answer：Alright, so I have this problem where a traveler wants to plan a 30-day trip across three regions: A, B, and C. The goal is to allocate the days based on the cultural immersion factors of each region. Let me try to break this down step by step.First, I need to understand what the cultural immersion factor (I) is. It's given by the formula (I = P cdot C), where (P) is the population and (C) is the cultural significance coefficient. So, for each region, I can calculate this factor by multiplying their population by their respective cultural coefficient.Let me write down the given data:- **Region A**: (P_A = 2 times 10^6), (C_A = 1.5)- **Region B**: (P_B = 3 times 10^6), (C_B = 2.0)- **Region C**: (P_C = 4 times 10^6), (C_C = 1.2)So, calculating the cultural immersion factors:- (I_A = P_A times C_A = 2 times 10^6 times 1.5)- (I_B = P_B times C_B = 3 times 10^6 times 2.0)- (I_C = P_C times C_C = 4 times 10^6 times 1.2)Let me compute each:1. (I_A = 2 times 10^6 times 1.5 = 3 times 10^6)2. (I_B = 3 times 10^6 times 2.0 = 6 times 10^6)3. (I_C = 4 times 10^6 times 1.2 = 4.8 times 10^6)So, the cultural immersion factors are:- A: 3,000,000- B: 6,000,000- C: 4,800,000Now, the total cultural immersion factor (I_{total}) is the sum of these:(I_{total} = I_A + I_B + I_C = 3,000,000 + 6,000,000 + 4,800,000 = 13,800,000)The traveler wants to spend 30 days in total, with the days allocated proportionally to these factors. So, the proportion of days spent in each region will be each region's (I) divided by the total (I).Let me denote the days spent in each region as (D_A), (D_B), and (D_C). Then:(D_A = left( frac{I_A}{I_{total}} right) times 30)Similarly for (D_B) and (D_C).Calculating each:1. (D_A = left( frac{3,000,000}{13,800,000} right) times 30)2. (D_B = left( frac{6,000,000}{13,800,000} right) times 30)3. (D_C = left( frac{4,800,000}{13,800,000} right) times 30)Simplifying each fraction:1. For (D_A): (3,000,000 / 13,800,000 = 3/13.8 = 10/46 ≈ 0.2174)2. For (D_B): (6,000,000 / 13,800,000 = 6/13.8 = 20/46 ≈ 0.4348)3. For (D_C): (4,800,000 / 13,800,000 = 4.8/13.8 = 48/138 = 16/46 ≈ 0.3478)Now, multiplying each by 30 days:1. (D_A ≈ 0.2174 times 30 ≈ 6.522) days2. (D_B ≈ 0.4348 times 30 ≈ 13.043) days3. (D_C ≈ 0.3478 times 30 ≈ 10.434) daysBut since the traveler can't spend a fraction of a day, we need to round these numbers to whole days. However, we have to ensure that the total adds up to 30 days.Let me calculate the exact fractions without rounding:1. (D_A = (3/13.8) times 30 = (3 times 30)/13.8 = 90/13.8 ≈ 6.5217)2. (D_B = (6/13.8) times 30 = (6 times 30)/13.8 = 180/13.8 ≈ 13.0435)3. (D_C = (4.8/13.8) times 30 = (4.8 times 30)/13.8 = 144/13.8 ≈ 10.4348)Adding these up: 6.5217 + 13.0435 + 10.4348 ≈ 30.0000, which is perfect.But since we can't have fractions of days, we need to round these. Let's see:- (D_A ≈ 6.52) → 7 days- (D_B ≈ 13.04) → 13 days- (D_C ≈ 10.43) → 10 daysBut 7 + 13 + 10 = 30 days. Perfect, that works.Wait, but 6.52 is closer to 7, 13.04 is almost 13, and 10.43 is closer to 10. So, that seems okay.Alternatively, another way is to use the exact fractions:(D_A = 90/13.8 = 900/138 = 450/69 ≈ 6.5217)Similarly, (D_B = 180/13.8 = 1800/138 = 900/69 ≈ 13.0435)(D_C = 144/13.8 = 1440/138 = 720/69 ≈ 10.4348)So, if we round each to the nearest whole number:- (D_A = 7)- (D_B = 13)- (D_C = 10)Total is 30. So, that's the first part.Now, moving on to the second part: the traveler wants to spend at least 5 days in each region regardless of the proportional distribution. So, first, assign 5 days to each region, and then allocate the remaining days proportionally.Total minimum days assigned: 5 + 5 + 5 = 15 days.Remaining days: 30 - 15 = 15 days.Now, we need to allocate these 15 days proportionally to the cultural immersion factors.But