🧠:<think>嗯，我现在要解决这个问题，首先得仔细理解题目。题目说的是一位农场主，他不滑雪，而是种植有机蔬菜，想要最大化产量和可持续性。他的农场总面积是10000平方米，种植四种蔬菜：胡萝卜、番茄、黄瓜和生菜。每种蔬菜的产量和灌溉需求不同。问题分为两部分，第一部分是建立一个优化模型，最大化总产量，同时满足水资源的限制。第二部分是增加菠菜，重新建立模型。首先，我得明确变量。设x1、x2、x3、x4分别为胡萝卜、番茄、黄瓜和生菜的种植面积，单位是平方米。每种蔬菜的种植面积必须是整数，这点很重要，可能后面需要用整数规划来解决。目标函数是总产量最大化。那每种蔬菜的产量是面积乘以每平方米的产量。所以目标函数应该是：最大化 Z = 5x1 + 4x2 + 3x3 + 2x4对吗？因为胡萝卜每平方米产5公斤，番茄4公斤，黄瓜3公斤，生菜2公斤。接下来是约束条件。首先，总面积不能超过10000平方米，所以：x1 + x2 + x3 + x4 ≤ 10000然后是水资源的限制。水库容量是10000升，每天补充500升水。那每天的总用水量不能超过水库容量加上每天补充的水量吗？或者水库每天只能提供500升水？这点需要仔细理解题目。题目说“水库容量为10000升，每天可以补充500升水。”这可能意味着水库每天可以提供500升水，而水库本身的容量可能不影响，因为每天补充500升，所以每天的用水量不能超过500升。或者，水库容量是10000升，可以存储，但每天只能补充500升。这可能意味着水库每天的可用水量是500升，因为补充的量是500升，而水库容量可能只是存储，不影响每天的用水量。或者，可能水库每天可以提供500升水，而水库容量是10000升，可能意味着水库可以存储，但每天只能补充500升，所以每天的用水量不能超过500升。这可能更合理，因为如果水库容量很大，而每天只能补充500升，那么每天的用水量不能超过500升，否则水库会被耗尽。所以，每天的总灌溉用水量不能超过500升。每种蔬菜的用水量是面积乘以每平方米每天的用水量。所以：0.8x1 + 1.2x2 + 1x3 + 0.6x4 ≤ 500另外，所有变量都是非负整数，因为种植面积不能为负，且必须是整数。所以，第一部分的优化模型应该是：最大化 Z = 5x1 + 4x2 + 3x3 + 2x4约束条件：1. x1 + x2 + x3 + x4 ≤ 100002. 0.8x1 + 1.2x2 + 1x3 + 0.6x4 ≤ 5003. x1, x2, x3, x4 ≥ 0 且为整数现在，第二部分，农场主增加菠菜，每平方米产6公斤，每天用水0.9升。同样，总面积不变，还是10000平方米，水资源限制还是每天500升。目标还是最大化总产量。所以，现在变量增加到x5，菠菜的种植面积。目标函数变为：最大化 Z = 5x1 + 4x2 + 3x3 + 2x4 + 6x5约束条件：1. x1 + x2 + x3 + x4 + x5 ≤ 100002. 0.8x1 + 1.2x2 + 1x3 + 0.6x4 + 0.9x5 ≤ 5003. x1, x2, x3, x4, x5 ≥ 0 且为整数这样，模型就建立好了。不过，我得再检查一下水资源的约束是否正确。水库容量是10000升，每天补充500升。这可能意味着水库每天的用水量不能超过500升，因为水库每天补充500升，所以每天只能用500升。或者，水库容量是10000升，可以存储，但每天只能补充500升，所以每天的用水量不能超过水库的可用水量加上补充的水量？这可能更复杂，但题目可能只是说每天的用水量不能超过500升，因为水库每天补充500升，所以总用水量不能超过500升，否则水库会被耗尽。或者，水库容量是10000升，每天补充500升，所以水库每天的可用水量是500升，因为补充的量是500升，而水库容量可能只是存储，不影响每天的用水量。所以，总用水量每天不能超过500升。所以，约束条件是正确的。总结一下：第一部分：目标函数：Z = 5x1 +4x2 +3x3 +2x4 → 最大化约束：x1 +x2 +x3 +x4 ≤100000.8x1 +1.2x2 +x3 +0.6x4 ≤500x1,x2,x3,x4 ≥0，整数第二部分：目标函数：Z =5x1 +4x2 +3x3 +2x4 +6x5 → 最大化约束：x1 +x2 +x3 +x4 +x5 ≤100000.8x1 +1.2x2 +x3 +0.6x4 +0.9x5 ≤500x1,x2,x3,x4,x5 ≥0，整数这样应该没错。</think>

🧠:<think>Alright, so I'm trying to solve these two sub-problems related to water demand in a city divided into three regions. Let me take it step by step because I want to make sure I understand everything correctly.Starting with Sub-problem 1: Determine the total water demand for the entire city at t = 10 days. Okay, so each region has its own quadratic function for water demand. That means I need to calculate the demand for each region separately when t is 10 and then add them all together to get the total demand.Let me write down the functions again to keep track:- Region A: ( D_A(t) = 2t^2 + 3t + 5 )- Region B: ( D_B(t) = -t^2 + 4t + 6 )- Region C: ( D_C(t) = 3t^2 - 2t + 4 )So, for t = 10, I need to plug in 10 into each of these functions.Starting with Region A:( D_A(10) = 2*(10)^2 + 3*(10) + 5 )Calculating each term:- ( 2*(10)^2 = 2*100 = 200 )- ( 3*(10) = 30 )- The constant term is 5.Adding them up: 200 + 30 + 5 = 235 million liters.Wait, that seems straightforward. Let me double-check:2*(10)^2 is 2*100 which is 200. 3*10 is 30. 