🧠:<think>Alright, so I have two problems to solve here, both related to population dynamics. The first one is about exponential growth with a carrying capacity, modeled by a logistic equation, and the second one is about exponential decay due to an epidemic. Let me tackle them one by one.Starting with the first problem: The population ( P(t) ) of a Tasmanian town is modeled by the differential equation ( frac{dP}{dt} = kP(1 - frac{P}{M}) ). The given values are ( P(0) = 1000 ), ( k = 0.03 ) per year, and ( M = 10,000 ). I need to find ( P(t) ) as a function of time ( t ).Hmm, okay. So this is the logistic growth model. I remember that the solution to this differential equation is given by:[P(t) = frac{M}{1 + left( frac{M - P_0}{P_0} right) e^{-k t}}]Where ( P_0 ) is the initial population. Let me verify that. Yes, the logistic equation is a common model for population growth with a carrying capacity, and the solution is indeed in that form.So, plugging in the given values: ( P_0 = 1000 ), ( M = 10,000 ), and ( k = 0.03 ).First, let's compute the term ( frac{M - P_0}{P_0} ). That would be ( frac{10,000 - 1000}{1000} = frac{9000}{1000} = 9 ).So, substituting back into the equation:[P(t) = frac{10,000}{1 + 9 e^{-0.03 t}}]Let me double-check the formula. Yes, the denominator is 1 plus (M - P0)/P0 times e^{-kt}, so that seems correct.Alternatively, sometimes the logistic equation is written as:[P(t) = frac{M}{1 + left( frac{M}{P_0} - 1 right) e^{-k t}}]Which is the same thing because ( frac{M}{P_0} - 1 = frac{M - P_0}{P_0} ). So that's consistent.Therefore, the population as a function of time is ( P(t) = frac{10,000}{1 + 9 e^{-0.03 t}} ). That should be the answer for the first part.Moving on to the second problem: The population ( Q(t) ) of another site is modeled by exponential decay due to an epidemic, given by ( Q(t) = Q_0 e^{-rt} ). The initial population ( Q_0 ) is 5000, and it decreased to 2000 in 5 years. I need to find the decay rate ( r ) and predict the population after 10 years.Okay, so we have ( Q(5) = 2000 ). Let's plug that into the equation:[2000 = 5000 e^{-5r}]We can solve for ( r ). Let's divide both sides by 5000:[frac{2000}{5000} = e^{-5r}]Simplify the fraction:[0.4 = e^{-5r}]To solve for ( r ), take the natural logarithm of both sides:[ln(0.4) = -5r]So,[r = -frac{ln(0.4)}{5}]Calculating ( ln(0.4) ). Let me recall that ( ln(0.4) ) is approximately... Well, ( ln(1) = 0 ), ( ln(0.5) approx -0.6931 ), so ( ln(0.4) ) should be a bit less than that, maybe around -0.9163? Let me confirm:Using a calculator, ( ln(0.4) approx -0.916291 ). So,[r = -frac{-0.916291}{5} = frac{0.916291}{5} approx 0.183258]So, approximately 0.1833 per year. Let me write that as ( r approx 0.1833 ).Now, to predict the population after 10 years, we can use the same formula:[Q(10) = 5000 e^{-10r}]Substituting ( r approx 0.1833 ):First, compute ( 10r approx 10 * 0.1833 = 1.833 ).So,[Q(10) = 5000 e^{-1.833}]Calculating ( e^{-1.833} ). Let me approximate that. I know that ( e^{-1} approx 0.3679 ), ( e^{-2} approx 0.1353 ). Since 1.833 is between 1 and 2, closer to 2.Let me use a calculator for a better approximation. ( e^{-1.833} approx e^{-1.8} * e^{-0.033} approx 0.1653 * 0.9677 approx 0.1600 ). Wait, is that right?Wait, let me compute 1.833:Alternatively, using a calculator:( e^{-1.833} approx 0.1600 ). Hmm, let's verify:Compute ( ln(0.16) approx -1.8326 ). Yes, so ( e^{-1.8326} = 0.16 ). So, ( e^{-1.833} approx 0.16 ).Therefore,[Q(10) = 5000 * 0.16 = 800]So, the population after 10 years would be approximately 800.Wait, but let me check my calculation for ( e^{-1.833} ). Maybe I should use a more precise method.Alternatively, using the formula:( e^{-1.833} = e^{-1.8} * e^{-0.033} ).Compute ( e^{-1.8} approx 0.1653 ), ( e^{-0.033} approx 1 - 0.033 + (0.033)^2/2 - (0.033)^3/6 approx 0.9677 ).Multiplying these: 0.1653 * 0.9677 ≈ 0.1600.So, yes, approximately 0.16.Therefore, 5000 * 0.16 = 800.Alternatively, using a calculator for more precision:Compute ( e^{-1.833} ):1.833 is approximately 1.833.Let me compute 1.833:We can write 1.833 as 1 + 0.8 + 0.033.Compute ( e^{-1} = 0.367879 ), ( e^{-0.8} approx 0.449329 ), ( e^{-0.033} approx 0.9677 ).Wait, but actually, ( e^{-1.833} = e^{-1} * e^{-0.8} * e^{-0.033} approx 0.367879 * 0.449329 * 0.9677 ).Compute 0.367879 * 0.449329 ≈ 0.1653.Then, 0.1653 * 0.9677 ≈ 0.1600.So, same result.Therefore, Q(10) ≈ 5000 * 0.16 = 800.Alternatively, to get a more precise value, maybe use a calculator:Compute 1.833:Using a calculator, e^{-1.833} ≈ e^{-1.833} ≈ 0.1600.Yes, so 5000 * 0.16 = 800.So, the population after 10 years is approximately 800.Wait, but let me think again. The initial population is 5000, after 5 years it's 2000, which is a decrease by a factor of 0.4. So, the decay factor is 0.4 over 5 years. So, the decay rate r is such that e^{-5r} = 0.4, which we solved as r ≈ 0.1833.Then, over 10 years, it's e^{-10r} = (e^{-5r})^2 = (0.4)^2 = 0.16. So, 5000 * 0.16 = 800. So, that's consistent.Therefore, the decay rate is approximately 0.1833 per year, and the population after 10 years is 800.Wait, but let me compute r more accurately. Earlier, I approximated ( ln(0.4) approx -0.916291 ), so r = 0.916291 / 5 ≈ 0.183258, which is approximately 0.1833.So, r ≈ 0.1833 per year.Alternatively, if I use more decimal places, maybe 0.183258, which is approximately 0.1833.So, that's the decay rate.Therefore, summarizing:1. The population function is ( P(t) = frac{10,000}{1 + 9 e^{-0.03 t}} ).2. The decay rate ( r ) is approximately 0.1833 per year, and the population after 10 years is 800.I think that's it. Let me just make sure I didn't make any calculation errors.For the first problem, plugging in t=0, P(0) should be 1000. Let's check:( P(0) = 10,000 / (1 + 9 e^{0}) = 10,000 / (1 + 9) = 10,000 / 10 = 1000. Correct.For the second problem, at t=5, Q(5) should be 2000:( Q(5) = 5000 e^{-5r} = 5000 * 0.4 = 2000. Correct.At t=10, Q(10) = 5000 e^{-10r} = 5000 * (e^{-5r})^2 = 5000 * (0.4)^2 = 5000 * 0.16 = 800. Correct.So, all checks out.</think>

🧠:<think>Alright, so I have this problem about a coffee grower who wants to intercrop cacao trees within a rectangular plot. The plot is 200 meters by 150 meters. The coffee plants are arranged in a grid where each plant takes up 4 square meters. The grower wants to add cacao trees in a hexagonal pattern around each cacao tree, with each side of the hexagon being 2 meters. Also, the distance between any two cacao trees needs to be at least 10 meters to prevent competition. First, I need to figure out the maximum number of cacao trees that can be intercropped without disturbing the coffee plant grid. Then, I have to calculate the total coffee yield considering that each cacao tree improves the surrounding coffee plants by 10%. Let me start with the first part. So, the plot is 200m by 150m. Each coffee plant occupies 4 square meters, which means each plant is spaced in a grid where each spacing is 2 meters apart because sqrt(4) is 2. So, the coffee plants are arranged in a grid with 2m spacing between them. Now, the cacao trees are to be placed such that each is surrounded by coffee plants in a hexagonal pattern. Each side of the hexagon is 2 meters. So, the distance from the cacao tree to each surrounding coffee plant is 2 meters. But wait, the distance between two cacao trees should be at least 10 meters. Hmm, so I need to figure out how to place these cacao trees within the existing coffee grid without violating the 10-meter distance rule.First, let me visualize the coffee grid. Since each coffee plant is 2m apart, the grid is like a lattice with points every 2 meters in both x and y directions. So, the number of coffee plants along the length (200m) would be 200 / 2 = 100 plants, and along the width (150m) it would be 150 / 2 = 75 plants. So, total coffee plants are 100 * 75 = 7500 plants.But the grower wants to intercrop cacao trees. Each cacao tree is surrounded by a hexagon of coffee plants, each side 2 meters. So, the hexagon around each cacao tree would have a radius of 2 meters. But the distance between two cacao trees needs to be at least 10 meters. Wait, so the distance between two cacao trees is 10 meters. So, if I place a cacao tree at a certain point, the next one should be at least 10 meters away. But where exactly can I place the cacao trees within the coffee grid? Since the coffee plants are on a grid, maybe the cacao trees can be placed at certain intervals within this grid.Let me think. If each cacao tree is surrounded by a hexagon of coffee plants with each side 2 meters, then the cacao tree is at the center of a hexagon with 2m sides. So, the distance from the cacao tree to each coffee plant is 2 meters. But the distance between two cacao trees should be at least 10 meters. So, how does this translate in terms of their positions on the grid?Perhaps the cacao trees can be placed in a grid themselves, spaced 10 meters apart. But since the coffee plants are on a 2m grid, the cacao trees need to be placed at points that are multiples of 2m apart, but at least 10m apart from each other.Wait, 10 meters is 5 times 2 meters. So, if the cacao trees are placed every 10 meters, that would be every 5th coffee plant in each direction.But let me confirm. If the coffee plants are 2m apart, then the distance between two cacao trees placed 5 coffee plants apart would be 5 * 2m = 10m. So, that satisfies the 10m distance requirement.So, the cacao trees can be placed in a grid where each cacao tree is 10m apart from the next in both x and y directions. But wait, is that the most efficient way? Because if we place them in a square grid, the diagonal distance would be sqrt(10^2 + 10^2) = ~14.14m, which is more than 10m, so that's fine. But maybe arranging them in a hexagonal grid would allow more cacao trees? But the problem says each cacao tree is surrounded by a hexagon of coffee plants, but the cacao trees themselves need to be spaced at least 10m apart.But the original coffee grid is square, so maybe the cacao trees can be placed in a square grid within the coffee grid.Alternatively, perhaps the cacao trees can be placed in a hexagonal pattern relative to each other, but given the coffee grid is square, it might complicate things. Maybe it's simpler to place them in a square grid.So, assuming a square grid for cacao trees, spaced 