1. **Problem 1: How many of them were beetles?** Given: Total insects = 88, Beetles = 3 The question asks how many were beetles, which is directly given as 3. **Answer:** 3 beetles. 2. **Problem 2: Calculate the length of ST in the garden shed diagram.** Given: - SV = 190 cm - TU = 230 cm - VU = 120 cm Assuming the quadrilateral is a trapezium or a shape where ST can be found using the Pythagorean theorem or coordinate geometry, but since no angles are given, we consider ST as the diagonal of a right triangle formed by sides SV and VU. Using the Pythagorean theorem: $$ST = \sqrt{SV^2 + VU^2} = \sqrt{190^2 + 120^2}$$ Calculate: $$190^2 = 36100$$ $$120^2 = 14400$$ $$ST = \sqrt{36100 + 14400} = \sqrt{50500}$$ Simplify: $$ST \approx 224.5 \text{ cm}$$ 3. **Problem 3: Class 5C money collection statistics** Amounts collected: 27, 26, 17, 27, 18, 21, 23, 19, 18, 27, 24, 20, 31, 28 (a) Find the median. Step 1: Sort the data: $$17, 18, 18, 19, 20, 21, 23, 24, 26, 27, 27, 27, 28, 31$$ Step 2: Number of data points = 14 (even) Median = average of 7th and 8th values: $$\frac{23 + 24}{2} = 23.5$$ (b) Find the range. Range = max - min = 31 - 17 = 14 (c) Compare with class 5M (median = 10, range = 17): - Class 5C has a higher median (23.5 vs 10), indicating generally higher amounts collected. - Class 5M has a larger range (17 vs 14), indicating more variability in amounts collected. 4. **Problem 4: Calculate the area being watered (semi-circle + right angled triangle).** Given: - Radius of semi-circle = 10 m - Triangle base = 8 m - Triangle height = 6 m Step 1: Area of semi-circle: $$A_{semi} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (10)^2 = 50\pi$$ Step 2: Area of triangle: $$A_{triangle} = \frac{1}{2} \times base \times height = \frac{1}{2} \times 8 \times 6 = 24$$ Step 3: Total area: $$A_{total} = 50\pi + 24 \approx 50 \times 3.1416 + 24 = 157.08 + 24 = 181.08 \text{ m}^2$$ 5. **Problem 5: Find average speed for the rest of Ross's journey.** Given: - Total distance = 190 miles - Total time = 3 hours 30 minutes = 3.5 hours - First part: 2 hours at 68 mph Step 1: Distance covered in first part: $$d_1 = speed \times time = 68 \times 2 = 136 \text{ miles}$$ Step 2: Remaining distance: $$d_2 = 190 - 136 = 54 \text{ miles}$$ Step 3: Remaining time: $$t_2 = 3.5 - 2 = 1.5 \text{ hours}$$ Step 4: Average speed for rest: $$v_2 = \frac{d_2}{t_2} = \frac{54}{1.5} = 36 \text{ mph}$$ **Final answers:** 1. 3 beetles 2. Length of ST is approximately 224.5 cm 3. (a) Median = 23.5 (b) Range = 14 (c) Class 5C collected generally more money with less variability than 5M 4. Area being watered is approximately 181.08 m² 5. Average speed for rest of journey is 36 mph