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The mode = 8 with the highest frequency of 4 times.

Mean of 20 observations = 4 All the observations = 20 x 4 = 80 Largest observed value = 23 Remaining observation = 80 - 23 = 57 Mean of observation = 57/19 = 3

Arranging the numbers in ascending order, we have 2.35, 2.50, 2.64, 2.72, 2.91. The median = 2.64

at least 10 => ≥ 10 6/36 = 1/6

2r, 6w, 5b balls Total balls = 2 + 6 + 5 = 13 balls p(black) = 5/13

\(3310_{5} - 1442_{5} = 1313_{5}\) You can either do a direct subtraction (which is easier), or convert both numbers in base 10, subtract and convert back to base 5 (if you're not sure).

3.1415926 \(\approxeq\) 3.14159 (to 5 d.p)

Error = (16.8cm - 15cm = 1.8cm\) \(\text{% error} = \frac{1.8}{15} \times 100% = \frac{180}{15}\) = \(12.00%\)

Present salary = N24,000. Increment after first year = \(\frac{10}{100} \times N24,000 = N2,400\) Salary at the beginning of second year = N(24,000 + 2,400) = N26,400. Increment after second year = \(\frac{10}{100} \times N26,400 = N2,640\) Salary at the beginning of third year = N(26,400 + 2,640) = N29,040. \(\therefore\) His salary at the beginning of the third year = N29,040.

\(0.0014 = 1.4 \times 10^{-3}\) \(0.011 = 1.1 \times 10^{-2}\) \(\therefore 0.0014 \times 0.011 = 1.4 \times 10^{-3} \times 1.1 \times 10^{-2}\) = \(1.54 \times 10^{-5}\)

\(\frac{(81)^{\frac{3}{4}} - (27)^{\frac{1}{3}}}{3 \times 2^3}\) = \(\frac{(3^{4})^{\frac{3}{4}} - (3^{3})^{\frac{1}{3}}}{3 \times 2^{3}}\) = \(\frac{3^{3} - 3^{1}}{24}\) = \(\frac{27 - 3}{24} = 1\)

\(\frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}}\) \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\) \(\therefore \frac{\sqrt{12} - \sqrt{3}}{\sqrt{12} + \sqrt{3}} = \frac{2\sqrt{3} - \sqrt{3}}{2\sqrt{3} + \sqrt{3}}\) = \(\frac{\sqrt{3}}{3\sqrt{3}}\) = \(\frac{1}{3}\)

Number of people playing both rugby and cricket = 4 Number that play cricket only = 11 - 4 = 7 Number that play rugby only = 14 - 4 = 10. Number that play for at least one of the teams = 4 + 7 + 10 = 21.

No expanded working has been attached for this question yet. The verified answer is Option C.

\(x^{3} + px^{2} + qx + 6 = 0\) f(x - 1) = 0; f(1) = 0. \(1^{3} + p(1^{2}) + q(1) + 6 = 0 \implies p + q = -7 ... (1)\) f(x + 1) = 0; f(-1) = 0. \((-1)^{3} + p(-1^{2}) + q(-1) + 6 = 0 \implies p - q = -5 ... (2)\) Subtract (2) from (1). \(2q = -2 \implies q = -1\) \(p - (-1) = -5 \implies p = -5 - 1 = -6\) \((p, q) = (-6, -1)\)