10m apart. Now, the plot is 200m by 150m. So, along the length (200m), the number of cacao trees would be 200 / 10 = 20 trees. But wait, since the first tree is at 0m, the last tree would be at 190m, leaving 10m remaining. So, actually, 20 positions.Similarly, along the width (150m), 150 / 10 = 15 trees. So, 15 positions.Therefore, the total number of cacao trees would be 20 * 15 = 300 trees.But wait, I need to make sure that placing a cacao tree doesn't interfere with the coffee plant grid. Since each cacao tree is surrounded by a hexagon of coffee plants, each side 2m, that means the cacao tree is at the center of a hexagon with radius 2m. So, the cacao tree itself doesn't occupy any coffee plant space, but it's just placed in the grid.But wait, each coffee plant is 2m apart, so the grid is 2m spacing. So, the cacao trees can be placed at every 10m, which is every 5th coffee plant spacing.But let me think about the exact placement. If the coffee plants are on a grid with coordinates (2i, 2j) where i and j are integers, then the cacao trees can be placed at (10k, 10l) where k and l are integers. So, the number of cacao trees along the length would be floor(200 / 10) = 20, and along the width floor(150 / 10) = 15. So, 20 * 15 = 300 cacao trees.But wait, is there a way to fit more cacao trees by offsetting the grid? For example, placing some cacao trees in between the 10m grid points? But the problem states that the distance between any two cacao trees should be at least 10 meters. So, if we try to place them in a hexagonal grid, the distance between adjacent cacao trees would be 10m, but the number might be higher.Wait, in a hexagonal packing, the number of points in a given area is higher than in a square grid. So, maybe we can fit more cacao trees by arranging them in a hexagonal pattern.But the coffee plants are on a square grid, so how would that affect the placement of cacao trees? Each cacao tree is surrounded by a hexagon of coffee plants, but the cacao trees themselves need to be spaced at least 10m apart.Alternatively, perhaps the cacao trees can be placed in a hexagonal grid relative to each other, but within the coffee grid. Let me think about the distance between cacao trees in a hexagonal grid. In a hexagonal grid, each point has six neighbors at equal distance. The distance between adjacent points is the same, say d. The number of points per unit area is higher than in a square grid.But in our case, the minimum distance between cacao trees is 10m, so d must be at least 10m. So, if we arrange cacao trees in a hexagonal grid with spacing 10m, how many can we fit in the 200m x 150m plot?In a hexagonal grid, the number of points along the x-axis would be similar to a square grid, but along the y-axis, the points are offset by half the spacing. The number of cacao trees along the length (200m) would still be 200 / 10 = 20, as before.Along the width (150m), in a hexagonal grid, the vertical distance between rows is d * sqrt(3)/2, where d is the horizontal spacing. So, with d = 10m, the vertical distance is 10 * sqrt(3)/2 ≈ 8.66m.So, the number of rows along the width would be 150 / 8.66 ≈ 17.3, so 17 rows.But wait, in a hexagonal grid, the number of points per row alternates between full and half. So, the total number of points would be approximately (number of rows) * (number of points per row). But since the plot is 150m wide, and each row is spaced 8.66m apart, starting from 0m, the first row is at 0m, the second at 8.66m, the third at 17.32m, and so on. The last row would be at (n-1)*8.66m ≤ 150m. So, n = floor(150 / 8.66) + 1 ≈ floor(17.3) +1 = 17 +1 = 18 rows.But wait, the first row is at 0m, the 18th row would be at (17)*8.66 ≈ 147.22m, leaving about 2.78m remaining, which is less than the spacing, so we can't fit another row.So, 18 rows. Now, in a hexagonal grid, the number of points per row alternates between full and half. So, if the first row has 20 points, the next row would have 19 points, then 20, and so on.But wait, actually, in a hexagonal grid, each row is offset by half a spacing, so the number of points per row alternates. However, in our case, the plot is 200m long, so the number of points per row would be 200 / 10 = 20 points.But when offsetting, the next row would start at 5m (half of 10m), so the number of points in the next row would be 20 as well, because 200m /10m = 20, even if starting at 5m, it would still end at 195m, which is within 200m.Wait, no, actually, if you start at 5m, the last point would be at 5 + 19*10 = 195m, which is still within 200m. So, each row can have 20 points regardless of the offset.Therefore, the total number of cacao trees would be 18 rows * 20 points = 360 trees.But wait, is that correct? Because in a hexagonal grid, the number of points per unit area is higher than a square grid. So, in a square grid, we had 20*15=300 trees, and in a hexagonal grid, we have 18*20=360 trees. But does this fit within the 200m x 150m plot? Let me check the vertical spacing. The vertical distance between rows is ~8.66m, so 18 rows would occupy (18-1)*8.66 ≈ 17*8.66 ≈ 147.22m, which is within the 150m width. So, yes, 18 rows fit.But wait, the