200 + 30 is 230, plus 5 is 235. Yep, that's correct.Moving on to Region B:( D_B(10) = -(10)^2 + 4*(10) + 6 )Calculating each term:- ( -(10)^2 = -100 )- ( 4*(10) = 40 )- The constant term is 6.Adding them up: -100 + 40 + 6 = (-100 + 40) is -60, plus 6 is -54.Wait, that can't be right. Water demand can't be negative. Did I do something wrong?Let me check the function again. Region B's demand is ( -t^2 + 4t + 6 ). So for t=10, it's -100 + 40 + 6. Hmm, that does give -54. But negative water demand doesn't make sense in this context. Maybe the function is only valid for a certain range of t where the demand is positive. Perhaps t=10 is beyond that range? Or maybe I made a mistake in interpreting the function.Wait, let me recalculate:( D_B(10) = -(10)^2 + 4*10 + 6 = -100 + 40 + 6 ). So, -100 + 40 is -60, plus 6 is -54. Hmm, same result. That's negative, which doesn't make sense. Maybe the function is supposed to be ( -t^2 + 4t + 6 ), so perhaps it's a downward opening parabola, and after a certain point, the demand becomes negative, which would imply that the model isn't valid beyond that point.But since the problem is asking for t=10, I guess we have to go with the mathematical result even if it's negative. Maybe in reality, the demand can't be negative, so perhaps the city would have to adjust, but the question is just asking for the total water demand as per the functions given. So, I'll proceed with -54 million liters for Region B.But wait, that would mean the total demand is less because of a negative contribution from Region B. That seems odd. Maybe I made a mistake in the calculation.Wait, let me check again:- ( t^2 = 100 )- So, -100 + 40 + 6. Yes, that's -54. Hmm.Alternatively, maybe the function is supposed to be ( -t^2 + 4t + 6 ), which is a quadratic that peaks at t = -b/(2a) = -4/(2*(-1)) = 2 days. So, the maximum demand for Region B is at t=2, and after that, it decreases. So, at t=10, it's definitely on the downward slope, but it's still possible that it becomes negative. So, perhaps the model is correct, and the demand is negative, but in reality, it can't be negative, so maybe the city would have to cap it at zero. But the question doesn't specify that, so I think I should just use the value as per the function, even if it's negative.So, moving on, Region C:( D_C(10) = 3*(10)^2 - 2*(10) + 4 )Calculating each term:- ( 3*(10)^2 = 3*100 = 300 )- ( -2*(10) = -20 )- The constant term is 4.Adding them up: 300 - 20 + 4 = 284 million liters.So, now, adding up all three regions:Region A: 235Region B: -54Region C: 284Total demand: 235 + (-54) + 284Calculating that:235 - 54 = 181181 + 284 = 465 million liters.So, the total water demand at t=10 is 465 million liters.Wait, but Region B's demand is negative, which seems odd. Maybe I should check if I interpreted the function correctly. Let me look back at the problem statement.The functions are given as:- Region A: ( 2t^2 + 3t + 5 )- Region B: ( -t^2 + 4t + 6 )- Region C: ( 3t^2 - 2t + 4 )Yes, that's correct. So, the function for Region B is indeed a downward opening parabola, which peaks at t=2, as I calculated earlier. So, after t=2, the demand decreases, and at t=10, it's negative. So, mathematically, that's correct, but in reality, the demand can't be negative, so perhaps the city would have to adjust, but the question is just asking for the total as per the functions. So, I think 