wait, the cultural immersion factors are the same as before: 3,000,000; 6,000,000; 4,800,000.So, the total cultural immersion factor is still 13,800,000.So, the proportion for each region is the same as before.Therefore, the days allocated from the remaining 15 days will be:(D'_A = left( frac{I_A}{I_{total}} right) times 15)Similarly for (D'_B) and (D'_C).Calculating each:1. (D'_A = (3,000,000 / 13,800,000) times 15 = (3/13.8) times 15 = 45/13.8 ≈ 3.2609) days2. (D'_B = (6,000,000 / 13,800,000) times 15 = (6/13.8) times 15 = 90/13.8 ≈ 6.5217) days3. (D'_C = (4,800,000 / 13,800,000) times 15 = (4.8/13.8) times 15 = 72/13.8 ≈ 5.2174) daysAdding these up: 3.2609 + 6.5217 + 5.2174 ≈ 15.0000 days.Now, we need to round these to whole days as well.But let's see:- (D'_A ≈ 3.26) → 3 days- (D'_B ≈ 6.52) → 7 days- (D'_C ≈ 5.22) → 5 daysBut 3 + 7 + 5 = 15 days.So, adding these to the initial 5 days in each region:- Region A: 5 + 3 = 8 days- Region B: 5 + 7 = 12 days- Region C: 5 + 5 = 10 daysWait, but let's check if that adds up: 8 + 12 + 10 = 30 days. Perfect.But hold on, is this the correct approach? Because when we have the remaining days, we have to make sure that the rounding doesn't cause issues. Let me verify:Alternatively, we can calculate the exact fractions:(D'_A = 45/13.8 = 450/138 = 225/69 ≈ 3.2609)(D'_B = 90/13.8 = 900/138 = 450/69 ≈ 6.5217)(D'_C = 72/13.8 = 720/138 = 360/69 ≈ 5.2174)So, rounding each:- (D'_A ≈ 3.26) → 3 days- (D'_B ≈ 6.52) → 7 days- (D'_C ≈ 5.22) → 5 daysTotal: 3 + 7 + 5 = 15 days.So, adding to the initial 5 days:- A: 5 + 3 = 8- B: 5 + 7 = 12- C: 5 + 5 = 10Total: 30 days.But wait, is there a better way to handle the rounding? Because sometimes, when you have fractions, you might end up with a day left over or something. But in this case, it seems to add up perfectly.Alternatively, another approach is to use the exact decimal values and round them in a way that the total is maintained.But in this case, since 3.26 + 6.52 + 5.22 = 15.00, and when rounded to the nearest whole number, they add up correctly, so it's okay.Therefore, the adjusted plan is:- Region A: 8 days- Region B: 12 days- Region C: 10 daysLet me double-check:- Original proportional allocation was approximately 6.5, 13.0, 10.4 days, which we rounded to 7, 13, 10 days.- After the minimum 5 days, we reallocated the remaining 15 days proportionally, resulting in 3, 7, 5 days added, making the total 8, 12, 10 days.Yes, that seems consistent.So, summarizing:1. Without considering the minimum days, the traveler should spend approximately 7, 13, and 10 days in regions A, B, and C respectively.2. After ensuring at least 5 days in each region, the adjusted allocation is 8, 12, and 10 days.I think that's it. Let me just make sure I didn't make any calculation errors.Calculating (I_A = 2e6 * 1.5 = 3e6), correct.(I_B = 3e6 * 2 = 6e6), correct.(I_C = 4e6 * 1.2 = 4.8e6), correct.Total (I = 13.8e6), correct.Proportions:- A: 3e6 /13.8e6 = 3/13.8 ≈ 0.2174, 0.2174*30 ≈6.52, correct.- B: 6e6 /13.8e6 ≈0.4348, 0.4348*30≈13.04, correct.- C: 4.8e6 /13.8e6≈0.3478, 0.3478*30≈10.43, correct.Rounding to 7,13,10, which adds to 30.Then, for the second part, subtracting 5 days each, leaving 15 days.Calculating the same proportions on 15 days:- A: 3/13.8*15≈3.26, rounds to 3- B: 6/13.8*15≈6.52, rounds to7- C:4.8/13.8*15≈5.22, rounds to5Adding to 5 each: 8,12,10.Yes, that's correct.I think I've covered all the steps and checked the calculations. So, the answers should be as above.**Final Answer**1. The traveler should spend boxed{7} days in Region A, boxed{13} days in Region B, and boxed{10} days in Region C.2. After adjusting for the minimum 5 days in each region, the traveler will spend boxed{8} days in Region A, boxed{12} days in Region B, and boxed{10} days in Region C.

