🧠:<think>Okay, so I have this problem where I need to figure out the combined time complexity of two algorithms, A and B, when they're run one after the other. Algorithm A has a time complexity of O(n log n), and Algorithm B has a time complexity of O(n²). The engineer wants to know the overall time complexity when these two are executed sequentially, with B processing the output of A. First, I need to recall how time complexities combine when algorithms are run sequentially. I remember that when you have two algorithms, their combined time complexity is just the sum of their individual time complexities. So, if Algorithm A takes O(n log n) time and Algorithm B takes O(n²) time, then the total time should be O(n log n) + O(n²). But wait, in Big-O notation, we usually keep the dominant term, right? So, between n log n and n², which one grows faster as n increases? I think n² grows much faster than n log n. For example, when n is 1000, n log n is about 1000 * 10 = 10,000, while n² is 1,000,000. So, n² is way bigger. That means the combined time complexity should be dominated by the O(n²) term. Therefore, the combined time complexity is O(n²). Let me write that down: T_combined(n) = O(n²). Now, moving on to the second part. I need to estimate the approximate number of operations for n = 10^6. Hmm, so n is one million. First, let's compute the operations for Algorithm A. Its time complexity is O(n log n). So, the number of operations would be roughly proportional to n log n. Let me compute that:n = 10^6log n is log base 2 of 10^6. I know that log2(10^6) is approximately log2(2^20) since 2^20 is about a million. Wait, 2^20 is 1,048,576, which is roughly 10^6. So, log2(10^6) ≈ 20. Therefore, n log n ≈ 10^6 * 20 = 20,000,000 operations for Algorithm A.Now, Algorithm B has a time complexity of O(n²). So, the number of operations is roughly n squared. n² = (10^6)^2 = 10^12 operations. So, Algorithm B alone would take about a trillion operations. Adding the operations from both algorithms: 20,000,000 + 1,000,000,000,000. But 20 million is negligible compared to a trillion. So, the total number of operations is approximately 1,000,000,000,000, which is 10^12 operations.Wait, but the question says to provide the answer in terms of n and simplify as much as possible. So, maybe I should express it in terms of n instead of plugging in n = 10^6. So, for the combined operations, it's O(n log n) + O(n²). As I thought earlier, the dominant term is n², so the total is O(n²). But if I have to write the approximate number in terms of n, it would be n² + n log n. But since n² is dominant, we can approximate it as n².But the problem says to estimate the approximate number of operations, so maybe I should write it as n² + n log n. But for n = 10^6, it's about 10^12 + 2*10^6, which is roughly 10^12. Wait, but the question says to provide the answer in terms of n and simplify as much as possible. So, maybe I should just write it as O(n²), but the question says "estimate the approximate number of operations", so perhaps they want the exact expression in terms of n, not the Big-O. So, the total operations would be approximately n log n + n². For n = 10^6, that's 10^6 * log2(10^6) + (10^6)^2. As I calculated, log2(10^6) ≈ 20, so it's 2*10^7 + 10^12, which is approximately 1.00002*10^12, which is roughly 10^12. But maybe I should write it as n² + n log n, which is the exact expression. However, since n² is much larger, it's approximately n². Wait, the question says "estimate the approximate number of operations required for the combined execution". So, they might accept either, but since it's an estimate, maybe just n² is sufficient. But let me think again. The exact number is n² + n log n. For n = 10^6, n² is 10^12, and n log n is about 2*10^7. So, 2*10^7 is 0.002% of 10^12, which is negligible. So, the approximate number is about 10^12 operations. But if I have to write it in terms of n, it's n² + n log n. But since n² is dominant, it's approximately n². So, to answer the sub-problems:1. The combined time complexity is O(n²).2. For n = 10^6, the approximate number of operations is about 10^12, which is n².But the question says to provide the answer in terms of n and simplify as much as possible. So, maybe I should write it as n² + n log n, but since n² is dominant, it's approximately n². Alternatively, since the combined