465 million liters is the correct answer for Sub-problem 1.Now, moving on to Sub-problem 2: Given the city's reservoir has a maximum capacity of 500 million liters, calculate the maximum number of days t before the total water demand exceeds the reservoir capacity.So, we need to find the largest t such that the total demand is less than or equal to 500 million liters.First, let's express the total demand as a function of t. Since the total demand is the sum of the three regions' demands, we can write:Total Demand ( D(t) = D_A(t) + D_B(t) + D_C(t) )Plugging in the functions:( D(t) = (2t^2 + 3t + 5) + (-t^2 + 4t + 6) + (3t^2 - 2t + 4) )Let me simplify this expression step by step.First, combine like terms:Quadratic terms (t^2):2t^2 - t^2 + 3t^2 = (2 - 1 + 3)t^2 = 4t^2Linear terms (t):3t + 4t - 2t = (3 + 4 - 2)t = 5tConstant terms:5 + 6 + 4 = 15So, the total demand function simplifies to:( D(t) = 4t^2 + 5t + 15 )Okay, so we have ( D(t) = 4t^2 + 5t + 15 ). We need to find the maximum t such that D(t) ≤ 500.So, we set up the inequality:4t^2 + 5t + 15 ≤ 500Subtract 500 from both sides:4t^2 + 5t + 15 - 500 ≤ 0Simplify:4t^2 + 5t - 485 ≤ 0Now, we need to solve the quadratic inequality 4t^2 + 5t - 485 ≤ 0.First, let's find the roots of the equation 4t^2 + 5t - 485 = 0.We can use the quadratic formula:t = [-b ± sqrt(b^2 - 4ac)] / (2a)Where a = 4, b = 5, c = -485.Calculating the discriminant:D = b^2 - 4ac = 5^2 - 4*4*(-485) = 25 + 4*4*485Wait, let me compute that step by step.First, 4ac = 4*4*(-485) = 16*(-485) = -7760So, D = 25 - (-7760) = 25 + 7760 = 7785So, sqrt(7785). Let me calculate that.I know that 88^2 = 7744 and 89^2 = 7921. So, sqrt(7785) is between 88 and 89.Let me compute 88^2 = 77447785 - 7744 = 41So, sqrt(7785) ≈ 88 + 41/(2*88) ≈ 88 + 41/176 ≈ 88 + 0.233 ≈ 88.233So, approximately 88.233.So, the roots are:t = [-5 ± 88.233]/(2*4) = [-5 ± 88.233]/8Calculating both roots:First root: (-5 + 88.233)/8 ≈ (83.233)/8 ≈ 10.404Second root: (-5 - 88.233)/8 ≈ (-93.233)/8 ≈ -11.654Since time t cannot be negative, we discard the negative root.So, the roots are approximately t ≈ 10.404 and t ≈ -11.654.Since the quadratic opens upwards (because the coefficient of t^2 is positive), the inequality 4t^2 + 5t - 485 ≤ 0 holds between the two roots.But since t cannot be negative, the relevant interval is from t = 0 to t ≈ 10.404.Therefore, the total demand is less than or equal to 500 million liters for t ≤ 10.404 days.But since t must be an integer (days are whole numbers), we need to find the maximum integer t such that D(t) ≤ 500.So, t can be 10 days because at t=10, we calculated the total demand as 465 million liters, which is less than 500.But wait, let's check at t=10.404, the demand is exactly 500. So, t=10.404 is the point where it exceeds 500. Therefore, the maximum integer t before exceeding is 10.But wait, let me verify by calculating D(10) and D(11).We already know D(10) = 465, which is less than 500.Now, let's compute D(11):Using the total demand function ( D(t) = 4t^2 + 5t + 15 )So, D(11) = 4*(11)^2 + 5*(11) + 15Calculating each term:- 4*(121) = 484- 5*11 = 55- 15Adding them up: 484 + 55 = 539 + 15 = 554 million liters.554 is