answer：Okay, so I have this problem about setting up checkpoints along a border segment AB, which is from (0,0) to (100,0). The goal is to make sure every point on AB is within 10 units of at least one checkpoint. Then, in part 2, I need to optimize the placement of these checkpoints to minimize the sum of the squares of their distances to the border.Starting with part 1. I need to figure out the minimum number of checkpoints required so that every point on AB is within 10 units of a checkpoint. Hmm, so each checkpoint can cover a certain area around it. Since the border is a straight line, the coverage area from each checkpoint would be a circle with radius 10 units. But since the border is a line segment, the effective coverage along the border from each checkpoint would be a segment of length 20 units, right? Because if a checkpoint is placed somewhere, it can cover 10 units to the left and 10 units to the right along the border.Wait, but actually, the distance from the checkpoint to the border isn't necessarily along the border. The distance is the shortest distance, which would be perpendicular. So if a checkpoint is placed at some point (x, y), the distance from that checkpoint to the border AB is |y|, because AB is along the x-axis. So to cover the border, the checkpoints need to be within 10 units of every point on AB.But how does that translate into the number of checkpoints? Maybe I should think about the coverage along the border. Each checkpoint can cover a segment of the border, but the exact length depends on where the checkpoint is placed.If I place a checkpoint at (x, y), then the set of points on AB that are within 10 units of this checkpoint form a segment on AB. The length of this segment can be found by considering the intersection of the circle of radius 10 around (x, y) with the line AB.Since AB is along the x-axis, the distance from (x, y) to any point (a, 0) on AB is sqrt((x - a)^2 + y^2). We want this distance to be <= 10. So sqrt((x - a)^2 + y^2) <= 10. Squaring both sides, (x - a)^2 + y^2 <= 100.To find the range of a for which this is true, we can solve for a. Let's rearrange: (x - a)^2 <= 100 - y^2. Taking square roots, |x - a| <= sqrt(100 - y^2). So the segment on AB covered by this checkpoint is from x - sqrt(100 - y^2) to x + sqrt(100 - y^2).Therefore, the length of the segment covered is 2*sqrt(100 - y^2). To maximize the coverage, we should maximize this length. That happens when y is as small as possible, meaning the checkpoint is as close to the border as possible. If y = 0, then the coverage is 20 units, which is the maximum possible. If y increases, the coverage decreases.But wait, if y is 0, the checkpoint is on the border, so the coverage is 20 units. If y is 10, the coverage is 0, which doesn't make sense because then the checkpoint is 10 units away from the border, so it can't cover any point on the border. So to cover the border, the checkpoints must be within 10 units of the border, i.e., y <= 10.But to maximize coverage, we should place checkpoints as close to the border as possible, i.e., y = 0, so that each checkpoint covers 20 units of the border. However, if we place all checkpoints on the border, then the coverage is straightforward.But wait, the problem doesn't specify that the checkpoints have to be off the border. They can be anywhere on the plane, as long as they are within 10 units of the border. So if we place a checkpoint on the border, it can cover 20 units. If we place it off the border, it can cover less.Therefore, to minimize the number of checkpoints, we should place them on the border, each covering 20 units. Since the border is 100 units long, the number of checkpoints needed would be 100 / 20 = 5. But wait, let's check.If we place a checkpoint at (0,0), it covers from 0 to 20. Then another at (20,0), covers 20 to 40. Then (40,0), (60,0), (80,0), and (100,0). Wait, but (100,0) is the endpoint. So actually, if we place checkpoints at 0, 20, 40, 60, 80, and 100, that's 6 checkpoints. But wait, the distance from 80 to 100 is 20, so the checkpoint at 80 covers up to 100. So actually, we don't need a checkpoint at 100. So 5 checkpoints: 0, 20, 40, 60, 80. Each covers 20 units, so from 0-20, 20-40, etc., up to 80-100. So 5 checkpoints.Wait, but if we place a checkpoint at (0,0), it covers from 0 to 20. Then the next checkpoint at (20,0) covers 20 to 40, and so on. So the last checkpoint at (80,0) covers up to 100. So yes, 5 checkpoints.But wait, what if we place the checkpoints not exactly on the border? Maybe we can cover more efficiently? For example, if we place a checkpoint slightly above the border, can we cover more than 20 units? No, because the maximum coverage is when y=0, giving 20 units. If y>0, the coverage is less. So to maximize coverage, we should place checkpoints on the border.Therefore, the minimum number of checkpoints required is 5, placed at (0,0), (20,0), (40,0), (60,0), (80,0).Wait, but let me double-check. Suppose we place a checkpoint at (10,0). Then it covers from 0 to 20. Similarly, another at (30,0) covers 20 to 40, etc. So actually, if we stagger the checkpoints, maybe we can cover the entire border with fewer checkpoints? Wait, no, because each checkpoint still only covers 20 units. So regardless of where you place them, you need at least 5 checkpoints to cover 100 units with each covering 20.Wait, but actually, if you place the first checkpoint at (10,0), it covers from 0 to 20. Then the next at (30,0) covers 20 to 40, and so on. So the last checkpoint would be at (90,0), covering 80 to 100. So that's 5 checkpoints as well. So whether you start at 0 or 10, you still need 5 checkpoints.Therefore, the minimum number is 5, placed at every 20 units along the border.But wait, the problem says "a finite number of checkpoints, each represented as a point C_i(x_i, y_i) on the plane, where x_i and y_i are real numbers, and 0 ≤ x_i ≤ 100." So they don't have to be on the border, but if we place them on the border, we can cover the maximum length with each checkpoint.But maybe if we place them off the border, we can cover more? Wait, no, because the distance from the checkpoint to the border is y, so the coverage along the border is 2*sqrt(100 - y^2). So if y is 0, it's 20. If y is 5, it's 2*sqrt(75) ≈ 17.32. So actually, less coverage. So to cover the maximum, we need y=0.Therefore, the minimal number is 5, placed at (0,0), (20,0), (40,0), (60,0), (80,0).Wait, but let me think again. Suppose we place a checkpoint not on the border but somewhere else. For example, if we place a checkpoint at (10, y), then the coverage on the border would be from 10 - sqrt(100 - y^2) to 10 + sqrt(100 - y^2). To cover from 0 to 20, we need 10 - sqrt(100 - y^2) ≤ 0 and 10 + sqrt(100 - y^2) ≥ 20. Let's see:10 - sqrt(100 - y^2) ≤ 0 => sqrt(100 - y^2) ≥ 10 => 100 - y^2 ≥ 100 => y^2 ≤ 0 => y=0. So only when y=0 can we cover from 0 to 20. Similarly, to cover up to 20, the next checkpoint would need to be at (20,0). So no, you can't cover more than 20 units unless you place the checkpoint on the border.Therefore, the minimal number is indeed 5, placed at (0,0), (20,0), (40,0), (60,0), (80,0).Wait, but let me think about the endpoints. The first checkpoint at (0,0) covers from 0 to 20. The last checkpoint at (80,0) covers from 80 to 100. So that's 20 units each, and the total is 100. So yes, 5 checkpoints.Alternatively, if we place checkpoints at (10,0), (30,0), (50,0), (70,0), (90,0), that also covers the entire border, with each covering 20 units. So either way, 5 checkpoints.So for part 1, the minimum number is 5, and one possible arrangement is placing them at (0,0), (20,0), (40,0), (60,0), (80,0).Now, moving on to part 2. We need to maximize coverage efficiency by minimizing the sum of the squares of the distances from every checkpoint to the nearest point on AB. Wait, but the nearest point on AB is just the projection of the checkpoint onto AB, which is (x_i, 0) for a checkpoint at (x_i, y_i). So the distance from the checkpoint to AB is |y_i|. Therefore, the sum we need to minimize is the sum of y_i^2 for all checkpoints.But we have the constraint from part 1, which is that every point on AB is within 10 units of at least one checkpoint. So we need to place the checkpoints such that this coverage condition is satisfied, and then minimize the sum of y_i^2.So the optimization problem is: minimize sum_{i=1}^n y_i^2, subject to the constraint that for every point (a, 0) on AB (0 ≤ a ≤ 100), there exists at least one checkpoint (x_i, y_i) such that sqrt( (x_i - a)^2 + y_i^2 ) ≤ 10.But since we are trying to minimize the sum of y_i^2, we want the checkpoints to be as close to the border as possible, i.e., y_i as small as possible. However, they still need to cover the entire border.Wait, but if we place all checkpoints on the border, y_i = 0, then the sum is 0, which is minimal. But does that satisfy the coverage condition? Yes, because each checkpoint covers 20 units, and 5 checkpoints cover the entire 100 units.But wait, in part 1, we concluded that 5 checkpoints on the border are sufficient. So in that case, the sum of squares is 0, which is the minimal possible. So is the optimal solution just placing all checkpoints on the border?But wait, maybe if we place some checkpoints off the border, we can have fewer checkpoints? But no, because in part 1, we already determined that 5 is the minimal number. So we can't have fewer than 5. Therefore, the minimal sum is 0, achieved by placing all 5 checkpoints on the border.Wait, but the problem says "maximize the coverage efficiency by minimizing the sum of the squares of the distances from every checkpoint to the nearest point on AB." So if we can achieve the coverage with some checkpoints off the border, but with a smaller sum of squares, that would be better. But since placing them on the border gives sum 0, which is the minimal possible, that's the optimal.But wait, maybe I'm misunderstanding. Maybe the checkpoints don't have to be on the border, but they can be anywhere, as long as they cover the border. So perhaps, by placing some checkpoints off the border, we can have the same coverage with a smaller sum of squares? But no, because the sum of squares is minimized when y_i is as small as possible, which is 0. So placing them on the border gives the minimal sum.Therefore, the optimal placement is to place all 5 checkpoints on the border, at (0,0), (20,0), (40,0), (60,0), (80,0), resulting in a sum of squares equal to 0.But wait, maybe the problem allows for a different number of checkpoints? No, part 1 specifies the minimal number, which is 5. So in part 2, we have to use 5 checkpoints, placed optimally to minimize the sum of squares.Therefore, the optimization problem is:Minimize sum_{i=1}^5 y_i^2Subject to:For all a in [0, 100], there exists i such that sqrt( (x_i - a)^2 + y_i^2 ) ≤ 10.And x_i ∈ [0, 100], y_i ∈ [0, 10].But as we saw, placing all y_i = 0 satisfies the constraint and gives the minimal sum. So the optimal solution is y_i = 0 for all i.Therefore, the coordinates are (0,0), (20,0), (40,0), (60,0), (80,0).Wait, but maybe there's a way to place some checkpoints off the border and still cover the entire border with 5 checkpoints, but with a smaller sum of squares? No, because if you place any checkpoint off the border, its y_i > 0, which increases the sum. So the minimal sum is achieved when all y_i = 0.Therefore, the answer to part 2 is the same as part 1, with all checkpoints on the border.But let me think again. Suppose we place some checkpoints slightly above the border. For example, if we place a checkpoint at (10, y), it can cover from 10 - sqrt(100 - y^2) to 10 + sqrt(100 - y^2). If we choose y such that sqrt(100 - y^2) = 10, then y=0. So no gain. If we choose y>0, the coverage is less, so we'd need more checkpoints, but we are constrained to 5 checkpoints.Wait, but if we place checkpoints off the border, maybe we can cover more efficiently? For example, if we place a checkpoint at (10, y), it can cover from 10 - sqrt(100 - y^2) to 10 + sqrt(100 - y^2). If we set y such that sqrt(100 - y^2) = 15, but that's impossible because sqrt(100 - y^2) can't exceed 10. So no, the maximum coverage is 20 units when y=0.Therefore, placing checkpoints on the border is optimal for both coverage and minimizing the sum of squares.So, to summarize:1. Minimum number of checkpoints: 5, placed at (0,0), (20,0), (40,0), (60,0), (80,0).2. The optimization problem is to minimize sum y_i^2, subject to coverage constraint, and the optimal solution is to place all checkpoints on the border, resulting in sum 0.But wait, the problem says "formulate an optimization problem". So I need to write it formally.Let me define variables:Let n be the number of checkpoints, which from part 1 is 5.Let x_i and y_i be the coordinates of checkpoint i, for i=1 to 5.We need to minimize sum_{i=1}^5 y_i^2.Subject to:For all a in [0, 100], there exists i such that sqrt( (x_i - a)^2 + y_i^2 ) ≤ 10.Additionally, 0 ≤ x_i ≤ 100, and y_i ≥ 0 (since distance is non-negative).But since we are minimizing the sum of y_i^2, the optimal solution will have y_i as small as possible, which is 0, as discussed.Therefore, the optimization problem is:Minimize ∑_{i=1}^5 y_i²Subject to:For each a ∈ [0, 100], ∃ i such that (x_i - a)² + y_i² ≤ 100,and 0 ≤ x_i ≤ 100, y_i ≥ 0.And the optimal solution is x_i = 0, 20, 40, 60, 80, and y_i = 0 for all i.So that's the answer.

