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Total surface area of hemisphere = half of TSA of sphere + circumference of circle = \(2\pi r^{2} + \pi r^{2} = 3\pi r^{2}\) = \( 3 \times \frac{22}{7} \times 7^{2} = 462cm^{2}\)

2y - ax + 4 = 0 \(\implies\) 2y = ax - 4. y + (x/4) - 7 = 0 \(\implies\) y = -x/4 + 7 Slope of line 1 = a/2 Slope of line 2 = -1/4 -1/4 x a/2 = -1 (product of the slope of two perpendicular lines = -1) a/8 = 1 \(\implies\) a = 8.

The locus of a point P that is equidistant from two points S and T is the perpendicular bisector of the segment ST. This means that every point on this line is the same distance from both S and T, effectively dividing the segment into two equal parts and forming a right angle with it.

\(Mean: \frac{7 + (-3) + 4 + (-2) + 5 + (-9) + 4 + 8 + (-6) + 12}{10} = \frac{20}{10}\) = 2

Range = Highest value - Lowest value = (k + 6) - (k - 5) = 11

P(not passing exam) = \(1 - \frac{2}{3} = \frac{1}{3}\) P(not passing any of the three exams) = \(\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}\) = \(\frac{1}{27}\)

Total: 2 + 5 + 3 + 11 + 9 = 30 Cassava: \(\frac{3}{30} \times 360^\circ = 36^\circ\)

To get the answer, we simply do \(\frac{5!}{(5 - 3)!}\) = \(\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}\) = 60

Let the sum of the first five numbers and the sixth number be x and t respectively. \(\frac{x + t}{6} = 60 \implies x + t = 360\) \(\frac{x}{5} = 50 \implies x = 250\) \(\therefore t = 360 - 250 = 110\)

x 7 3 14 9 7 8 Total \(x - \bar{x}\) -1 -5 6 1 -1 0 \(|x - \bar{x}|\) 1 5 6 1 1 0 14 Mean : \(\frac{7 + 3 + 14 + 9 + 7 + 8}{6} = \frac{48}{6} = 8\) Mean deviation : \(\frac{\sum |x - \bar{x}|}{n} = \frac{14}{6} = \frac{7}{3}\)

Since radius = \(\frac{15}{2}\) diameter = 2 * (\(\frac{15}{2}\)) = 15cm XYZ is a right-angled triangle (since

Source: Myschool Past Questions Archive (Record #7728)