greater than 500, so at t=11, the demand exceeds the reservoir capacity.Therefore, the maximum number of days before the total demand exceeds 500 million liters is t=10.Wait, but the quadratic solution gave t≈10.404, which is between 10 and 11. So, the demand exceeds 500 at t≈10.404, meaning that on day 10, it's still under, and on day 11, it's over. So, the maximum integer t is 10.Therefore, the answer to Sub-problem 2 is t=10 days.But let me double-check my calculations to be sure.First, the total demand function: I combined the three regions correctly.Region A: 2t² + 3t +5Region B: -t² +4t +6Region C: 3t² -2t +4Adding them:2t² - t² +3t² = 4t²3t +4t -2t =5t5+6+4=15So, D(t)=4t²+5t+15. Correct.Then, setting 4t² +5t +15 ≤5004t² +5t -485 ≤0Quadratic formula:t = [-5 ± sqrt(25 + 4*4*485)] /8Wait, 4ac is 4*4*485=16*485=7760So, discriminant is 25 +7760=7785sqrt(7785)≈88.233So, t=( -5 +88.233)/8≈83.233/8≈10.404Yes, correct.So, t≈10.404 is when D(t)=500.Therefore, the maximum integer t is 10.But just to be thorough, let me check D(10.404) to see if it's exactly 500.Using D(t)=4t²+5t+15t≈10.404Compute t²: 10.404²≈108.2334*108.233≈432.9335*10.404≈52.02Adding up: 432.933 +52.02≈484.953 +15≈499.953≈500. So, correct.Therefore, the maximum number of days is 10.So, summarizing:Sub-problem 1: Total demand at t=10 is 465 million liters.Sub-problem 2: Maximum t before exceeding 500 million liters is 10 days.</think>

🧠:<think>Okay, so I'm trying to solve these two problems related to CorpX's market share and a proposed tax model. Let me take them one at a time.Starting with the first problem: I need to find the rate of change of CorpX's market share at t = 10 years. The market share is given by the function M(t) = 100 / (1 + e^{-0.5(t - 10)}). Hmm, this looks like a logistic growth model, which makes sense because it's an S-shaped curve that approaches an asymptote. To find the rate of change, I need to compute the derivative of M(t) with respect to t, which is M'(t). Let me recall how to differentiate such a function. Since it's a function of the form 100 divided by something, I can use the quotient rule or recognize it as a composition of functions and use the chain rule.Let me write M(t) as 100 * [1 + e^{-0.5(t - 10)}]^{-1}. So, using the chain rule, the derivative M'(t) will be 100 * (-1) * [1 + e^{-0.5(t - 10)}]^{-2} * derivative of the inside function.The inside function is 1 + e^{-0.5(t - 10)}. The derivative of that with respect to t is e^{-0.5(t - 10)} * (-0.5). So putting it all together:M'(t) = 100 * (-1) * [1 + e^{-0.5(t - 10)}]^{-2} * (-0.5) * e^{-0.5(t - 10)}.Simplifying the negatives: the two negatives make a positive, so:M'(t) = 100 * 0.5 * e^{-0.5(t - 10)} / [1 + e^{-0.5(t - 10)}]^2.That simplifies to:M'(t) = 50 * e^{-0.5(t - 10)} / [1 + e^{-0.5(t - 10)}]^2.Now, I need to evaluate this at t = 10. Let's plug in t = 10:M'(10) = 50 * e^{-0.5(10 - 10)} / [1 + e^{-0.5(10 - 10)}]^2.Simplify the exponent: 10 - 10 is 0, so e^0 is 1. Therefore:M'(10) = 50 * 1 / (1 + 1)^2 = 50 / 4 = 12.5.So the rate of change at t = 10 is 12.5 percentage points per year. Hmm, that seems significant. Since the derivative is positive, it means the market