Correct Answer: D. 225

Explanation

s = 120t - 16t\(^2\) - 16t\(^2\)

\(\frac{ds}{dt} = 120 - 32t\)

120 - 32t = 0

\(\frac{32t}{32}\) = \(\frac{120}{32}\)

t = 3.75

s = 120(3.75) - 16(3.75)\(^2\)

= 450 - 225

= 225

Explanation

\(\frac{625(\frac{3x}{4} - 1) + 125^{(x - 1)}}{5^{(3x - 2)}}\)

\(\frac{5^4(\frac{3x}{4} - 1) + 5^{3(x - 1)}}{5^{(3x - 2)}}\)

\(\frac{5^{3x - 4} + 5^{3x - 3}}{5^{3x} \times 5^{-2}}\)

= \(\frac{5^{3x}(5^{-4} + 5^{-3})}{5^{3x} \times 5{-2}}\)

= \(\frac{5^{-3}(5^{-1} + 1)}{5^{-2}}\)

= 5\(^{-1} (5^{-1} + 1)\)

= \(\frac{1}{5}(\frac{1}{5} + \frac{1}{1}\))

= \(\frac{1}{5} \times \frac{6}{5} = \frac{6}{25}\)

Correct Answer: C. 2 square units

Explanation

\(\int^0_{-2} x + 1\) x\(^2 + x|^0_{-2}\)

0 - (2\(^2 - 2\))

= 2 square units

Correct Answer: C. 45°

Explanation

\(2x + y = 4 \equiv y = 4 - 2x \implies m_{1} = -2\)

\(-3x + y + 7 = 0 \equiv y = -7 + 3x \implies m_{2} = 3\)

\(\tan \theta = \frac{m_{1} - m_{2}}{1 - m_{1}m_{2}} = \frac{-2 - 3}{1 - (-2)(3)} = \frac{-5}{-5} = 1\)

\(\tan \theta = 1 \implies \theta = 45°\)

Correct Answer: D. \(x = 4, \frac{-3}{2}\)

Explanation

\(\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x \end{vmatrix} = -3 \implies (1+2x)(3-x) - (-6) = -3\)

\(3 - x + 6x - 2x^{2} + 6 = -3\)

\(-2x^{2} + 5x + 3 + 6 + 3 = 0\)

Multiplying through with -1,

\(2x^{2} - 5x -12 = 0\)

\((2x + 3)(x - 4) = 0 \implies x = \frac{-3}{2} , 4\)

Correct Answer: C. - 0.01

Explanation

(1 + 3x )\(^6\) = (0.97)\(^6\)

1 + 3x = 0.97

3x = 0.97.1

\(\frac{3x}{3} = \frac{0.03}{3}\)

x = -0.01

Explanation

(b) The model score = 57

Correct Answer: D. \(\frac{8}{15}\)

Explanation

Prime number in the experiment are 2,3 and 5 with frequencies 10,14 and 8

P(obtaining a prime number) = \(\frac{10+14+8}{60}\)

= \(\frac{32}{60}\) or \(\frac{8}{15}\)

Explanation

\(\frac{log_2 ^8 + log_2 ^ {16} - 4 log_2^ 2}{log_4^{16}}\)

= \(\frac{3log_2 ^2 + 4 log_2^ 2 - 4 log_2 2}{2 log_4^4}\)

= \(\frac{3 + 4 - 4}{2}\)

= \(\frac{3}{2}\)

(b)

T\(_1\) + T\(_2\) = 6

a + a + d = 6

2a + d = 6.....(i)

a = a

a + 2d = ar, a + 6d = ar\(^2\)

\(\frac{a + 2d}{a} = \frac{a + 6d}{a + 2d}\)

(a + 2d)\(^2\) = \(\frac{a + 6d}{a + 2d}\)

(a + 2d)\(^2\) = a(a + 6d)

a\(^2\) + 4d\(^2\) + 4ad = a\(^2\) + 6ad

4d\(^2\) = 6ad - 4ad

4d\(^2\) = 2ad

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

a = 2d ......(ii)

from;

2a + d = 6

2(2d) + d = 6

4d + d = 6

5d = 6

d = \(\frac{6}{5}\)

= 1\(\frac{1}{5}\)

from; a = 2d

a = \(\frac{2 \times 6}{5}\)

= \(\frac{12}{5}\)

= 2\(\frac{2}{5}\)

First term = 2\(\frac{2}{5}\)

Common difference = 1\(\frac{4}{5}\)

Explanation

(a)

(b)(i) 80th percentile = \(\frac{80 \times 60th}{100} = \text{48th mark}\).

= 31.5.

(ii) Median = \(\frac{60th}{2} = \text{30th mark} = 32.3\)

(iii) Upper quartile = \(\frac{3 \times 60th}{4} = \text{45th mark} = 35.1\)

Lower quartile = \(\frac{60th}{4} = \text{15th mark} = 29.7\)

Semi-interquartile range = \(\frac{35.1 - 29.7}{2} = 2.7\)