problem is that the coffee plants are on a square grid, so the cacao trees need to be placed in such a way that they don't interfere with the coffee plants. Each cacao tree is surrounded by a hexagon of coffee plants with each side 2m. So, the cacao tree is at the center of a hexagon with radius 2m. But if we place cacao trees in a hexagonal grid spaced 10m apart, does that interfere with the coffee plants? Because each cacao tree is surrounded by coffee plants, but the coffee plants are already on a 2m grid. So, as long as the cacao trees are placed at points that are multiples of 2m apart, they can fit into the grid.Wait, but in a hexagonal grid, the horizontal spacing is 10m, but the vertical spacing is 8.66m, which is not a multiple of 2m. So, the cacao trees would not align with the coffee grid in the vertical direction. This might cause some coffee plants to be displaced or not fit properly. Because the coffee plants are on a 2m grid, if the cacao trees are placed at positions that are not multiples of 2m, it might disrupt the grid.So, perhaps the cacao trees need to be placed at points that are multiples of 2m in both x and y directions. That would mean that the cacao trees are on a square grid, spaced 10m apart, which is 5*2m. So, that way, they fit into the coffee grid without disturbing it.Therefore, the maximum number of cacao trees would be 20 along the length and 15 along the width, totaling 300 trees.But wait, is there a way to place more cacao trees by using a different pattern? Maybe placing them in a staggered grid within the coffee grid, but ensuring that the distance between any two cacao trees is at least 10m.Alternatively, perhaps the cacao trees can be placed in a hexagonal pattern relative to each other, but within the coffee grid, which is square. But the problem is that the coffee grid is square, so the cacao trees have to be placed at points that are on the coffee grid. So, their coordinates must be multiples of 2m in both x and y directions. Therefore, the cacao trees can only be placed at points (2i, 2j), where i and j are integers.Given that, the distance between any two cacao trees must be at least 10m. So, the minimum distance between two cacao trees is 10m. So, to maximize the number of cacao trees, we need to place them as densely as possible on the 2m grid, with each pair at least 10m apart.This is similar to placing points on a grid with a minimum distance constraint.In such cases, the most efficient way is to place them in a square grid where each is spaced 10m apart, but since the coffee grid is 2m, the cacao trees can be placed every 5 coffee plants (since 5*2=10m).So, along the length of 200m, the number of cacao trees would be 200 /10 = 20, and along the width of 150m, 150 /10 =15. So, 20*15=300 cacao trees.But wait, is there a way to place more cacao trees by offsetting every other row? For example, in a staggered grid, where the cacao trees in adjacent rows are offset by 5m, which is half of 10m. In such a case, the vertical distance between rows would be sqrt(10^2 -5^2)=sqrt(75)=~8.66m, as in a hexagonal grid. But since the coffee grid is 2m, the offset of 5m is 2.5*2m, which is not an integer multiple. So, the cacao trees would not align with the coffee grid in the vertical direction. Therefore, the cacao trees would have to be placed at points that are multiples of 2m, so the offset would have to be a multiple of 2m as well. If we try to stagger the rows by 2m, then the vertical distance between rows would be sqrt(10^2 -2^2)=sqrt(96)=~9.798m, which is less than 10m. Therefore, the distance between cacao trees in adjacent rows would be less than 10m, which violates the minimum distance requirement.Therefore, we cannot stagger the rows by 2m. If we try to stagger by 4m, then the vertical distance would be sqrt(10^2 -4^2)=sqrt(84)=~9.165m, still less than 10m.If we stagger by 6m, sqrt(10^2 -6^2)=sqrt(64)=8m, still less than 10m.Wait, if we stagger by 8m, sqrt(10^2 -8^2)=sqrt(36)=6m, still less than 10m.Wait, actually, if we stagger by 10m, then the vertical distance would be sqrt(10^2 -10^2)=0, which doesn't make sense.Wait, perhaps I'm approaching this incorrectly. The offset in the x-direction would be 5m (half of 10m), but since the coffee grid is 2m, 5m is 2.5*2m, which is not an integer multiple. Therefore, the cacao trees cannot be placed at 5m offset because that would not align with the coffee grid.Therefore, we cannot stagger the rows in a way that maintains the 10m distance between cacao trees while keeping them on the coffee grid.Therefore, the maximum number of cacao trees is achieved by placing them in a square grid, 10m apart, resulting in 20*15=300 trees.But wait, let me double-check. If we place them in a square grid, 10m apart, then the distance between adjacent cacao trees is exactly 10m, which meets the requirement. But is there a way to fit more cacao trees by using a hexagonal packing within the coffee grid? Wait, in a hexagonal packing, the number of points per unit area is higher, but since the coffee grid is square, the cacao trees have to be placed on the square grid. Therefore, the most efficient packing is the square grid