answer：Alright, so I've got these two probability questions about a correctional officer monitoring a 10x10 grid of cells. Let me try to figure them out step by step.Starting with the first question: The probability of detecting contraband in any single random cell is 0.05. I need to find the expected number of cells containing contraband in the entire grid. Hmm, okay. I remember that expectation is like the average outcome we'd expect over many trials. So, for each cell, the expected value of it containing contraband is just the probability, right? So, for one cell, it's 0.05. Since expectation is linear, I can just add up the expectations for all 100 cells. That should give me the total expected number.Let me write that down. The expected number E is the sum of the expected values for each cell. Each cell has an expectation of 0.05, and there are 100 cells. So, E = 100 * 0.05. Calculating that, 100 times 0.05 is 5. So, the expected number is 5. That seems straightforward.Moving on to the second question. Now, the officer notices that contraband is more likely in prime-numbered positions. So, if the cells are numbered from 1 to 100 row-wise, the prime-numbered cells have a higher probability of 0.10. I need to find the new expected number, considering both prime and non-prime cells.Okay, so first, I need to figure out how many cells are prime-numbered between 1 and 100. Then, for those prime-numbered cells, the probability is 0.10, and for the rest, it's still 0.05. Then, the expected number will be the sum of (number of prime cells * 0.10) plus (number of non-prime cells * 0.05).So, step one: count the number of prime numbers between 1 and 100. Hmm, primes are numbers greater than 1 that have no divisors other than 1 and themselves. Let me list them out or recall how many there are.I remember that there are 25 prime numbers between 1 and 100. Let me verify that. Starting from 2, which is the first prime, then 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Let me count these: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97. That's 25 primes. Okay, so 25 prime-numbered cells.Therefore, the number of non-prime cells is 100 - 25 = 75.Now, for the prime cells, the expected number is 25 * 0.10, and for non-prime, it's 75 * 0.05. Let me compute each part.25 * 0.10 is 2.5, and 75 * 0.05 is 3.75. Adding those together, 2.5 + 3.75 = 6.25. So, the new expected number is 6.25.Wait, let me double-check my calculations. 25 primes, each with 0.10, so 25 * 0.10 = 2.5. Non-primes: 75 * 0.05. 75 * 0.05 is the same as 75 / 20, which is 3.75. Yeah, that's correct. So, 2.5 + 3.75 is indeed 6.25.So, the expected number increases from 5 to 6.25 when considering the prime-numbered cells have a higher probability.Just to make sure I didn't make a mistake in counting primes. Let me recount the primes between 1 and 100. Starting from 2:2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97. That's 25 numbers. Yep, that's correct.So, I think my calculations are solid. The first expected number is 5, and the second is 6.25.**Final Answer**1. The expected number of cells containing contraband is boxed{5}.2. The new expected number of cells containing contraband is boxed{6.25}.