time complexity is O(n²), the number of operations is on the order of n². Wait, but the question says "estimate the approximate number of operations", so maybe they want the exact expression. Alternatively, perhaps they want the sum, but simplified. Since n² is much larger, it's approximately n². I think the answer they are looking for is that the combined time complexity is O(n²), and the number of operations is approximately n², which for n=10^6 is 10^12 operations.So, putting it all together:1. T_combined(n) = O(n²)2. The approximate number of operations is n², which is 10^12 when n=10^6.But wait, the second part says "estimate the approximate number of operations required for the combined execution of Algorithm A and Algorithm B. Provide your answer in terms of n and simplify as much as possible."So, in terms of n, the number of operations is n² + n log n. But since n² is dominant, it's approximately n². So, the approximate number is n².Alternatively, if they want the exact expression, it's n² + n log n, but for estimation, n² is sufficient.So, I think the answer is:1. The combined time complexity is O(n²).2. The approximate number of operations is n², which for n=10^6 is 10^12 operations.But the question says to provide the answer in terms of n and simplify as much as possible. So, maybe just n².Alternatively, perhaps they want the exact sum, but since n² is dominant, it's approximately n².I think that's the way to go.</think>

🧠:<think>Okay, so Alex is dealing with two math problems here. The first one is about regular polygons inscribed in a unit circle, and the second one is about derangements of the polygon's vertices. Let me try to work through each problem step by step.Starting with the first problem: Alex has a regular n-sided polygon inscribed in a unit circle. Each vertex is labeled from 0 to n-1 in a counterclockwise direction. The coordinates of the k-th vertex are given by (cos(2kπ/n), sin(2kπ/n)). He wants to find the sum of the distances from each vertex to the center of the circle and show that this sum is equal to n.Hmm, okay. So the polygon is inscribed in a unit circle, which means the radius is 1. The distance from each vertex to the center is just the radius, right? Because all vertices lie on the circumference of the circle. So each distance is 1.Therefore, if there are n vertices, each at a distance of 1 from the center, the sum of all these distances should be n * 1 = n. That seems straightforward. But let me make sure I'm not missing something.Wait, maybe Alex is considering the distance from each vertex to the center in a different way? But no, in a unit circle, the distance from any point on the circumference to the center is always 1. So adding that up n times should just give n. Yeah, that makes sense.So, for the first problem, the sum is indeed n. I think that's it.Moving on to the second problem: Alex is thinking about permutations of the vertices of a regular n-gon. Specifically, he's interested in derangements, which are permutations where no vertex is in its original position. He defines D(n) as the number of such derangements and wants a formula for D(n) and the value of D(6).Alright, derangements. I remember that derangements are a classic problem in combinatorics. The number of derangements of n objects is often denoted by !n or D(n). The formula for derangements is:D(n) = n! * (1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n / n!)Alternatively, it can also be expressed using the inclusion-exclusion principle. Let me recall how that works.The total number of permutations is n!. To find the number of derangements, we subtract the permutations where at least one element is fixed. Using inclusion-exclusion, we alternate adding and subtracting the number of permutations fixing 1, 2, ..., n elements.So, the formula is:D(n) = n! - C(n,1)(n-1)! + C(n,2)(n-2)! - ... + (-1)^n C(n,n)(n-n)! Simplifying each term:C(n,k)(n - k)! = n! / k!So, substituting back:D(n) = n! - n!/1! + n!/2! - n!/3! + ... + (-1)^n n!/n!Factor out n!:D(n) = n! [1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n /n!]Yes, that's the same formula as before. So, that's the general formula for derangements.Alternatively, another way to express derangements is using the nearest integer function. I think it's something like D(n) = floor(n!