answer:Since the lengths of the three sides of the triangle are 3, 1-2a, and 8, we have 8-3 < 1-2a < 8+3, which simplifies to -5 < a < -2. Therefore, the correct choice is boxed{text{C}}. This problem requires setting up inequalities based on the relationship between the sides of a triangle. The key to solving this problem is to be familiar with the relationship between the sides of a triangle, which states that the sum of any two sides must be greater than the third side, and the difference between any two sides must be less than the third side.

answer:# Solution: (Ⅰ) Calculation of E(X) Given the probabilities p_0=0.4, p_1=0.3, p_2=0.2, and p_3=0.1, we calculate the expected value E(X) as follows: [ E(X) = 0 times 0.4 + 1 times 0.3 + 2 times 0.2 + 3 times 0.1 ] [ = 0 + 0.3 + 0.4 + 0.3 ] [ = 1 ] Thus, E(X) = 1. Therefore, we have E(X) = boxed{1}. (Ⅱ) Proof for the Probability of Extinction p Given that p_0+p_1+p_2+p_3=1, we can express E(X) as E(X)=p_1+2p_2+3p_3. The equation p_0+p_1x+p_2x^2+p_3x^3=x can be rearranged as: [ p_0-(1-p_1)x+p_2x^2+p_3x^3=0 ] [ Rightarrow p_0+p_2x^2+p_3x^3-(p_0+p_2+p_3)x=0 ] [ Rightarrow p_0(1-x)+p_2x(x-1)+p_3x(x-1)(x+1)=0 ] [ Rightarrow left(x-1right)[p_3x^2+(p_2+p_3)x-p_0]=0 ] Let f(x)=p_3x^2+(p_2+p_3)x-p_0. If p_3 neq 0, the axis of symmetry of f(x) is x=-frac{p_2+p_3}{2p_3} < 0. Note that f(0)=-p_0 < 0 and f(1)=2p_3+p_2-p_0=p_1+2p_2+3p_3-1=E(X)-1. If p_3=0, then f(1)=E(X)-1. - When E(X) leq 1, f(1) leq 0, implying the positive real root x_0 of f(x)=0 is x_0 geq 1. Thus, the smallest positive real root of the original equation is p=1. - When E(X) > 1, f(1) > 0, implying the positive real root x_0 of f(x)=0 is x_0 < 1. Thus, the smallest positive real root of the original equation is p < 1. Therefore, we have p=1 when E(X) leq 1, and p < 1 when E(X) > 1. (Ⅲ) Practical Implications The practical implications of the conclusion in question (Ⅱ) are as follows: - When the expected number of offspring E(X) leq 1, the microorganism is on the verge of extinction after multiple generations of reproduction, indicating a population that cannot sustain itself over time. - When E(X) > 1, the microorganism has the possibility of continued reproduction after multiple generations, suggesting a population that can grow or at least maintain its numbers over time. This analysis highlights the critical role of the reproductive rate in determining the long-term viability of a population.

answer:To find the number of moles of Methanol formed, we need to follow the stoichiometry of the reactions given. Let's tackle each reaction step by step. **Step 1: Methane reacts with Oxygen** The balanced chemical equation for the reaction of methane with oxygen is: CH₄(g) + 2O₂(g) -> CO₂(g) + 2H₂O(g) From the equation, 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O. We have 5 moles of CH₄, so it would require 5 * 2 = 10 moles of O₂ to react completely. However, we have 15 moles of O₂ available, which is more than enough. Therefore, O₂ is in excess, and CH₄ is the limiting reactant. The reaction will produce 5 moles of CO₂ (since 1 mole of CH₄ produces 1 mole of CO₂). **Step 2: Carbon Dioxide reacts with Hydrogen** The balanced chemical equation for the reaction of carbon dioxide with hydrogen is: CO₂(g) + 3H₂(g) -> CH₃OH(g) + H₂O(g) From the equation, 1 mole of CO₂ reacts with 3 moles of H₂ to produce 1 mole of CH₃OH. We have 5 moles of CO₂ from the first reaction, and we need 5 * 3 = 15 moles of H₂ to react with all the CO₂. However, we only have 10 moles of H₂ available, which means H₂ is the limiting reactant in this step. To find out how many moles of CH₃OH we can produce with the available H₂, we use the stoichiometry of the reaction. Since 3 moles of H₂ produce 1 mole of CH₃OH, 10 moles of H₂ will produce 10/3 moles of CH₃OH. Therefore, the number of moles of Methanol (CH₃OH) formed is boxed{10/3} moles, which is approximately 3.33 moles.

answer:We first determine the area of the base of the pyramid, which is a regular octagon. The octagon can be divided into 8 isosceles triangles by drawing lines from the center O to each vertex. The side length s of each isosceles triangle is the same as the side of the octagon. To find the area of each isosceles triangle, we need the length of the apothem, which is also the altitude of each triangle. Given the side length s of the octagon is 10, we can use the formula for the side length of a regular octagon in terms of its circumradius R: [ s = 2R sin(45^circ) = Rsqrt{2} ] Here, R is the circumradius of the octagon, which is also s divided by sqrt{2}, giving us R = frac{s}{sqrt{2}} = frac{10}{sqrt{2}} = 5sqrt{2}. The area of each triangle is given by: [ text{Area} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 10 times 5sqrt{2} = 25sqrt{2} ] The total area of the octagon (base of the pyramid) is 8 times 25sqrt{2} = 200sqrt{2}. The next step is to find the height PO of the pyramid. Since PAE is an equilateral triangle with side length 10, this implies that PO lies along the altitude of this triangle, splitting PAE into two 30^circ-60^circ-90^circ triangles. The length PO is opposite the 60^circ angle, giving: [ PO = frac{s sqrt{3}}{2} = frac{10 sqrt{3}}{2} = 5sqrt{3} ] Finally, the volume of the pyramid is calculated by: [ V = frac{1}{3} times text{Area}_{text{base}} times text{height} = frac{1}{3} times 200sqrt{2} times 5sqrt{3} = frac{1000 sqrt{6}}{3} ] Putting it all together, we find the volume of the pyramid is: [ boxed{frac{1000 sqrt{6}}{3}} ]