share is increasing at that point. But since t = 10 is the midpoint of the logistic curve, this should be the point of maximum growth rate. That makes sense because the logistic curve has its steepest slope at the midpoint.So, interpreting this, at t = 10 years, CorpX's market share is growing at the fastest rate, 12.5% per year. This indicates that around this time, the corporation is expanding its market share most rapidly, which could have significant implications on the local economy and social equity.Moving on to the second problem: I need to find the constants a, b, and c for the tax model T(M) = aM² + bM + c. The conditions given are:- T(40) = 10- T(60) = 20- T(80) = 35So, plugging these into the equation, I get three equations:1. 10 = a*(40)^2 + b*(40) + c2. 20 = a*(60)^2 + b*(60) + c3. 35 = a*(80)^2 + b*(80) + cLet me write these out numerically:1. 10 = 1600a + 40b + c2. 20 = 3600a + 60b + c3. 35 = 6400a + 80b + cNow, I have a system of three equations with three unknowns. I can solve this using substitution or elimination. Let me subtract the first equation from the second to eliminate c:Equation 2 - Equation 1:20 - 10 = (3600a - 1600a) + (60b - 40b) + (c - c)10 = 2000a + 20bSimplify: 10 = 2000a + 20bDivide both sides by 10: 1 = 200a + 2bLet me call this Equation 4: 200a + 2b = 1Similarly, subtract Equation 2 from Equation 3:35 - 20 = (6400a - 3600a) + (80b - 60b) + (c - c)15 = 2800a + 20bSimplify: 15 = 2800a + 20bDivide both sides by 5: 3 = 560a + 4bLet me call this Equation 5: 560a + 4b = 3Now, I have two equations:Equation 4: 200a + 2b = 1Equation 5: 560a + 4b = 3I can solve these two equations for a and b. Let me multiply Equation 4 by 2 to make the coefficients of b the same:Equation 4 * 2: 400a + 4b = 2Now, subtract this from Equation 5:Equation 5 - (Equation 4 * 2):(560a - 400a) + (4b - 4b) = 3 - 2160a = 1So, a = 1 / 160 = 0.00625Now, plug a back into Equation 4:200*(1/160) + 2b = 1200/160 = 5/4 = 1.25So, 1.25 + 2b = 1Subtract 1.25: 2b = 1 - 1.25 = -0.25Thus, b = -0.25 / 2 = -0.125Now, with a and b known, plug into Equation 1 to find c:10 = 1600*(1/160) + 40*(-0.125) + cCalculate each term:1600 / 160 = 1040 * (-0.125) = -5So, 10 = 10 - 5 + cSimplify: 10 = 5 + cThus, c = 5So, the constants are:a = 1/160 = 0.00625b = -1/8 = -0.125c = 5Let me double-check these values with the original equations.First equation: T(40) = 0.00625*(40)^2 + (-0.125)*40 + 5= 0.00625*1600 - 5 + 5= 10 - 5 + 5 = 10. Correct.Second equation: T(60) = 0.00625*(60)^2 + (-0.125)*60 + 5= 0.00625*3600 - 7.5 + 5= 22.5 - 7.5 + 5 = 20. Correct.Third equation: T(80) = 0.00625*(80)^2 + (-0.125)*80 + 5= 0.00625*6400 - 10 + 5= 40 - 10 + 5 = 35. Correct.Looks good. So, the tax model is T(M) = (1/160)M² - (1/8)M + 5.Final Answer1. The rate of change of the market share at ( t = 10 ) years is boxed{12.5} percentage points per year.2. The values of the constants are ( a = boxed{dfrac{1}{160}} ), ( b = boxed{-dfrac{1}{8}} ), and ( c = boxed{5} ).</think>