/e + 0.5), where e is the base of the natural logarithm. But for exact values, especially for small n, it's better to use the formula above.So, for n = 6, let's compute D(6).First, let's compute 6! which is 720.Then, compute each term in the series:1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5! + 1/6!Compute each term:1 = 11/1! = 11/2! = 1/2 = 0.51/3! = 1/6 ≈ 0.16666671/4! = 1/24 ≈ 0.04166671/5! = 1/120 ≈ 0.00833331/6! = 1/720 ≈ 0.0013889Now, plug into the formula:1 - 1 + 0.5 - 0.1666667 + 0.0416667 - 0.0083333 + 0.0013889Let me compute step by step:Start with 1.1 - 1 = 00 + 0.5 = 0.50.5 - 0.1666667 ≈ 0.33333330.3333333 + 0.0416667 ≈ 0.3750.375 - 0.0083333 ≈ 0.36666670.3666667 + 0.0013889 ≈ 0.3680556So, the series inside the brackets is approximately 0.3680556.Multiply by 6! = 720:720 * 0.3680556 ≈ 720 * 0.3680556Let me compute that:0.3680556 * 700 = 257.638920.3680556 * 20 = 7.361112Adding together: 257.63892 + 7.361112 ≈ 265Wait, but let me do it more accurately.Compute 720 * 0.3680556:First, 700 * 0.3680556 = 257.6389220 * 0.3680556 = 7.361112Total: 257.63892 + 7.361112 = 265 (exactly, since 257.63892 + 7.361112 = 265.000032, which is approximately 265).But wait, let me check if 720 * 0.3680556 is exactly 265.Compute 720 * 0.3680556:0.3680556 * 720Let me compute 0.3680556 * 700 = 257.638920.3680556 * 20 = 7.361112Total: 257.63892 + 7.361112 = 265.000032So, approximately 265.000032, which is 265 when rounded to the nearest integer.But wait, derangements for n=6 should be 265? Let me verify that.Alternatively, I can compute D(6) using the recursive formula or another method.I remember that derangements satisfy the recurrence relation:D(n) = (n - 1) * (D(n - 1) + D(n - 2))With base cases D(1) = 0, D(2) = 1.Let me compute D(3) to D(6) step by step.D(1) = 0D(2) = 1D(3) = (3 - 1)*(D(2) + D(1)) = 2*(1 + 0) = 2D(4) = (4 - 1)*(D(3) + D(2)) = 3*(2 + 1) = 3*3 = 9D(5) = (5 - 1)*(D(4) + D(3)) = 4*(9 + 2) = 4*11 = 44D(6) = (6 - 1)*(D(5) + D(4)) = 5*(44 + 9) = 5*53 = 265Yes, so D(6) is indeed 265. So, that matches with the previous calculation.Therefore, the formula for D(n) is:D(n) = n! * [1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n /n!]And for n=6, D(6) = 265.So, summarizing:1. The sum of distances from each vertex to the center is n, since each distance is 1 and there are n vertices.2. The number of derangements D(n) is given by the formula above, and D(6) is 265.Final Answer1. The sum of the distances is boxed{n}.2. The number of derangements ( D(6) ) is boxed{265}.</think>

🧠:<think>Okay, so I have this problem about a business consultant who gives advice over meals. There are two parts to the problem. Let me try to tackle them one by one.Starting with the first part: The consultant charges 200 per hour for advice. They typically spend 1.5 hours per session. But here's the catch: for every 20 increase in the meal cost, they provide an additional 0.1 hour of consultation time for free, up to a maximum of 2 hours in total. I need to find the meal cost at which the total value (consultation fee plus the value of the free additional time) reaches its maximum. The value of the free time is equivalent to the regular hourly rate, which is 200 per hour.Alright, let's break this down. The consultant's regular fee is based on 1.5 hours, so that's 1.5 * 200 = 300. But when the meal cost increases, they get extra time for free. So, the total value is the regular fee plus the value of the extra time.Let me define some variables. Let x be the number of 20 increases in the meal cost. So, each x corresponds to a 20 increase, and for each x, the consultant gives an additional 0.1 hours. But the total consultation time can't exceed 2 hours. So, the initial time is 1.5 hours, and the extra time is 0.1x hours. The total time is 1.5 + 0.1x, but it can't go beyond 2 hours. So, 1.5 + 0.1x ≤ 2. Solving for x, 0.1x ≤ 0.5, so x ≤ 5. So, x can be at most 5. That means the maximum number of 20 increases is 5, which would make the meal cost increase by 100, but let's check that.Wait, actually, the meal cost is variable, and x is the number of 20 increments. So, the meal cost is 20x dollars? Or is the meal cost something else? Hmm, the problem says "for every 20 increase in the meal cost," so it's not that the meal cost is 20x, but rather, if the meal cost increases by 20, then x increases by 1. So, the meal cost is variable, and x is how many times 