with 10m spacing.Alternatively, perhaps we can place cacao trees in a way that some are offset by 2m in one direction, but then the distance between them would still have to be at least 10m. But if we place a cacao tree at (0,0), the next one could be at (10,0), (20,0), etc. Then, in the next row, we could place them at (5, y), but y has to be such that the distance from (0,0) to (5,y) is at least 10m. So, sqrt(5^2 + y^2) >=10. Therefore, y >= sqrt(75)≈8.66m. But since the coffee grid is 2m, y has to be a multiple of 2m. The smallest y that satisfies y >=8.66m is 10m. So, the next row would be at y=10m, but then the distance between (0,0) and (5,10) is sqrt(5^2 +10^2)=sqrt(125)=~11.18m, which is more than 10m. Wait, but if we place cacao trees at (0,0), (10,0), (20,0), etc., and then in the next row at (5,10), (15,10), (25,10), etc., then the distance between (0,0) and (5,10) is ~11.18m, which is acceptable. Similarly, the distance between (5,10) and (10,0) is sqrt(5^2 +10^2)=~11.18m, which is also acceptable. So, in this way, we can fit more cacao trees. How many rows can we fit in 150m? Each row is spaced 10m apart vertically, but the first row is at y=0, the second at y=10, the third at y=20, etc. Wait, but if we stagger the rows, the vertical spacing is 10m, but the horizontal offset is 5m. Wait, no, in this case, the vertical distance between rows is 10m, but the horizontal offset is 5m. So, the vertical spacing between rows is 10m, but the horizontal offset is 5m. Therefore, the number of rows along the width (150m) would be 150 /10 =15 rows. But wait, the first row is at y=0, the second at y=10, the third at y=20, etc., up to y=140m, which is the 15th row. But in this case, the horizontal offset alternates between 0 and 5m for each row. So, the first row is at (0,0), (10,0), (20,0),..., the second row is at (5,10), (15,10), (25,10),..., the third row is at (0,20), (10,20),..., and so on.Therefore, in each row, the number of cacao trees would be 200 /10 =20 per row. But in the rows that are offset by 5m, the first cacao tree is at 5m, so the last one would be at 5 +19*10=195m, which is within 200m. So, each row has 20 cacao trees.Therefore, the total number of cacao trees would be 15 rows *20 trees =300 trees, same as the square grid.Wait, but actually, in this staggered grid, the number of rows is 15, but each row has 20 trees, so total is 300. But wait, in reality, in a hexagonal packing, the number of points per unit area is higher. So, why isn't it higher here? Because we are constrained by the coffee grid. Wait, in a hexagonal packing without grid constraints, the number of points would be higher, but here, since the cacao trees have to be placed on the coffee grid, which is square, the maximum number we can fit is 300, whether we use a square grid or a staggered grid.Wait, but in the staggered grid, we have the same number of cacao trees as the square grid, but the distance between some cacao trees is more than 10m, which is acceptable, but we don't gain any additional cacao trees because of the grid constraints.Therefore, the maximum number of cacao trees is 300.Wait, but let me think again. If we stagger the rows, can we fit more rows? Because the vertical distance between rows is 10m, but the actual vertical distance in the staggered grid is 10m, so the number of rows is the same as the square grid.Wait, no, in a hexagonal grid, the vertical distance between rows is less, which allows more rows in the same vertical space. But in our case, because of the grid constraint, the vertical distance between rows is 10m, same as the square grid. Therefore, the number of rows is the same.Therefore, the maximum number of cacao trees is 300.But wait, let me confirm with another approach. The area of the plot is 200*150=30,000 square meters.Each cacao tree requires a hexagon of coffee plants around it, but the cacao tree itself doesn't occupy any area, just the surrounding coffee plants. But the distance between cacao trees is 10m, so the area per cacao tree is a circle of radius 5m (since the distance between centers is 10m, the radius for non-overlapping circles would be 5m). But actually, the distance between cacao trees is 10m, so the area per cacao tree is a circle with diameter 10m, area π*(5)^2=25π≈78.54 square meters.But the plot area is 30,000, so the maximum number of cacao trees would be 30,000 /78.54≈381. But this is without considering the grid constraint.But since the cacao trees have to be placed on a 2m grid, the maximum number is limited by the grid spacing.Therefore, the grid-based calculation of 300 is more accurate.Therefore, the maximum number of cacao trees is 300.Now, moving on to the second part. Each cacao tree improves the yield of the surrounding coffee plants by 10%. Each undisturbed coffee plant yields 2kg per season.So, first, I need to figure out how many coffee plants are affected by each cacao tree. Since each cacao tree is surrounded by a hexagon of coffee plants with each side 2m, the hexagon has a radius of 2m.A hexagon with side length 2m has 6 coffee plants around the cacao tree, each at a distance of 2m. But wait, actually, a hexagon with side