answer：Alright, so I have this problem about Mr. X moving around a grid-like neighborhood. It's divided into two sub-problems. Let me try to tackle them one by one.Starting with Sub-problem 1: Mr. X's house is at (3,4). On the first day, he walks to (7,1), and on the second day to (11,-2). They want to know where he'll be on the 10th day. Hmm, so it's a geometric pattern, which probably means it's some kind of sequence or maybe a linear progression.First, let me figure out the pattern in his movements. So, starting from (3,4), he goes to (7,1) on day 1. Let me calculate the change in x and y coordinates from his house to day 1.Change in x: 7 - 3 = 4Change in y: 1 - 4 = -3So, from his house, he moves 4 units right and 3 units down on the first day.Then, on day 2, he goes to (11,-2). Let me see the change from day 1 to day 2.Change in x: 11 - 7 = 4Change in y: -2 - 1 = -3Wait, so from day 1 to day 2, he also moves 4 units right and 3 units down. So, is this a consistent movement each day? Let me check.If on day 1, he moves from (3,4) to (7,1), which is +4x, -3y.On day 2, he moves from (7,1) to (11,-2), which is again +4x, -3y.So, if this pattern continues, each day he moves 4 units right and 3 units down from the previous day's position.Therefore, each day, his position is increasing by 4 in the x-direction and decreasing by 3 in the y-direction.So, on day n, his position can be modeled as:x-coordinate: 3 + 4ny-coordinate: 4 - 3nWait, hold on. Let me check that.Wait, on day 1, he is at (7,1). So, plugging n=1 into the equations:x = 3 + 4*1 = 7, which is correct.y = 4 - 3*1 = 1, which is correct.On day 2:x = 3 + 4*2 = 11, correct.y = 4 - 3*2 = -2, correct.So, yes, the general formula seems to be:x = 3 + 4ny = 4 - 3nWhere n is the day number.Therefore, on day 10, plugging n=10:x = 3 + 4*10 = 3 + 40 = 43y = 4 - 3*10 = 4 - 30 = -26So, on day 10, he will be at (43, -26).Wait, let me verify this again.Starting from (3,4):Day 1: +4, -3 => (7,1)Day 2: +4, -3 => (11,-2)Day 3: +4, -3 => (15,-5)...Each day, adding 4 to x and subtracting 3 from y.So, on day n, x = 3 + 4n, y = 4 - 3n.Therefore, on day 10, x = 3 + 40 = 43, y = 4 - 30 = -26. Seems correct.So, Sub-problem 1 answer is (43, -26).Moving on to Sub-problem 2: Calculate the total Euclidean distance Mr. X has walked from his house from day 1 to day 10.So, each day, he walks from his previous day's position to the next day's position. So, each day, he walks a segment of 4 units right and 3 units down, which is a straight line.Wait, but the problem says "from his house from day 1 to day 10." Hmm, does that mean starting from his house each day, or does it mean the cumulative distance from his house over 10 days?Wait, no, the problem says: "the total Euclidean distance Mr. X has walked from his house from day 1 to day 10."Wait, so does that mean each day he starts from his house and walks to the point, or does he walk from the previous day's position?Wait, the problem says: "Mr. X takes daily walks that form a geometric pattern on the plane. On the first day, he walks to the point (7,1). On the second day, he walks to the point (11,-2)."So, it seems like each day, he walks from his house to a new point, which is a different location each day. So, each day, he starts at (3,4) and walks to a new point, which is (7,1) on day 1, (11,-2) on day 2, etc.Wait, but that contradicts the initial thought. Because if he starts at (3,4) each day, then each day's walk is from (3,4) to (7,1), then from (3,4) to (11,-2), etc. But that would mean each day he's making a different walk, not moving from the previous day's position.But in the first sub-problem, it says "assuming Mr. X continues this pattern," which implies that each day he moves from the previous day's position. So, perhaps I need to clarify.Wait, let me re-examine the problem statement."Mr. X takes daily walks that form a geometric pattern on the plane. On the first day, he walks to the point (7,1). On the second day, he walks to the point (11,-2). Assuming Mr. X continues this pattern, determine the coordinates of the point where he will be on the 10th day."So, the key here is that each day, he walks to a new point, which is part of a geometric pattern. So, the movement from day 1 to day 2 is from (7,1) to (11,-2). So, each day, he moves from the previous day's position to the next point in the pattern.Therefore, the total distance walked from day 1 to day 10 is the sum of the distances between each consecutive day's positions.So, starting from (3,4):Day 1: walks to (7,1)Day 2: walks to (11,-2)Day 3: walks to (15,-5)...Up to Day 10: walks to (43,-26)Therefore, the total distance is the sum of the distances between each pair of consecutive points from day 1 to day 10.But wait, actually, starting from his house on day 1, he walks to (7,1). Then on day 2, he walks from (7,1) to (11,-2). So, each day, he walks from the previous day's endpoint to the next point.Therefore, the total distance is the sum of the distances from day 1 to day 2, day 2 to day 3, ..., day 9 to day 10.So, we need to compute the distance between each consecutive pair of points and sum them up.But since the movement each day is consistent: each day, he moves 4 units in x and -3 units in y. So, the displacement each day is a vector of (4, -3). Therefore, the distance each day is the magnitude of this vector.So, the distance each day is sqrt(4^2 + (-3)^2) = sqrt(16 + 9) = sqrt(25) = 5 units.Therefore, each day, he walks 5 units. So, over 10 days, the total distance is 10 * 5 = 50 units.Wait, but hold on. Wait, on day 1, he walks from (3,4) to (7,1), which is 5 units.On day 2, he walks from (7,1) to (11,-2), which is another 5 units.Similarly, each day, he walks 5 units. So, over 10 days, it's 10 * 5 = 50 units.But wait, the problem says "from his house from day 1 to day 10." Hmm, does that mean starting from his house each day, so each day he walks 5 units, so total is 50 units? Or does it mean the straight-line distance from his house to the 10th day's position?Wait, the problem says: "Calculate the total Euclidean distance Mr. X has walked from his house from day 1 to day 10."Hmm, the wording is a bit ambiguous. It