20 has been added to the base meal cost.Wait, actually, the problem doesn't specify a base meal cost. Hmm. Maybe I need to assume that the meal cost starts at some base, but since it's not given, perhaps I can let the meal cost be represented as 20x, where x is the number of 20 increments. So, if x=0, meal cost is 0, which doesn't make sense. Maybe the base meal cost is separate. Hmm, the problem is a bit unclear.Wait, actually, the problem says "for every 20 increase in the meal cost," so perhaps the meal cost can be any amount, and for each 20 increment, you get 0.1 hours extra. So, if the meal cost is M dollars, then the number of 20 increments is M / 20, but only in whole numbers? Or is it continuous? Hmm, the problem doesn't specify, so maybe it's continuous. So, x = M / 20, where M is the meal cost. So, the extra time is 0.1 * (M / 20) hours, but subject to the total time not exceeding 2 hours.So, the total consultation time is 1.5 + 0.1*(M / 20). But this can't exceed 2 hours. So, 1.5 + 0.1*(M / 20) ≤ 2. Let's solve for M.0.1*(M / 20) ≤ 0.5Multiply both sides by 20: 0.1*M ≤ 10So, M ≤ 100.So, if the meal cost M is more than 100, the extra time doesn't increase anymore because the total time is capped at 2 hours.So, the total value is the consultation fee plus the value of the extra time. The consultation fee is based on the total time, right? Wait, no. Wait, the consultant charges a fee of 200 per hour for their advice. They typically spend 1.5 hours per session. So, the fee is 200 * 1.5 = 300. But when they increase the meal cost, they give extra time for free. So, the total value is the fee plus the value of the extra time.Wait, is the fee fixed at 300 regardless of the time? Or is the fee based on the total time? Hmm, the problem says "the total value (consultation fee plus the value of the free additional time)." So, the consultation fee is fixed at 300, and the extra time is additional value. So, the total value is 300 + (extra time * 200).Alternatively, maybe the fee is based on the total time. Let me read the problem again."The consultant charges a fee of 200 per hour for their advice. They typically spend 1.5 hours per session with a client. However, they noticed that for every 20 increase in the meal cost, they provide an additional 0.1 hour of consultation time for free, up to a maximum of 2 hours in total. Determine the meal cost at which the total value (consultation fee plus the value of the free additional time) reaches its maximum. Assume the value of the free consultation time is equivalent to the regular hourly rate."So, the fee is 200 per hour. They typically spend 1.5 hours, so fee is 1.5 * 200 = 300. But when the meal cost increases, they give extra time for free. So, the total value is the fee plus the value of the extra time. So, the fee is fixed at 300, and the extra time is 0.1*(M / 20) hours, up to 0.5 hours (since 1.5 + 0.5 = 2). So, the total value is 300 + 200*(extra time).But wait, is the fee fixed or variable? If the fee is based on the time, then if they give extra time, the fee would increase. But the problem says "the total value (consultation fee plus the value of the free additional time)." So, it seems that the fee is fixed, and the extra time is a bonus, so the total value is fee + (extra time * rate).So, total value V = 300 + 200*(0.1*(M / 20)).Simplify that: 300 + 200*(M / 200) = 300 + M.Wait, that can't be right because if M increases, V increases without bound, but the extra time is capped at 0.5 hours. So, when M is less than or equal to 100, V = 300 + M. When M is greater than 100, V = 300 + 200*0.5 = 300 + 100 = 400.So, the total value increases linearly with M until M = 100, after which it remains constant at 400.Therefore, the maximum total value is 400, achieved when M is 100 or more. But since the extra time is capped at 2 hours total, which requires M = 100. So, the meal cost at which the total value reaches its maximum is 100.Wait, but let me double-check. If M is 100, then extra time is 0.1*(100 / 20) = 0.1*5 = 0.5 hours. So, total time is 2 hours, and total value is 300 + 200*0.5 = 300 + 100 = 400.If M is more than 100, say 120, then extra time would be 0.1*(120 / 20) = 0.6 hours, but total time can't exceed 2 hours, so extra time is capped at 0.5 hours. So, total value is still 400.Therefore, the maximum total value is achieved when