1. **Problem 1: Median, Upper Quartile, and Number of Students Above a Mark** The box and whisker plot shows marks of 32 students with: - Minimum = 1 - Lower quartile (Q1) = 2 - Median (Q2) = 3 - Upper quartile (Q3) = 6 - Maximum = 7 (a) The median mark is the middle value, so **median = 3**. (b) The upper quartile is the value below which 75% of data lie, so **upper quartile = 6**. (c) To estimate the number of students with marks greater than 6: - Total students = 32 - Marks greater than 6 lie between 6 and 7 (right whisker). - Since Q3 = 6, 25% of students have marks above 6. - Number of students = 25% of 32 = $0.25 \times 32 = 8$. 2. **Problem 2: Area of Triangle PQR** Given: - Side $PQ = 13$ cm - Side $PR = 4$ cm - Angle at $P = 70^\circ$ Formula for area of triangle using two sides and included angle: $$\text{Area} = \frac{1}{2} ab \sin C$$ where $a=13$, $b=4$, and $C=70^\circ$. Calculate: $$\text{Area} = \frac{1}{2} \times 13 \times 4 \times \sin 70^\circ = 26 \times \sin 70^\circ$$ Using $\sin 70^\circ \approx 0.9397$: $$\text{Area} \approx 26 \times 0.9397 = 24.43 \text{ cm}^2$$ 3. **Problem 3: Function Equation of the Graphs** - Left graph: Sinusoidal wave with amplitude 3, max at $x=0$, period $2\pi$. - This matches $y = 3 \cos x$. - Right graph: Sinusoidal wave oscillating between -2 and 4. - Amplitude $= \frac{4 - (-2)}{2} = 3$. - Midline $= \frac{4 + (-2)}{2} = 1$. - Higher frequency, period $= \pi$ (since two full cycles between $-2\pi$ and $2\pi$). - Function: $y = 3 \cos 2x + 1$. 4. **Problem 4: Depth of Water in Harbor** Given: $$d = P + Q \cos \left( \frac{\pi}{6} t \right), 0 \leq t \leq 24$$ From graph: - Max depth $= 14.6$ at $t=0$ (cosine max = 1) - Min depth $= 8.2$ at $t=6$ (cosine min = -1) (a) Find $P$ and $Q$: $$P + Q = 14.6$$ $$P - Q = 8.2$$ Adding: $$2P = 22.8 \Rightarrow P = 11.4$$ Subtracting: $$2Q = 6.4 \Rightarrow Q = 3.2$$ (b) Find first time $t$ when $d=11.4$: $$11.4 = 11.4 + 3.2 \cos \left( \frac{\pi}{6} t \right)$$ $$0 = 3.2 \cos \left( \frac{\pi}{6} t \right) \Rightarrow \cos \left( \frac{\pi}{6} t \right) = 0$$ Cosine zero at $\frac{\pi}{2}$, so: $$\frac{\pi}{6} t = \frac{\pi}{2} \Rightarrow t = 3$$ (c) Depth less than 11.4 means: $$d < 11.4 \Rightarrow 11.4 + 3.2 \cos \left( \frac{\pi}{6} t \right) < 11.4$$ $$\cos \left( \frac{\pi}{6} t \right) < 0$$ Cosine is negative in intervals: $$\frac{\pi}{2} < \frac{\pi}{6} t < \frac{3\pi}{2}$$ Multiply by 6/$\pi$: $$3 < t < 9$$ Similarly for next period: $$15 < t < 21$$ 5. **Problem 5: Find Missing Angles and Sides in Triangles** **First triangle:** - Angles: $40^\circ$, $94^\circ$, and $x$. - Sum of angles in triangle = $180^\circ$. Calculate $x$: $$x = 180^\circ - 40^\circ - 94^\circ = 46^\circ$$ Use Law of Sines to find side $x$ opposite $94^\circ$: $$\frac{x}{\sin 94^\circ} = \frac{10.5}{\sin 40^\circ}$$ $$x = \frac{10.5 \sin 94^\circ}{\sin 40^\circ}$$ Using $\sin 94^\circ \approx 0.9986$, $\sin 40^\circ \approx 0.6428$: $$x \approx \frac{10.5 \times 0.9986}{0.6428} = 16.3$$ **Second triangle:** - Angles: $19^\circ$, unknown angle $x$, and third angle. - Sides: 65 cm adjacent to $19^\circ$, 86 cm adjacent to $x$. Sum of angles: $$x = 180^\circ - 19^\circ - \text{third angle}$$ Use Law of Cosines or Sines depending on given info. Assuming third angle is $x$'s complement: Use Law of Sines: $$\frac{65}{\sin x} = \frac{86}{\sin 19^\circ}$$ $$\sin x = \frac{65 \sin 19^\circ}{86}$$ Using $\sin 19^\circ \approx 0.3256$: $$\sin x = \frac{65 \times 0.3256}{86} = 0.246$$ $$x = \arcsin(0.246) \approx 14.2^\circ$$