\(\frac{1}{54}\)
\(\frac{13}{54}\)
\(\frac{20}{27}\)
\(\frac{41}{54}\)
Correct answer is B
P(only one hit target) = P(Kofi not Ama) + P(Ama not Kofi)
P(Kofi not Ama) = P(Kofi and Ama') = \(\frac{1}{6} \times \frac{8}{9} = \frac{8}{54}\)
P(Ama not Kofi) = P(Ama and Kofi') = \(\frac{1}{9} \times \frac{5}{6} = \frac{5}{54}\)
P(only one hit target) = \(\frac{8}{54} + \frac{5}{54} = \frac{13}{54}\)
The mean of 2, 5, (x + 2), 7 and 9 is 6. Find the median.
5.5
6.0
6.5
7.0
Correct answer is D
\(\frac{2 + 5+ (x + 2) + 7 + 9}{5} = 6 \implies 25 + x = 30\)
\(x = 5 \therefore x + 2 = 5 + 2 = 7\)
Arranging the numbers in ascending order: 2, 5, 7, 7, 9.
Median = 7.0
Determine the coefficient of \(x^{2}\) in the expansion of \((a + 3x)^{6}\)
\(18a^{2}\)
\(45a^{4}\)
\(135a^{4}\)
\(1215a^{2}\)
Correct answer is C
\((a + 3x)^{6}\).
The coefficient of \(x^{2}\) is:
\(^{6}C_{4}(a)^{6 - 2} (3x)^{2} = \frac{6!}{(6 - 4)! 4!} (a^{4})(9x^{2})\)
\(15 \times a^{4} \times 9 = 135a^{4}\)
Find the equation of a circle with centre (-3, -8) and radius \(4\sqrt{6}\)
\(x^{2} - y^{2} - 6x + 16y + 23 = 0\)
\(x^{2} + y^{2} + 6x + 16y - 23 = 0\)
\(x^{2} + y^{2} + 6x - 16y + 23 = 0\)
\(x^{2} + y^{2} - 6x + 16y + 23 = 0\)
Correct answer is B
Equation of a circle: \((x - a)^{2} + (y - b)^{2} = r^{2}\)
where (a, b) and r are the coordinates of the centre and radius respectively.
Given : \((a, b) = (-3, -8); r = 4\sqrt{6}\)
\((x - (-3))^{2} + (y - (-8))^{2} = (4\sqrt{6})^{2}\)
\(x^{2} + 6x + 9 + y^{2} + 16y + 64 = 96\)
\(x^{2} + y^{2} + 6x + 16y + 9 + 64 - 96 = 0\)
\(\implies x^{2} + y^{2} + 6x + 16y - 23 = 0\)
Evaluate \(\frac{1}{1 - \sin 60°}\), leaving your answer in surd form.
\(1 - \sqrt{3}\)
\(2 - \sqrt{3}\)
\(4 - 2\sqrt{3}\)
\(4 + 2\sqrt{3}\)
Correct answer is D
No explanation has been provided for this answer.