M is 100. So, the meal cost is 100.Wait, but let me think again. If the meal cost is 100, the extra time is 0.5 hours, so total value is 300 + 100 = 400. If the meal cost is less than 100, say 80, then extra time is 0.4 hours, so total value is 300 + 80 = 380. If it's 100, it's 400. If it's more than 100, it's still 400. So, the maximum is at M=100.Therefore, the answer to the first part is 100.Now, moving on to the second part: The restaurant offers a special loyalty program where for every 100 spent, a customer receives a 10 discount on their next meal. The consultant estimates that they will have 10 sessions in a month. If the average meal cost they pay is determined from the solution of the first sub-problem, calculate the total cost of the meals after applying all possible discounts from the loyalty program.So, from the first part, the average meal cost is 100. They have 10 sessions, so total spending before discounts is 10 * 100 = 1000.Now, the loyalty program gives 10 discount for every 100 spent. So, for every 100, they get 10 off on the next meal. So, how does this work over 10 meals?Let me think. Each time they spend 100, they get a 10 discount on the next meal. So, the first meal costs 100, then the next meal is 100 - 10 = 90. Then the third meal is 100, because they spent 90 on the second meal, which is less than 100, so no discount. Wait, no, the discount is based on the previous meal's spending.Wait, actually, the problem says "for every 100 spent, a customer receives a 10 discount on their next meal." So, it's per 100 spent, not per meal. So, if they spend 100 in total, they get a 10 discount on the next meal. But if they spend more than 100, say 200, they get two 10 discounts on the next meal? Or is it per 100 spent in total?Wait, the problem says "for every 100 spent, a customer receives a 10 discount on their next meal." So, it's per 100 spent, regardless of how many meals. So, if they spend 100 in total, they get 10 off on the next meal. If they spend 200, they get 20 off on the next meal, etc.But in this case, the consultant is having 10 meals, each costing 100 on average. So, total spending is 1000. So, for every 100 spent, they get 10 off on the next meal. So, how does this apply?Wait, but the discount is on the next meal. So, the first meal is 100, then the next meal gets a 10 discount because they spent 100 on the first meal. So, second meal is 90. Then, the third meal: they spent 90 on the second meal, which is less than 100, so no discount. Wait, no, the discount is based on the previous meal's spending. So, for each meal, if the previous meal was 100, they get 10 off on the next meal.Wait, maybe it's better to model it step by step.Let me list the 10 meals:Meal 1: 100 (no discount, since it's the first meal)Meal 2: 100 - 10 = 90 (because Meal 1 was 100)Meal 3: 100 - 0 = 100 (because Meal 2 was 90, which is less than 100, so no discount)Meal 4: 100 - 10 = 90 (because Meal 3 was 100)Meal 5: 100 - 0 = 100 (Meal 4 was 90)Meal 6: 100 - 10 = 90 (Meal 5 was 100)Meal 7: 100 - 0 = 100 (Meal 6 was 90)Meal 8: 100 - 10 = 90 (Meal 7 was 100)Meal 9: 100 - 0 = 100 (Meal 8 was 90)Meal 10: 100 - 10 = 90 (Meal 9 was 100)Wait, so in this case, every even-numbered meal gets a 10 discount because the previous meal was 100. The odd-numbered meals (except the first) are either 100 or 90 depending on the previous meal.Wait, but Meal 2 is 90 because Meal 1 was 100. Meal 3: Meal 2 was 90, so no discount, so Meal 3 is 100. Meal 4: Meal 3 was 100, so Meal 4 is 90. And so on.So, in total, how many meals get a discount? Meals 2,4,6,8,10: 5 meals. Each discounted by 10. So, total discount is 5*10 = 50.Therefore, total cost is 10*100 - 50 = 1000 - 50 = 950.Wait, but let me check again. Meal 1: 100. Meal 2: 90. Meal 3: 100. Meal 4: 90. Meal 5: 100. Meal 6: 90. Meal 7: 100. Meal 8: 90. Meal 9: 100. Meal 10: 90.So, adding them up: 100 + 90 + 100 + 90 + 100 + 90 + 100 + 90 + 100 + 90.Let's compute this:100*5 = 50090*5 = 450Total = 500 + 450 = 950.Yes, that's correct.Alternatively, since every pair of meals (Meal 1 & 2, Meal 3 & 4, etc.) costs 100 + 90 = 190. There are 5 such pairs in 10 meals. So, 5*190 = 950.So, the total cost after discounts is 950.Therefore, the answers are:1. Meal cost: 1002. Total cost after discounts: 950But let me just make sure I didn't make a mistake in the first part. The total value is fee + extra time value. Fee is 300, extra time is 0.5 hours, so 100. Total value 400. If meal cost is higher than 100, extra time doesn't increase, so total value remains 400. So, the maximum is at 100.Yes, that seems correct.For the second part, the loyalty program gives a 10 discount on the next meal for every 100 spent. So, every time they spend 100, the next meal is 10 cheaper. Since they have 10 meals, starting with 100, then 90, then 100, etc., leading to 5 discounts of 10 each, totaling 50 off. So, total cost is 950.I think that's solid.</think>