length 2m would have 6 coffee plants at the vertices, each 2m away from the cacao tree.But in reality, the hexagon is formed by the coffee plants, so each cacao tree is at the center of a hexagon with 6 coffee plants around it.But wait, the problem says each cacao tree is surrounded by coffee plants in a hexagonal pattern, with each side of the hexagon being 2 meters in length. So, the distance from the cacao tree to each coffee plant is 2 meters.Therefore, each cacao tree affects 6 coffee plants, each 2m away.But wait, actually, in a hexagonal grid, each cacao tree would have 6 neighboring coffee plants, but also, coffee plants further out might be affected. But the problem states that each cacao tree is surrounded by coffee plants in a hexagonal pattern, with each side of the hexagon being 2 meters. So, it's just the immediate surrounding hexagon, meaning 6 coffee plants.But wait, actually, a hexagon with side length 2m would have 6 coffee plants at the vertices, each 2m away from the cacao tree. But the area around the cacao tree is a hexagon with side 2m, so the number of coffee plants affected is 6.But wait, actually, in a hexagonal grid, each cacao tree would have 6 neighboring coffee plants, but also, coffee plants in the next layer would be affected. But the problem says each cacao tree is surrounded by coffee plants in a hexagonal pattern, with each side of the hexagon being 2 meters. So, it's just the immediate surrounding hexagon, meaning 6 coffee plants.But wait, actually, the hexagon with side length 2m would have 6 coffee plants at the vertices, each 2m away from the cacao tree. So, each cacao tree affects 6 coffee plants.But wait, let me think again. If the hexagon has side length 2m, then the distance from the center to each vertex is 2m. So, each cacao tree is surrounded by 6 coffee plants, each 2m away.Therefore, each cacao tree affects 6 coffee plants.But wait, in reality, the coffee plants are on a square grid, not a hexagonal grid. So, the distance from the cacao tree to the coffee plants is 2m, but in a square grid, the surrounding coffee plants would be at (2,0), (0,2), (-2,0), (0,-2), (2,2), (-2,2), etc. But wait, in a square grid, the distance from the cacao tree to the surrounding coffee plants is 2m, but the number of surrounding coffee plants is 4 (up, down, left, right), but diagonally, they are 2.828m away, which is more than 2m.Wait, this is getting confusing. Let me clarify.The coffee plants are on a square grid with 2m spacing. So, each coffee plant is at (2i, 2j). The cacao tree is at some point, say (x,y). The distance from (x,y) to each coffee plant is 2m.But in a square grid, the distance from (x,y) to the nearest coffee plants would be 2m in the cardinal directions (up, down, left, right), but diagonally, it's sqrt(2^2 +2^2)=sqrt(8)=2.828m.But the problem states that each cacao tree is surrounded by coffee plants in a hexagonal pattern, with each side of the hexagon being 2 meters. So, the distance from the cacao tree to each surrounding coffee plant is 2 meters.Therefore, in a square grid, the cacao tree must be equidistant (2m) from 6 coffee plants arranged in a hexagon. But in a square grid, it's not possible to have 6 coffee plants at exactly 2m distance from a central point unless the cacao tree is at the intersection of the grid lines.Wait, actually, in a square grid, the maximum number of coffee plants at exactly 2m distance from a point is 4 (the cardinal directions). The diagonal coffee plants are further away.Therefore, perhaps the problem is assuming that the coffee plants are arranged in a hexagonal grid, but the problem states it's a rectangular plot with a grid where each plant occupies 4 square meters. So, it's a square grid.Therefore, perhaps the cacao trees are placed at points where they are equidistant to 6 coffee plants, but in a square grid, that's not possible. Therefore, maybe the cacao tree is at the center of a hexagon formed by 6 coffee plants, but in a square grid, that would require the coffee plants to be arranged in a hexagon, which is not possible.Wait, perhaps the problem is assuming that the coffee plants are arranged in a hexagonal grid, but the plot is rectangular. But the problem says it's a rectangular plot with a grid where each plant occupies 4 square meters. So, it's a square grid.Therefore, perhaps the cacao tree is surrounded by 4 coffee plants at 2m distance (up, down, left, right), and the other two coffee plants are at diagonal positions, but further away. But the problem says each side of the hexagon is 2m, so the distance from the cacao tree to each surrounding coffee plant is 2m.Therefore, perhaps the cacao tree is at the center of a hexagon formed by 6 coffee plants, each 2m away. But in a square grid, this is not possible because the coffee plants cannot form a regular hexagon around the cacao tree.Therefore, perhaps the problem is assuming that the coffee plants are arranged in a hexagonal grid, but the plot is rectangular. But the problem states it's a rectangular plot with a grid where each plant occupies 4 square meters. So, it's a square grid.This is a bit confusing. Maybe I need to proceed with the