| 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.
18.75 N
15.75 N
9.50 N
8.66 N
Correct answer is D
\(F_{1} = (5 N, 060°) = 5 \cos 60 i + 5 \sin 60 j = 2.5 i + \frac{5\sqrt{3}}{2}j\)
\(F_{2} = (10 N, 180°) = 10 \cos 180 i + 10 \sin 180 j = -10i\)
\(R = F_{1} + F_{2} = -7.5 i + \frac{5\sqrt{3}}{2}j\)
\(|R| = \sqrt{(-7.5)^{2} + (\frac{5\sqrt{3}}{2})^{2}}\)
= \(\sqrt{56.25 + 18.75} = \sqrt{75}\)
= 8.66 N