🧠:<think>Alright, so I have this problem about a college student analyzing support for two candidates in the student government election. The student has gathered data on how different groups within the college are likely to vote. The problem has two parts, and I need to solve both. Let me take it step by step.First, let me parse the information given. The probability density functions for Candidates A and B are given as normal distributions. Specifically, ( P_A(x) ) and ( P_B(x) ) are both normal distributions with the same standard deviation ( sigma = 1 ), but different means: ( mu_A = 4 ) and ( mu_B = 6 ). The demographic characteristic score ( x ) ranges from 0 to 10.Problem 1: I need to find the value of ( x ) where the support for both candidates is equal, meaning ( P_A(x) = P_B(x) ).Okay, so since both are normal distributions, their probability density functions (PDFs) are given by:[P_A(x) = frac{1}{sqrt{2pi}sigma} e^{-frac{(x - mu_A)^2}{2sigma^2}}][P_B(x) = frac{1}{sqrt{2pi}sigma} e^{-frac{(x - mu_B)^2}{2sigma^2}}]Given that ( sigma = 1 ), ( mu_A = 4 ), and ( mu_B = 6 ), let me plug these values into the equations:[P_A(x) = frac{1}{sqrt{2pi}} e^{-frac{(x - 4)^2}{2}}][P_B(x) = frac{1}{sqrt{2pi}} e^{-frac{(x - 6)^2}{2}}]Since the denominators are the same, setting ( P_A(x) = P_B(x) ) simplifies to setting the exponents equal because the exponential function is one-to-one. So:[e^{-frac{(x - 4)^2}{2}} = e^{-frac{(x - 6)^2}{2}}]Taking the natural logarithm of both sides to eliminate the exponentials:[-frac{(x - 4)^2}{2} = -frac{(x - 6)^2}{2}]Multiply both sides by -2 to simplify:[(x - 4)^2 = (x - 6)^2]Now, let's expand both sides:Left side: ( (x - 4)^2 = x^2 - 8x + 16 )Right side: ( (x - 6)^2 = x^2 - 12x + 36 )Set them equal:[x^2 - 8x + 16 = x^2 - 12x + 36]Subtract ( x^2 ) from both sides:[-8x + 16 = -12x + 36]Now, bring all terms to one side:Add 12x to both sides:[4x + 16 = 36]Subtract 16 from both sides:[4x = 20]Divide by 4:[x = 5]So, the point where the support for both candidates is equal is at ( x = 5 ). That makes sense because the means are at 4 and 6, so the midpoint is 5. Since both distributions have the same variance, the point where they cross should be exactly halfway between the means.Problem 2: Now, the student uses a weight function ( w(x) = x^2 ) to calculate the weighted expected support for each candidate. I need to compute the integrals ( int_{0}^{10} x^2 P_A(x) , dx ) and ( int_{0}^{10} x^2 P_B(x) , dx ) and determine which candidate has a higher weighted expected support.Hmm, okay. So, these integrals represent the expected value of ( x^2 ) with respect to each candidate's distribution. In other words, it's like calculating ( E[x^2] ) for both distributions, but only over the interval [0,10], not the entire real line. However, since the normal distributions are defined over the entire real line, but we're integrating from 0 to 10, which is a finite interval. Hmm, that complicates things a bit because the expected value over the entire real line is straightforward, but over a finite interval, it's more involved.Wait, but let me think. The weight function is ( x^2 ), so the student is essentially calculating the expected value of ( x^2 ) for each candidate's support distribution, but only considering the interval from 0 to 10. Since the PDFs are defined over all real numbers, but the student is only considering the range 0 to 10, perhaps we need to adjust the PDFs to be conditional on ( x ) being between 0 and 10.Alternatively, maybe the student is just truncating the distributions at 0 and 10, meaning we're only considering the part of the normal distribution between 0 and 10, and the rest is ignored. So, in that case, the PDFs would be normalized over [0,10]. But the problem statement doesn't specify