assumption that each cacao tree is surrounded by 6 coffee plants, each 2m away, arranged in a hexagon. But in reality, in a square grid, the cacao tree can only have 4 coffee plants at 2m distance, and the other two would be further away. But the problem states that each cacao tree is surrounded by coffee plants in a hexagonal pattern, with each side of the hexagon being 2 meters. So, perhaps the coffee plants are arranged in a hexagonal grid, but the plot is rectangular. Wait, the problem says the plot is rectangular, and the coffee plants are arranged in a grid where each plant occupies 4 square meters. So, it's a square grid with 2m spacing.Therefore, perhaps the cacao tree is placed at the center of a hexagon formed by 6 coffee plants, but in a square grid, this would require the cacao tree to be at a point where 6 coffee plants are at 2m distance, which is not possible. Therefore, perhaps the cacao tree is placed at a point where it is surrounded by 4 coffee plants at 2m distance (up, down, left, right), and the other two coffee plants are at 2m distance diagonally, but in a square grid, the diagonal distance is 2.828m, which is more than 2m. Therefore, perhaps the problem is assuming that the coffee plants are arranged in a hexagonal grid, but the plot is rectangular. Alternatively, perhaps the cacao tree is surrounded by 6 coffee plants, each 2m away, but in a square grid, this would require the cacao tree to be at a point where 6 coffee plants are at 2m distance, which is not possible. Therefore, perhaps the problem is considering that each cacao tree is surrounded by 6 coffee plants, each 2m away, but in a square grid, this would require the cacao tree to be at a point where 6 coffee plants are at 2m distance, which is not possible. Therefore, perhaps the problem is assuming that the coffee plants are arranged in a hexagonal grid, but the plot is rectangular. Alternatively, perhaps the cacao tree is surrounded by 6 coffee plants, each 2m away, but in a square grid, this would require the cacao tree to be at a point where 6 coffee plants are at 2m distance, which is not possible. Therefore, perhaps the problem is considering that each cacao tree is surrounded by 6 coffee plants, each 2m away, but in a square grid, this would require the cacao tree to be at a point where 6 coffee plants are at 2m distance, which is not possible. Therefore, perhaps the problem is assuming that the coffee plants are arranged in a hexagonal grid, but the plot is rectangular. But since the problem states it's a rectangular plot with a grid where each plant occupies 4 square meters, it's a square grid. Therefore, perhaps the cacao tree is surrounded by 4 coffee plants at 2m distance, and the other two coffee plants are at a greater distance, but the problem states that each side of the hexagon is 2m, so the distance from the cacao tree to each surrounding coffee plant is 2m. Therefore, perhaps the cacao tree is at the center of a hexagon formed by 6 coffee plants, each 2m away, but in a square grid, this is not possible. Therefore, perhaps the problem is assuming that the coffee plants are arranged in a hexagonal grid, but the plot is rectangular. But given the problem statement, I think I need to proceed with the assumption that each cacao tree is surrounded by 6 coffee plants, each 2m away, arranged in a hexagon. Therefore, each cacao tree affects 6 coffee plants.Therefore, each cacao tree improves the yield of 6 coffee plants by 10%. So, each cacao tree adds 0.1*2kg=0.2kg per coffee plant. So, 6*0.2kg=1.2kg per cacao tree.But wait, actually, each cacao tree improves the yield of the surrounding coffee plants by 10%, so each affected coffee plant yields 2kg*1.1=2.2kg instead of 2kg.Therefore, the total yield increase per cacao tree is 6*(2.2kg -2kg)=6*0.2kg=1.2kg.Therefore, each cacao tree adds 1.2kg to the total yield.But wait, actually, the total yield would be the sum of all coffee plants, some of which are improved by the cacao trees.But we have to be careful not to double-count the coffee plants. Because if two cacao trees are close enough, their improved coffee plants might overlap.But in our case, the cacao trees are spaced at least 10m apart. The distance between cacao trees is 10m, and each cacao tree affects coffee plants within a 2m radius. So, the distance between two cacao trees is 10m, which is greater than the sum of their radii (2m +2m=4m). Therefore, the areas of influence do not overlap. Therefore, the improved coffee plants around each cacao tree are unique and do not overlap with those of other cacao trees.Therefore, the total number of improved coffee plants is 6 per cacao tree, and since there are 300 cacao trees, the total number of improved coffee plants is 300*6=1800.But wait, the total number of coffee plants is 7500. So, 1800 are improved, and 7500-1800=5700 are not improved.Therefore, the total yield would be 5700*2kg + 1800*2.2kg.Calculating that:5700*2 = 11,400kg1800*2.2 = 3,960kgTotal yield = 11,400 + 3,960 = 15,360kgTherefore, the total expected yield is 15,360kg per season.But wait, let me double-check. Each cacao tree affects 6 coffee plants, each yielding 2.2kg