Explanation

(a) From the diagram,

\(|AC|^{2} = 24^{2} + 7^{2}\)

= \(576 + 49\)

\(|AC|^{2} = 625 \implies |AC| = \sqrt{625} = 25 cm\)

\(Area = \frac{1}{2} \times b \times h\)

= \(\frac{7 \times 24}{2}\)

= \(84 cm^{2}\).

(b) Total Surface Area of prism

= Area of two right angles ( ABC and EFD) + Area of 3 rectangles (ACDF, ABEF and BCDE)

= \(( 2 \times 84 cm^{2}) + (25 \times 40)cm^{2} + (24 \times 40)cm^{2} + (7 \times 40)cm^{2}\)

= \((168 + 1000 + 960 + 280) cm^{2}\)

= \(2408 cm^{2}\)

Correct Answer: C. N4.00

Explanation

Ratio = 6 + 7 + 8 = 21

21 = N10.50

1 = 1050K/21 = 50K

6 = 6 x 50K = N3.00

7 = 7 x 50K = N3.50

8 = 8 x 50K = N4.00

the Largest share = N4.00

Explanation

(a) Area of?PQR = \(\frac{1}{2}\) * b * 12 = 20

b = \(\frac{2 x 20}{12}\) = 3.33cm

|QR| = |UT| = 3.33cm

IPQ| = √(12\(^2\) + 5\(^2\))\(^2\) = 13cm

IRT| = |QU| = 12cm

From?RTS

sin 50°= \(\frac{12}{|RS|}\)

|RS| = \(\frac{12}{sin 50°}\) = 15.6649 cm

|TS| = \(\frac{12}{tan 50°}\) = 10.069cm

Perimeter = 13 + 3.33 + 15.6649 + 18.399

= 50.3939cm = 50cm (nearest whole number)

(b)Area= \(\frac{1}{2}\) (3.33+ 18.4004)12

6 (21.7304) = 130.3824cm\(^2\)

= 130 cm\(^2\) (nearest whole number)

Correct Answer: D. \(PQ\)

Explanation

A \(1 \times 2\) matrix can be multiplied by \(2 \times 2\) matrix to give a \(1 \times 2\) matrix.

Correct Answer: C. 4i - j

Explanation

\(\overrightarrow{OA} = (3, 4)\)

\(\overrightarrow{OB} = (5, -6)\)

\(\overrightarrow{OM} = (\frac{3 + 5}{2}, \frac{4 + (-6)}{2})\)

= \((4, -1) = 4i - j\)

Correct Answer: B. x< -2 or x > 4

Explanation

P(a sum of 7) = \(\frac{6}{36}\) and P(a sum of 4) = \(\frac{3}{36}\)

P(a sum of 4) = \(\frac{3}{36}\)

P(a sum of 7 and a sum of 4) = \(\frac{6}{36} \times \frac{3}{36}\) = \(\frac{1}{72}\)

Explanation

P(a family has a car) = p = 65% = 0.65

P(a family has no car) = q = 35% = 0.35

For seven families and using the binomial probability distribution, we have

\((p + q)^{7} = p^{7} + 7p^{6} q + 21p^{5} q^{2} + 35p^{4} q^{3} + 35p^{3} q^{4} + 21p^{2} q^{5} + 7p q^{6} + q^{7}\)

(a) P(exactly 5 have cars) = \(21p^{5} q^{2} = 21(0.65)^{5} (0.35)^{2}\)

= \(21 \times 0.116029 \times 0.1225 = 0.29848\)

\(\approxeq 0.298\) (to 3 d.p)

(b) P(3 or 4 of them have cars) = \(35p^{4} q^{3} + 35p^{3} q^{4}\)

= \(35(0.65)^{4} (0.35)^{3} + 35(0.65)^{3} (0.35)^{4}\)

= \(35(0.27463)(0.015006) + 35(0.178506)(0.042875)\)

= \(0.144238 + 0.267871\)

\(\approxeq 0.412\) (to 3 d.p)

(c) P(at most 2 of them have cars) = P(0 cars) + P(1 car) + P(2 cars)

= \(q^{7} + 7p q^{6} + 21p^{2} q^{5}\)

= \((0.35)^{7} + 7(0.65) (0.35)^{6} + 21 (0.65)^{2} (0.35)^{5}\)

= \(0.000643 + 0.00836 + 0.04658\)

= \(0.5558 \approxeq 0.556\) (3 d.p)

Correct Answer: C. 96%

Explanation

We find the initial velocity using the formula \(v = u + at\)

\(33 = u + 3(6) \implies 33 = u + 18\)

\(u = 15ms^{-1}\)

Distance \(s = ut + \frac{1}{2} at^{2}\)

= \(15 \times 6 + \frac{1}{2} (3)(6^{2})\)

= \(90 + 54 = 144m\)