that. It just says ( x ) ranges from 0 to 10, but the PDFs are given as normal distributions. So, perhaps we are to integrate over [0,10] without normalizing, meaning the integrals might not sum to 1, but that's okay because we're just calculating the expected value.Wait, but in the problem statement, it says "the weighted expected support for each candidate by calculating ( int_{0}^{10} x^2 P_A(x) , dx ) and ( int_{0}^{10} x^2 P_B(x) , dx )." So, it's just the integral, not necessarily normalized. So, perhaps we don't need to worry about the fact that the integral over [0,10] isn't 1, because we're just calculating the expected value as per the weight function.But actually, in probability, the expected value is calculated as ( E[g(X)] = int_{-infty}^{infty} g(x) f(x) dx ). So, if we're only integrating over [0,10], that's equivalent to ( E[g(X) | X in [0,10]] times P(X in [0,10]) ). But since the problem doesn't specify conditioning, perhaps we can assume that the entire support is within [0,10], but the PDFs are given as normal distributions. So, maybe we need to consider the truncated normal distributions.Wait, this is getting complicated. Let me think again.Given that ( x ) is a demographic characteristic score that ranges from 0 to 10, it's possible that the PDFs are defined only over [0,10], but the given expressions are normal distributions. So, perhaps the student is using normal distributions truncated at 0 and 10. That would make sense because otherwise, the PDFs would have non-zero values outside of [0,10], which might not be meaningful in this context.Therefore, to compute the expected value ( E[x^2] ) for each candidate, we need to compute the integral of ( x^2 P_A(x) ) from 0 to 10, where ( P_A(x) ) is the truncated normal distribution.But wait, the problem statement doesn't specify whether the PDFs are truncated or not. It just says ( x ) ranges from 0 to 10. Hmm. So, perhaps we need to assume that the PDFs are defined over [0,10], but given as normal distributions. However, normal distributions extend to infinity, so that might not be the case.Alternatively, maybe the student is just using the normal distributions as models, even though ( x ) is bounded between 0 and 10. So, perhaps we can proceed by integrating over [0,10], even though the PDFs are defined over the entire real line.But in that case, the integrals would not represent the true expected values, because the normal distributions have tails beyond 0 and 10. However, since the problem statement says ( x ) ranges from 0 to 10, perhaps we can assume that the PDFs are truncated at 0 and 10. So, in that case, we need to normalize the PDFs over [0,10].Wait, but the problem doesn't mention anything about normalization. It just gives the PDFs as normal distributions. So, perhaps we can proceed by integrating over [0,10] without normalizing, and then compare the two integrals.But let's think about what that would mean. If we don't normalize, then the integrals ( int_{0}^{10} x^2 P_A(x) dx ) and ( int_{0}^{10} x^2 P_B(x) dx ) would not be expected values, but rather some weighted sums. However, the problem refers to them as "weighted expected support," so perhaps that's acceptable.Alternatively, maybe the student is using the entire normal distribution, but since ( x ) is only from 0 to 10, the rest is negligible. Let me check the values of the normal distributions at 0 and 10.Given ( mu_A = 4 ), ( sigma = 1 ), so 10 is 6 standard deviations above the mean. The probability beyond 10 is extremely small. Similarly, for ( mu_B = 6 ), 10 is 4 standard deviations above the mean, which is also very small. So, perhaps the contributions beyond 10 are negligible, and we can approximate the integrals from 0 to 10 as almost the entire distribution.But wait, actually, the PDFs are defined over the entire real line, but the problem restricts ( x ) to [0,10]. So, perhaps the student is considering only the part of the distribution within [0,10], meaning the PDFs are effectively truncated at 0 and 10, but not normalized. So, in that case, the integrals would not sum to 1, but we can still compute them as is.Alternatively, perhaps the student is considering the entire distribution, but since ( x ) is only from 0 to 10, the rest is ignored, but the PDFs are not normalized over [0,10]. So, in that case, the integrals would just be the expected