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\(\frac{10 + 8 + 5 + 11 + 12 + 9 + 6 + 3 + 15 + 23}{10}\) = \(\frac{102}{10}\) = 10.2

< PRO = 180^\circ - (110 + 50)^\circ = 180 - 160 = 20^\circ

An ogive is constructed using the cumulative frequencies from a frequency distribution table. Collect Data: Organize your data into a frequency distribution table. Calculate Cumulative Frequency: Add frequencies cumulatively for each class interval. Determine Class Boundaries: Identify the upper boundaries of the class intervals. Plot Points: Plot cumulative frequencies against the upper class boundaries. Draw the Ogive: Connect the points with a smooth curve, starting from (0, 0). Label the Axes: Clearly label the x-axis (upper boundaries) and y-axis (cumulative frequency).

\(\frac{\sqrt{6} - \sqrt{4}}{\sqrt{6} + \sqrt{4}}\) = \(\frac{6-2\sqrt{6} - 2\sqrt{6} + 4}{6 - 4}\) \(\frac{10 - 4\sqrt{6}}{2}\) = 5 - 2\(\sqrt{6}\)

3 + 2 + 5 + 5 + + 5 + 10 + 4 + 3 + 1 + 2 = 40

(3 x 0) + (1 x 2) + (5 x 2) + (5 x 3) + (5 x 4) + (10 x 5) + (4 x 6) + (3 x 7) + (1 x 8) + (2 x 9) \(\div\) 40 = \(\frac{168}{40}\) = 4.2

\(\theta\) = 180o -(90 + 60) \(\theta\) = 180o - 150o = 30o Fraction of shaded position = \(\frac{30}{360}\) + \(\frac{30}{360}\) = \(\frac{1}{12}\) + \(\frac{1}{12}\) = \(\frac{1}{6}\)

The required roots/region of x is -1\(\leq\)x\(\leq\)4

23x + 101x = 130x 2 x X1 + 3 x Xo + 1 x X2 + 0 x X1 + 1 x Xo = 1 x Xo = 1 x X2 + 3 x X1 + 0 x Xo = X2 + 3x + 0 2x + 3 = x2 + 0 + 1 + x2 + 3x 2x - 3x + x2 - x2 = -3 - 1 - x = -4 x = 4

(\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\) = (\(\frac{9 - 8}{12} \times \frac{6}{5}\)) = \(\frac{1}{12} \times \frac{6}{5}\) = \(\frac{1}{10}\)

Note that \(\frac{10\sqrt{3}}{\sqrt{5}} = \frac{10\sqrt{3}}{\sqrt{5}} \times - \frac{\sqrt{5}}{\sqrt{5}}\) = \(\frac{10\sqrt{15}}{\sqrt{5}} = 2\sqrt{15}\) hence, (\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))2 = (\(2\sqrt{15} - \sqrt{15}\))2 = (\(2\sqrt{15} - \sqrt{15}\))(\(2\sqrt{15} - \sqrt{15}\)) = 4\(\sqrt{15 \times 15} - 2\sqrt{15 \times 15} - 2\sqrt{15 x 15} + \sqrt{15 \times 15}\) = 4 x 15 - 2 x 15 - 2 x 15 + 15 = 60 - 30 - 30 + 15 = 15

d \(\alpha\) t2 d = t2 k where k is a constant. d = 45cm, when t = 3s; thus 45 = 32 x t k = \(\frac{45}{9}\) = 5 thus equation connecting d and t is d = 5t2 when t = 6s, d = 5 x 62 = 5 x 36 = 180cm

From the venn diagram, Nigeria footballers from a subset of good footballers.

On a map, 1cm represents 5km. Then it follows that 1cm2 represents 25km2. Acm2 represents 100km2. By apparent cross-multiplication, 1cm2 x 100km2 = Acm2x 25km2 therefore A = \(\frac{100}{25}\) = 4cm2

\(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}\) = \(\frac{3^{n - 1} \times 3^{3(n + 1)}}{3^{4n}}\) = 3\(^{n - 1 + 3n + 3 - 4n}\) = 3\(^{4n - 4n - 1 + 3}\) = 32 = 9