Correct Answer: A. TEAR

Correct Answer: D. 6cm

Explanation

. Total bulbs: n(Red)=5, n(B) = 7, n(G) = 4

(5+7+4) = 16

p(R)= 5/16, p(B) = 7/16, p(G) = 4/16

(a) p(All same colour of bulbs) = p(RR) Or p(BB) or p(GG)

\(\frac{5}{16}\) * \(\frac{4}{15}\) + \(\frac{7}{16}\) * \(\frac{6}{15}\) + \(\frac{4}{16}\) * \(\frac{3}{15}\)

= \(\frac{1}{12}\) + \(\frac{7}{40}\) + \(\frac{1}{20}\)

= \(\frac{10+21+6}{120}\) = \(\frac{37}{120}\)

(b) All different colors = 1-p (AIl the same colour) =

1 - \(\frac{37}{120}\) = \(\frac{83}{120}\)

(c) p(At least one red) = p(RR) + p(RB) + p(RG)

\(\frac{5}{16}\) * \(\frac{4}{15}\) + \(\frac{5}{16}\) * \(\frac{7}{15}\) + \(\frac{5}{16}\) * \(\frac{4}{15}\)

= \(\frac{1}{12}\) + \(\frac{7}{48}\) + \(\frac{1}{12}\)

= \(\frac{8+7}{48}\) = \(\frac{15}{48}\)

= \(\frac{5}{16}\)

Correct Answer: C. 37\(^?\)

Explanation

y = 2x + 5

m1 = 2

2y = x - 6

tanθ = \(\frac{3}{2}\) ÷ (1+1)

tanθ = \(\frac{3}{2}\) ÷ 2

θ = \(tan^{-1} (\frac{3}{4})\)

θ = 36.87º

θ = 37º

Correct Answer: B. \(16 \sqrt{3}\) N

Explanation

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

\(\frac{\mathrm d y}{\mathrm d x} = 4x + 1\)

At (2, 7), the gradient is \(4(2) + 1 = 9\)

The gradient of normal = \(-1 \div 9 = \frac{-1}{9}\)

Equation of the normal = \(y - 7 = \frac{-1}{9}(x - 2)\)

\(9y - 63 = 2 - x\)

At the point of meeting the x- axis, y = 0

\(0 - 63 = 2 - x \implies x = 65\)

The coordinate of P = (65, 0).

Explanation

(a) \(2x - y = \frac{9}{2} .... (1)\)

\(x + 4y = 0 ....... (2)\)

From (2), x = - 4y. Put into (1), we have

\(2(- 4y) - y = \frac{9}{2}\)

\(-8y - y = \frac{9}{2} \implies - 9y = \frac{9}{2}\)

\(y = \frac{\frac{9}{2}}{-9} = -\frac{1}{2}\)

\(x = - 4y = -4(-\frac{1}{2})\)

\(x = 2\)

\((x, y) = (2, -\frac{1}{2})\)

(b)

In \(\Delta ACD, \stackrel\frown{ACD} = 52° \) \(\stackrel\frown{ACD} = \stackrel\frown{DAT}\)

\(\therefore \stackrel\frown{BCD} = 40° + 52° = 92°\)

Note \(\stackrel\frown{BCD}\) and \(\stackrel\frown{BAD}\) are supplementary = 180°

\(\therefore \stackrel\frown{BAD} = 180° - 92° = 88°\)

Correct Answer: D. 4

Explanation

An equation can be written as \(x^{2} - (\alpha + \beta)x + (\alpha\beta) = 0\)

Making the coefficient of \(x^{2}\) = 1 in the given equation, we have

\(x^{2} + \frac{5}{2}x + \frac{n}{2} = 0\)

Comparing, we have \(\alpha\beta = \frac{n}{2} = 2 \implies n = 4\)

Correct Answer: B. 0.014

Correct Answer: D. (-1, 3)

Explanation

x\(^2\) - 2x - 3 = 0

x\(^2\) - 3x + x - 3 = 0

x(x - 3) + 1(x - 3) = 0

(x + 1)(x - 3) = 0

x = (-1, 3)