🧠:<think>Alright, so I've got these two sub-problems to solve for the comedian's world tour. Let me take them one at a time.Starting with Sub-problem 1. The comedian has a model where the effectiveness of a joke is calculated by summing up the products of weights and functions for each cultural factor. There are five cultural factors with given weights: w1=0.2, w2=0.3, w3=0.1, w4=0.25, w5=0.15. The total sum of weights is 1, which makes sense because they're probably normalized.Now, the problem is that in a particular country, only cultural factors 1 and 4 are significant. So, the comedian wants to adjust the weights to maximize the effectiveness in this country. But he still needs the weights to sum to 1. Hmm, okay. So, if only factors 1 and 4 matter, he should probably increase their weights and decrease the others. But how much?I think the idea is to redistribute the weights from the less significant factors (2, 3, 5) to the significant ones (1 and 4). Let me calculate the total weight currently allocated to the insignificant factors. That would be w2 + w3 + w5 = 0.3 + 0.1 + 0.15 = 0.55. So, 55% of the weight is on factors that aren't significant in this country.To maximize effectiveness, the comedian should allocate as much weight as possible to factors 1 and 4. But how should he distribute that 0.55 between w1 and w4? Maybe he should consider how much each contributes or if there's a way to maximize the effectiveness function. But since the functions f_i are not given, maybe it's just about giving equal weight or something else.Wait, the problem says "maximize the effectiveness." Since effectiveness is a linear combination, to maximize it, he should allocate more weight to the factors that contribute more. But without knowing the functions f_i, it's tricky. However, since only factors 1 and 4 are significant, perhaps he should set the weights for the other factors to zero and redistribute their weights to 1 and 4.So, if he sets w2, w3, w5 to zero, he can add their weights to w1 and w4. But how to distribute 0.55 between w1 and w4? Maybe equally? Or perhaps proportionally based on their original weights?Wait, the original weights are w1=0.2 and w4=0.25. So, w4 is more important than w1 in the original model. Maybe he should give more weight to w4. Alternatively, since both are significant, maybe he can set w1 and w4 to 0.5 each? But that would make the total weights 1. Let me check:If he sets w1=0.5 and w4=0.5, then w2=w3=w5=0. That sums to 1. Alternatively, he could keep the same ratio as before. Originally, w1 and w4 were 0.2 and 0.25, which is a ratio of 4:5. So, if he wants to keep that ratio, he can distribute the 0.55 in the same 4:5 ratio.So, total parts = 4 + 5 = 9 parts. So, 0.55 divided into 9 parts: each part is 0.55 / 9 ≈ 0.0611. Then, w1 would get 4 parts: 4 * 0.0611 ≈ 0.2444, and w4 would get 5 parts: 5 * 0.0611 ≈ 0.3056. Then, adding to their original weights: w1 becomes 0.2 + 0.2444 ≈ 0.4444, and w4 becomes 0.25 + 0.3056 ≈ 0.5556. But wait, that would make the total weights 0.4444 + 0.5556 = 1, but we also have w2, w3, w5 set to zero. So, that would work.Alternatively, maybe he should just set w1 and w4 to 0.5 each, ignoring the original ratio. But I think keeping the original ratio makes sense because if w4 was more important, it should still be more important after redistribution.So, the adjusted