could mean the total distance walked each day, starting from his house, which would be 10 walks, each 5 units, totaling 50 units.Alternatively, it could be interpreted as the straight-line distance from his house to his position on day 10, which would be the distance from (3,4) to (43,-26). Let me calculate that.Distance = sqrt((43 - 3)^2 + (-26 - 4)^2) = sqrt(40^2 + (-30)^2) = sqrt(1600 + 900) = sqrt(2500) = 50 units.Wait, that's interesting. So, both interpretations give 50 units. But that can't be a coincidence.Wait, actually, if each day he walks 5 units, over 10 days, the total distance is 50 units. Also, the straight-line distance from start to finish is 50 units. That's because he's moving in a straight line each day, with each segment being 5 units, and the entire path is a straight line from (3,4) to (43,-26), which is 50 units. So, the total distance walked is equal to the straight-line distance because he's moving in a straight line each day, accumulating to the same path.Wait, but actually, no. Wait, each day he moves in the same direction, so the entire path is a straight line. So, the total distance walked is the same as the straight-line distance from start to finish.But in reality, if he moves in a straight line each day, but each day is a separate walk, starting from his house, then the total distance would be 10 times the daily distance, which is 50 units.But if he moves each day from the previous day's position, then the total distance is the same as the straight-line distance from start to finish, which is 50 units.Wait, so which interpretation is correct?The problem says: "Calculate the total Euclidean distance Mr. X has walked from his house from day 1 to day 10."The phrase "from his house" might mean that each walk starts from his house, so each day he walks 5 units, totaling 50 units.But in the first sub-problem, it's clear that he moves each day from the previous day's position, because it's a geometric pattern. So, the movement is cumulative.Wait, but the problem says "from his house from day 1 to day 10." Hmm.Wait, let me think again.If each day he starts from his house, then each day's walk is independent, and the total distance is 10 * 5 = 50.But if he starts each day from where he ended the previous day, then the total distance is the distance from (3,4) to (43,-26), which is 50 units.But in the first sub-problem, it's clear that he is moving each day from the previous day's position, because the pattern is cumulative.So, in that case, the total distance walked is the straight-line distance from (3,4) to (43,-26), which is 50 units.But wait, that seems contradictory because each day he is moving 5 units, so over 10 days, it's 50 units. But if he's moving in a straight line each day, then the total path is a straight line, so the total distance is 50 units.Wait, actually, both interpretations lead to the same answer here, which is 50 units. So, regardless of whether he walks each day from his house or from the previous day's position, the total distance is 50 units.But that seems a bit odd. Let me verify.If he walks each day from his house, then each day he walks 5 units, so 10 days is 50 units.If he walks each day from the previous day's position, then the total path is a straight line of 50 units, so the total distance is 50 units.So, in both cases, the total distance is 50 units.Therefore, regardless of interpretation, the answer is 50 units.But wait, let me think again. If he walks each day from his house, then each day he walks 5 units, so total is 50. If he walks each day from the previous day's position, then the total distance is the sum of each day's movement, which is 10 * 5 = 50. So, either way, it's 50.Therefore, the total distance is 50 units.But let me double-check the math.Each day, he moves 4 in x and -3 in y, so the distance each day is 5 units.From day 1 to day 10, he makes 10 such moves, so total distance is 10 * 5 = 50 units.Alternatively, the straight-line distance from (3,4) to (43,-26) is sqrt((43-3)^2 + (-26-4)^2) = sqrt(40^2 + (-30)^2) = sqrt(1600 + 900) = sqrt(2500) = 50 units.So, both ways, it's 50 units.Therefore, the total Euclidean distance is 50 units.So, to summarize:Sub-problem 1: On day 10, he is at (43, -26).Sub-problem 2: Total distance walked is 50 units.**Final Answer**Sub-problem 1: boxed{(43, -26)}Sub-problem 2: boxed{50}

answer：First, I need to determine the number of books in each genre that Mary has in her collection. She has a total of 120 books, and her reading preference ratio for Historical Fiction to Classic Literature to Non-Thriller Mysteries is 3:5:2. To find the number of books in each genre, I'll start by adding up the parts of the ratio: 3 + 5 + 2 = 10 parts in total.Next, I'll calculate how many books correspond to one part of the ratio by dividing the total number of books by the total number of parts: 120 books ÷ 10 parts = 12 books per part.Now, I can find the number of books in each genre by multiplying the number of parts for each genre by the value of one part:- Historical Fiction: 3 parts × 12 books/part = 36 books- Classic Literature: 5 parts × 12 books/part = 60 books- Non-Thriller Mysteries: 2 parts × 12 books/part = 24 booksThis completes the first sub-problem, determining the number of books in each genre.For the second sub-problem, Mary wants to place an equal number of books on each of the 6 shelves while maintaining the genre ratio on each shelf. Since there are 120 books in total, each shelf will have 120 books ÷ 6 shelves = 20 books per shelf.To maintain the genre ratio of 3:5:2 on each shelf, I'll first determine the total number of parts in the ratio: 3 + 5 + 2 = 10 parts.Next, I'll find out how many books correspond to one part on a shelf by dividing the number of books per shelf by the total number of parts: 20 books ÷ 10 parts = 2 books per part.Finally, I'll calculate the number of books of each genre per shelf:- Historical Fiction: 3 parts × 2 books/part = 6 books- Classic Literature: 5 parts × 2 books/part = 10 books- Non-Thriller Mysteries: 2 parts × 2 books/part = 4 booksThis ensures that each shelf has 20 books while maintaining the desired genre ratio.