values multiplied by the probability that ( x ) is in [0,10]. But since the probability beyond 10 is negligible, as we saw, maybe we can approximate the integrals as the expected values.Wait, but let's think about the expected value of ( x^2 ) for a normal distribution. For a normal distribution ( N(mu, sigma^2) ), the expected value ( E[x^2] ) is ( mu^2 + sigma^2 ). So, for Candidate A, ( E_A[x^2] = 4^2 + 1^2 = 16 + 1 = 17 ). For Candidate B, ( E_B[x^2] = 6^2 + 1^2 = 36 + 1 = 37 ). So, if we were integrating over the entire real line, the expected values would be 17 and 37, respectively.But since we're only integrating over [0,10], and the contributions beyond 10 are negligible, perhaps the integrals are approximately 17 and 37. However, I need to verify that.Alternatively, perhaps the integrals are exactly 17 and 37 because the tails beyond 10 are negligible. But wait, actually, no. The expected value is over the entire distribution, but if we're only integrating over [0,10], it's not exactly the same as the expected value. However, given that the tails beyond 10 are so small, the difference might be negligible.But let me think again. If we have a normal distribution with mean 4 and standard deviation 1, the probability that ( x ) is greater than 10 is ( P(X > 10) ). Let's compute that.For ( mu = 4 ), ( sigma = 1 ), ( z = (10 - 4)/1 = 6 ). The probability that Z > 6 is effectively 0, as standard normal tables usually go up to about 3.49, beyond which the probability is less than 0.0001. Similarly, for ( mu = 6 ), ( z = (10 - 6)/1 = 4 ), which is also a very small probability, about 0.0000317.So, the probability that ( x ) is beyond 10 is extremely small for both distributions. Therefore, the integrals from 0 to 10 of ( x^2 P_A(x) dx ) and ( x^2 P_B(x) dx ) are approximately equal to the expected values ( E[x^2] ) for each distribution, which are 17 and 37, respectively.Therefore, the weighted expected support for Candidate A is approximately 17, and for Candidate B, it's approximately 37. Therefore, Candidate B has a higher weighted expected support.But wait, let me make sure. The problem says "the weighted expected support for each candidate by calculating ( int_{0}^{10} x^2 P_A(x) dx ) and ( int_{0}^{10} x^2 P_B(x) dx )." So, if we take the integrals as is, without considering the normalization, then the integrals would be approximately equal to the expected values, because the tails beyond 10 are negligible.Alternatively, if we were to compute the exact integrals, we would have to compute:For Candidate A:[int_{0}^{10} x^2 cdot frac{1}{sqrt{2pi}} e^{-frac{(x - 4)^2}{2}} dx]Similarly for Candidate B:[int_{0}^{10} x^2 cdot frac{1}{sqrt{2pi}} e^{-frac{(x - 6)^2}{2}} dx]But since the integrals from 0 to 10 are almost the entire distribution, we can approximate these integrals as the expected values ( E[x^2] ), which are 17 and 37.But wait, actually, the expected value ( E[x^2] ) is equal to ( mu^2 + sigma^2 ). So, for A, it's 16 + 1 = 17, and for B, it's 36 + 1 = 37.Therefore, the student should support Candidate B, as the weighted expected support is higher.But let me double-check. Is there a way to compute these integrals exactly without approximating?Well, integrating ( x^2 ) times a normal PDF is a standard integral, and it's equal to ( mu^2 + sigma^2 ) over the entire real line. However, since we're only integrating over [0,10], it's not exactly that, but as we saw, the contribution beyond 10 is negligible. So, the integrals are approximately 17 and 37.Alternatively, if we were to compute the exact integrals, we would have to use the error function or some other method, but given the negligible tails, it's safe to approximate.Therefore, the conclusion is that Candidate B has a higher weighted expected support.Summary of Thoughts:1. For Problem 1, setting the two PDFs equal leads to solving for ( x ) where the exponents are equal, resulting in ( x = 5 ).2. For Problem 2, calculating the weighted expected support involves integrating ( x^2 ) times the PDFs over [0,10]. Given the negligible tails beyond 10, these integrals approximate the expected values of ( x^2 ), which are 17 for A and 37 for B. Therefore, the student should support Candidate B.Final Answer1. The demographic characteristic score where support is equal is boxed{5}.2. The student should support Candidate boxed{B}.</think>