weights would be w1 ≈ 0.4444 and w4 ≈ 0.5556, with the rest zero. Let me verify the math:Original weights: w1=0.2, w2=0.3, w3=0.1, w4=0.25, w5=0.15.Total weight to redistribute: 0.3 + 0.1 + 0.15 = 0.55.Ratio of w1:w4 is 0.2:0.25 = 4:5.So, 4 parts for w1 and 5 parts for w4, total 9 parts.Each part is 0.55 / 9 ≈ 0.0611.So, w1 gets 4 * 0.0611 ≈ 0.2444, so new w1 = 0.2 + 0.2444 ≈ 0.4444.w4 gets 5 * 0.0611 ≈ 0.3056, so new w4 = 0.25 + 0.3056 ≈ 0.5556.Yes, that seems correct. So, the adjusted weights would be w1≈0.4444, w4≈0.5556, and the rest zero.Alternatively, to make it exact, 0.55 divided by 9 is 11/200, so 0.055 each part? Wait, 0.55 /9 is approximately 0.0611, which is 11/180 ≈ 0.0611.But maybe it's better to express it as fractions. 0.55 is 11/20. Divided by 9 is 11/180. So, 4 parts: 44/180 = 11/45 ≈ 0.2444, and 5 parts: 55/180 = 11/36 ≈ 0.3056.So, w1 becomes 0.2 + 11/45 ≈ 0.2 + 0.2444 ≈ 0.4444, and w4 becomes 0.25 + 11/36 ≈ 0.25 + 0.3056 ≈ 0.5556.Yes, that seems precise.So, the adjusted weights are w1≈0.4444, w4≈0.5556, and the rest zero.Moving on to Sub-problem 2. The comedian is performing in two countries, A and B, with laughter responses following normal distributions. Country A has μ_A=7, σ_A=1.5, and Country B has μ_B=6, σ_B=2. He wants to ensure that the probability of laughter response ≥8 is greater in A than in B.So, we need to calculate P(L ≥8) for both countries and compare them.For a normal distribution, P(X ≥ x) can be found by calculating the z-score and then using the standard normal distribution table or calculator.For Country A:μ_A =7, σ_A=1.5.x=8.z_A = (8 - 7)/1.5 = 1/1.5 ≈ 0.6667.Looking up z=0.6667 in the standard normal table, the area to the left is approximately 0.7486. Therefore, the area to the right (P(L≥8)) is 1 - 0.7486 ≈ 0.2514 or 25.14%.For Country B:μ_B=6, σ_B=2.x=8.z_B = (8 -6)/2 = 2/2 =1.Looking up z=1, the area to the left is approximately 0.8413. So, the area to the right is 1 - 0.8413 ≈ 0.1587 or 15.87%.Comparing the two probabilities: 25.14% in A vs 15.87% in B. So, the probability is indeed greater in Country A.Therefore, the comedian can be assured that the probability of receiving a laughter response of at least 8 is higher in Country A than in Country B.Wait, let me double-check the z-scores and the probabilities.For Country A:z = (8-7)/1.5 = 0.6667. Using a z-table, 0.6667 corresponds to about 0.7486 cumulative probability. So, 1 - 0.7486 ≈ 0.2514.For Country B:z = (8-6)/2 =1. The cumulative probability is 0.8413, so 1 - 0.8413 ≈ 0.1587.Yes, that seems correct. So, Country A has a higher probability.I think that's all. So, summarizing:1. Adjust weights by redistributing 0.55 from factors 2,3,5 to 1 and 4 in the ratio 4:5, resulting in w1≈0.4444 and w4≈0.5556.2. Calculate probabilities for laughter ≥8 in both countries, finding higher probability in Country A.Final Answer1. The adjusted weights should be ( w_1 = boxed{frac{11}{25}} ) and ( w_4 = boxed{frac{11}{20}} ) with the rest set to zero.2. The probability of laughter response of at least 8 is higher in Country A than in Country B, with probabilities approximately ( boxed{25.14%} ) and ( boxed{15.87%} ) respectively.</think>