} + bx + c\) is a perfect square if \(b^{2} = 4ac\).
\(x^{2} - x + p\)
\((-1)^{2} = 4(1)(p)\)
\(1 = 4p \implies p = \frac{1}{4}\)
\(k = \frac{8}{25}\)
\(k = \frac{25}{8}\)
\(k < \frac{25}{8}\)
\(k > \frac{25}{8}\)
Correct answer is C
No explanation has been provided for this answer.