1. **Problem 3.1:** Calculate the cost of producing one cylindrical can given the surface area formula and cost per m². The surface area $A$ of a cylinder is given by: $$A = 2 \pi r (r + h)$$ where $r$ is the radius and $h$ is the height. The cost per m² is 55.60. 2. **Problem 3.2:** Determine the length of the branding paper which includes an extra 0.8 cm for gluing. The length of the branding paper equals the circumference of the cylinder plus 0.8 cm: $$L = 2 \pi r + 0.8\text{ cm}$$ 3. **Problem 3.3:** Calculate the area of the branding paper. The branding paper covers the curved surface area: $$A_{paper} = L \times h$$ where $L$ is the length from 3.2 and $h$ is the height. 4. **Problem 3.4:** Calculate the weight of the branding paper. Given weight per area is 30 g per 10,000 cm², convert $A_{paper}$ to cm² and use: $$\text{Weight} = \frac{30}{10000} \times A_{paper}$$ 5. **Problem 3.5:** Calculate the volume of soup each can contains, filled to 90% capacity. Volume of cylinder: $$V = \pi r^2 h$$ Filled volume: $$V_{filled} = 0.9 \times V$$ 6. **Problem 4.1:** Probability of drawing two cylindrical tins without replacement. First draw cylindrical tin probability: $$P(C_1) = \frac{4}{10}$$ Second draw cylindrical tin given first was cylindrical: $$P(C_2|C_1) = \frac{3}{9}$$ Combined probability: $$P(C_1 \cap C_2) = P(C_1) \times P(C_2|C_1) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$$ 7. **Problem 4.2:** Probability of drawing a rectangular tin then a cylindrical tin without replacement. First draw rectangular tin probability: $$P(R_1) = \frac{6}{10}$$ Second draw cylindrical tin given first was rectangular: $$P(C_2|R_1) = \frac{4}{9}$$ Combined probability: $$P(R_1 \cap C_2) = P(R_1) \times P(C_2|R_1) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}$$