answer：Okay, so I have this problem where I need to analyze the cost efficiency of a company with two production lines, A and B. The task has two main parts. Let me try to break it down step by step.First, I need to find the breakeven point for each production line. The breakeven point is where the total revenue equals the total cost, right? So, for each line, I have fixed costs and variable costs. The formula for breakeven in units is:Breakeven Units = Fixed Costs / (Selling Price per Unit - Variable Cost per Unit)But wait, the problem doesn't mention the selling price in the first part. Hmm, actually, looking back, it does mention the selling price is 1,000 in part 2. So maybe I can assume that the selling price is 1,000 for both parts? Or is it only in part 2? The first part doesn't specify, but since part 2 mentions it, maybe I should use 1,000 for both. I think that's a safe assumption unless stated otherwise.So, for Production Line A:Fixed Cost (FC) = 500,000Variable Cost (VC) = 300 per unitSelling Price (SP) = 1,000Breakeven Units for A = 500,000 / (1,000 - 300) = 500,000 / 700 ≈ 714.29 units. Since you can't produce a fraction of a unit, it would be 715 units.For Production Line B:Fixed Cost (FC) = 300,000Variable Cost (VC) = 500 per unitSelling Price (SP) = 1,000Breakeven Units for B = 300,000 / (1,000 - 500) = 300,000 / 500 = 600 units.Okay, that seems straightforward. Now, part 1 also asks to analyze the sensitivity of the breakeven point for Production Line A if the variable cost per unit increases by 10%. So, the variable cost would go up by 10% of 300, which is 30. So new VC = 330.New Breakeven Units for A = 500,000 / (1,000 - 330) = 500,000 / 670 ≈ 746.27 units, so 747 units.So, the breakeven point increases from 715 to 747 units when the variable cost increases by 10%. That means the company needs to produce and sell more units to break even, which is not ideal. It makes Production Line A less cost-efficient because the breakeven point is higher, so they have to sell more to cover their costs.Moving on to part 2. The company is implementing a cost-control initiative that reduces fixed costs by 15% for both lines but increases the variable cost of Production Line B by 5%. The selling price remains 1,000.First, let's calculate the new fixed and variable costs.For Production Line A:Original FC = 500,000. Reducing by 15%: 500,000 * 0.15 = 75,000. So new FC = 500,000 - 75,000 = 425,000.Variable Cost remains the same unless specified otherwise. Wait, the problem says only Production Line B's variable cost increases. So for A, VC is still 300.For Production Line B:Original FC = 300,000. Reducing by 15%: 300,000 * 0.15 = 45,000. So new FC = 300,000 - 45,000 = 255,000.Variable Cost was 500. Increasing by 5%: 500 * 0.05 = 25. So new VC = 500 + 25 = 525.Now, calculate the new breakeven points.For Production Line A:New FC = 425,000VC = 300SP = 1,000Breakeven Units = 425,000 / (1,000 - 300) = 425,000 / 700 ≈ 607.14 units, so 608 units.For Production Line B:New FC = 255,000New VC = 525SP = 1,000Breakeven Units = 255,000 / (1,000 - 525) = 255,000 / 475 ≈ 536.84 units, so 537 units.Now, comparing the original breakeven points to the new ones:Original:A: 715 unitsB: 600 unitsNew:A: 608 units (decrease)B: 537 units (decrease)So both breakeven points have decreased, which is good because the company needs to sell fewer units to break even. This suggests that the cost-control initiative has improved cost efficiency for both lines.But wait, let me think about the changes. For Line A, fixed costs went down, which directly reduces the breakeven point. Variable costs stayed the same, so the decrease is solely due to lower fixed costs.For Line B, fixed costs went down, but variable costs went up. However, the decrease in fixed costs had a bigger impact on the breakeven point than the increase in variable costs. So overall, the breakeven point still decreased.In terms of fiscal responsibility, reducing fixed costs by 15% is a significant saving, which is good. However, increasing variable costs for Line B by 5% could affect profitability if the volume doesn't increase enough to offset the higher per-unit cost. But since the breakeven point decreased, it means that even with the higher variable cost, the lower fixed costs make it more efficient.I should also consider the contribution margin. Contribution margin per unit is SP - VC. For Line A, it's still 700, so the same as before. For Line B, it's now 475 instead of 500, which is a slight decrease. But since fixed costs are lower, the overall breakeven is still lower.So, overall, the cost-control initiative seems beneficial because both breakeven points decreased, indicating improved cost efficiency. However, the company should monitor the variable costs for Line B to ensure that the increase doesn't erode profits if production volumes don't meet expectations.I think that's about it. Let me just recap:1. Breakeven points originally: - A: ~715 units - B: 600 units - Sensitivity for A: increases to ~747 units if VC up 10%2. After cost-control: - A: ~608 units - B: ~537 units - Both improved, so better efficiency and fiscal responsibility.I should present this clearly, maybe in a table or something, but since it's text, I'll just outline the numbers.