Explanation

Assumed mean = 1.45

Mean \(\bar{x} = A + \frac{\sum fd}{\sum f}\)

= \(1.45 + \frac{0.4}{20}\)

= \(1.45 + 0.02\)

= \(1.47\)

Correct Answer: D. \((6 - 4\sqrt{3})N\)

Explanation

The vertical component = \(8 \sin 300 + 6 \sin 90 + 4 \sin 180\)

= \(-4\sqrt{3} + 6 + 0 \)

= \((6 - 4\sqrt{3}) N\)

Explanation

(a) p(Kofi passes) = \(\frac{9}{10}\), p(Kofi fails) = \(\frac{1}{10}\)

p(Kwasi passes) = \(\frac{4}{5}\), p(Kwasi fails) = \(\frac{1}{5}\)

p(Ama passes) = x, p(Ama fails) = 1 - x.

p(only one passes) = p(only Kofi passes) + p(only Kwasi passes) + p(only Ama passes)

= \((\frac{9}{10} \times \frac{1}{5} \times (1 - x)) + (\frac{4}{5} \times \frac{1}{10} \times (1 - x)) + (x \times \frac{1}{10} \times \frac{1}{5})\)

= \(\frac{9(1 - x)}{50} + \frac{4(1 - x)}{50} + \frac{x}{50}\)

\(\implies \frac{9}{50} = \frac{9 - 9x + 4 - 4x + x}{50} = \frac{13 - 12x}{50}\)

\(\therefore 13 - 12x = 9 \implies 12x = 4\)

\(x = \frac{4}{12} = \frac{1}{3}\).

(b) p(at least one passes) = 1 - p(none passes).

p(none passes) = \(\frac{1}{10} \times \frac{1}{5} \times \frac{2}{3}\)

= \(\frac{1}{75}\)

p(at least one passes) = \(1 - \frac{1}{75}\)

= \(\frac{74}{75}\)

Correct Answer: C. \(\frac{1}{5}\)(3i - 4j)

Explanation

Resultant

(- 2i - 3j) + (5i - j)

= 3i - 4j

Unit vector

= \(\frac{3i - 4j}{|3i - 4j|} = \frac{3i - 4j}{\sqrt{3i } + (-4)}\)

= \(\frac{3i - 4j}{\sqrt{25}}\)

= \(\frac{3i - 4j}{5}\)

Explanation

(a)(i) \(T_{n} = ar^{n - 1}\) (terms of a G.P)

\(T_{1} + T_{2} + T_{3} = a + ar + ar^{2} = 7 ... (1)\)

\(a(ar)(ar^{2}) = 8 .... (2)\)

\((a^{3} r^{3}) = 8 \implies (ar)^{3} = 2^{3}\)

\(\therefore T_{2} = ar = 2\)

\(a + ar^{2} = 5 \implies a(1 + r^{2}) = 5\)

\(a = \frac{5}{1 + r^{2}}\)

\((\frac{5}{1 + r^{2}})(\frac{5r^{2}}{1 + r^{2}}) = 4 = 2^{2}\)

\(\frac{25r^{2}}{(1 + r^{2})^{2}} = 2^{2}\)

Taking square root of both sides, we have

\(\frac{5r}{1 + r^{2}} = 2 \implies 5r = 2 + 2r^{2}\)

\(2r^{2} - 5r + 2 = 0 \implies 2r^{2} - 4r - r + 2 = 0\)

\(2r(r - 2) - 1(r - 2) = 0 \implies r = \frac{1}{2} ; r = 2\)

Since it is a decreasing sequence, the common ratio \(r = \frac{1}{2}\).

(ii) \(T_{2} = 2 = ar\)

\(\frac{a}{2} = 2 \implies a = 4\)

\(T_{3} = ar^{2} = 4(\frac{1}{2})^{2}\)

= \(4(\frac{1}{4}) = 1\)

\(\therefore \text{The first 3 terms of the sequence are } 4, 2, 1.\)

(b) \(\int_{1} ^{5} (x + \frac{2}{x^{2}}) \mathrm {d} x\)

\(Height (h) = 1 ; y_{1} = 3.0 ; y_{5} = 5.08\)

\(y_{2} = 2.5 ; y_{3} = 3.222 ; y_{4} = 4.125\)

\(y_{1} + y_{5} = 3.0 + 5.08 = 8.08\)

\(y_{2} + y_{3} + y_{4} = 2.5 + 3.222 + 4.125 = 9.847\)

\(2(y_{2} + y_{3} + y_{4}) = 2(9.847) = 19.694\)

\(Area = \frac{1}{2} \times 1[8.08 + 19.694]\)

= \(\frac{1}{2} [27.774]\)

= \(13.887 \approxeq 13.89\) (to 2 d.p.)

Correct Answer: D. -3

Explanation

\(2^{x} = 0.125 = \frac{125}{1000}\)

\(2^{x} = \frac{1}{8} = 8^{-1}\)

\(2^{x} = 2^{-3}\)

\(x = -3\)

Correct Answer: B. \(1\frac{1}{6}\)

Explanation

\(\int^1_0 x(x^2-2)^2 dx\)

\((x^2-2)^2=x^4-2x^2-2x^2+4\)

=\(x^4-4x^2+4\)

\(x(x^2-2)^2=x(x^4-4x^2+4)\)

=\(x^5-4x^3+4x\)

\(\int^1_0 x(x^2-2)^2 dx = \int^1_0 x^5 - 4x^3 + 4x dx\)

=\((\frac{x^6}{6} - x^4+2x^2)^1_0\)

= \((\frac{(1)^6}{6} - (1)^4 +2(1)^2)-(\frac{(0)^6}{6} - (0)^4+2(0)^2)\)

=\(\frac{7}{6} - 0 =\frac{7}{6}\)

\(\therefore 1\frac{1}{6}\)

Explanation

Let w be the adjacent side, then w\(^2\) = (P + q)\(^2\) = 4pq so that w = 2\(\sqrt{pq}\)

Since tan x \(\frac{p - q}{2\sqrt{pq}}\),

substituting; 1 - tan\(^2\)x = 1 - (\(\frac{p - q}{2\sqrt{pq}}\))\(^2\)

= 1 - \(\frac{P^2 - 2pq + q^2}{4pq}\)

= \(\frac{4pq - p^2 + 2pq - q^2}{4pq}\)

Therefore, 1 - tan\(^2\)x = \(\frac{6pq - p^2 - q^2}{4pq}\)

Correct Answer: A. -12

Explanation

Using remainder theorem, since x - 3 is a factor, then

given \(2x^{2} - 2x + p\), f(3) = 0

\(2(3^{2}) - 2(3) + p = 0 \implies 18 - 6 = -p\)

\(p = -12\)

Explanation

(a) \(u=6ms^{-1};s=70m;v=20ms^{-1}a=?\)

\(v^2=u^2+2as\)

⇒\(a=\frac{v^2 - u^2}{2s}=\frac{20^2 - 6^2}{2(70)}\)

\(a=\frac{400 - 36}{140}=\frac{364}{140}\)

\(∴a=2.6ms^{-2}\)(to 1 d.p)

(b) \(u=6ms^{-1};s=70m;v=20ms^{-1}t=?\)

v=u+at

⇒\(t=\frac{v - u}{a}=\frac{20 - 6}{2.6}\)

\(∴t=\frac{14}{2.6}=5.4s\)