instead of 2kg. So, per cacao tree, the additional yield is 6*(0.2kg)=1.2kg. Therefore, 300 cacao trees would add 300*1.2=360kg. The total yield without cacao trees would be 7500*2=15,000kg. Therefore, with cacao trees, the total yield is 15,000 +360=15,360kg.Yes, that matches.Therefore, the answers are:1. Maximum number of cacao trees: 3002. Total expected yield: 15,360kgBut wait, let me think again about the number of affected coffee plants. Each cacao tree is surrounded by 6 coffee plants, each 2m away. But in a square grid, the cacao tree is at a point where 4 coffee plants are at 2m distance (up, down, left, right), and the other two are at diagonal positions, which are 2.828m away, which is more than 2m. Therefore, in reality, only 4 coffee plants are at exactly 2m distance, and the other two are further away. But the problem states that each cacao tree is surrounded by coffee plants in a hexagonal pattern, with each side of the hexagon being 2 meters. So, perhaps the cacao tree is at the center of a hexagon with side length 2m, meaning that the distance from the cacao tree to each surrounding coffee plant is 2m. But in a square grid, this is not possible because the coffee plants cannot form a regular hexagon around the cacao tree. Therefore, perhaps the problem is assuming that the coffee plants are arranged in a hexagonal grid, but the plot is rectangular. Alternatively, perhaps the cacao tree is surrounded by 6 coffee plants, each 2m away, but in a square grid, this would require the cacao tree to be at a point where 6 coffee plants are at 2m distance, which is not possible. Therefore, perhaps the problem is considering that each cacao tree is surrounded by 6 coffee plants, each 2m away, but in reality, in a square grid, only 4 are at 2m, and the other two are further away. But the problem states that each side of the hexagon is 2m, so the distance from the cacao tree to each surrounding coffee plant is 2m. Therefore, perhaps the problem is assuming that the coffee plants are arranged in a hexagonal grid, but the plot is rectangular. But given the problem statement, I think I need to proceed with the assumption that each cacao tree is surrounded by 6 coffee plants, each 2m away, arranged in a hexagon, even though in a square grid, this is not possible. Therefore, each cacao tree affects 6 coffee plants, and since the cacao trees are spaced 10m apart, their areas of influence do not overlap. Therefore, the total number of improved coffee plants is 300*6=1800, as before.Therefore, the total yield is 15,360kg.But wait, another thought: the cacao trees are placed on the coffee grid, so their positions are at points where coffee plants could be. But since the cacao trees are interspersed, they don't replace coffee plants. Therefore, the total number of coffee plants remains 7500, with 1800 of them having their yield increased by 10%.Therefore, the total yield is 7500*2 + 1800*(0.1*2)=15,000 + 360=15,360kg.Yes, that's correct.Therefore, the answers are:1. 300 cacao trees2. 15,360kg of coffee beans per season</think>

🧠:<think>Okay, so I have this problem about modeling the changes in voter preferences over time in Australian electoral districts. It's split into two parts. Let me tackle them one by one.Starting with part 1: The distribution of voter preferences is initially Gaussian with mean μ₀ and variance σ₀². Over time, the mean changes according to the differential equation dμ/dt = α(μ - μ_c). I need to find μ(t) given that μ(0) = μ₀.Hmm, this looks like a linear differential equation. The standard form for such an equation is dμ/dt + P(t)μ = Q(t). In this case, it's dμ/dt = α(μ - μ_c). Let me rewrite that:dμ/dt = αμ - αμ_cSo, bringing the αμ term to the left:dμ/dt - αμ = -αμ_cThis is a linear ordinary differential equation (ODE). The integrating factor method should work here. The integrating factor, I think, is e^(∫-α dt) which is e^(-αt). Multiply both sides by this:e^(-αt) dμ/dt - α e^(-αt) μ = -α μ_c e^(-αt)The left side is the derivative of (μ e^(-αt)) with respect to t. So, integrating both sides with respect to t:∫ d/dt (μ e^(-αt)) dt = ∫ -α μ_c e^(-αt) dtThis simplifies to:μ e^(-αt) = ∫ -α μ_c e^(-αt) dt + CCompute the integral on the right. The integral of e^(-αt) is (-1/α) e^(-αt), so:μ e^(-αt) = -α μ_c * (-1/α) e^(-αt) + Cμ e^(-αt) = μ_c e^(-αt) + CNow, solve for μ(t):μ(t) = μ_c + C e^(αt)Apply the initial condition μ(0) = μ₀:μ₀ = μ_c + C e^(0) => μ₀ = μ_c + C => C = μ₀ - μ_cSo, substituting back:μ(t) = μ_c + (μ₀ - μ_c) e^(αt)Wait, hold on. That seems a bit odd because if α is positive, then as t increases, the exponential term grows, which might not make sense if μ is supposed to approach μ_c. Maybe I made a mistake in the sign.Let me double-check the integrating factor. The standard form is dμ/dt + P(t)μ = Q(t). Here, it's dμ/dt - αμ = -α μ_c. So P(t) is -α, and the integrating factor is e^(∫-α dt) = e^(-αt). That part is correct.Multiplying through:e^(-αt) dμ/dt - α e^(-αt) μ = -α μ_c e^(-αt)Left side is d/dt [μ e^(-αt)]. So integrating:μ e^(-αt) = ∫ -α μ_c e^(-αt) dt + CCompute the integral:∫ -α μ_c e^(-αt) dt = μ_c ∫ -α e^(-αt) dt = μ_c [e^(-αt)] + CSo,μ e^(-αt) = μ_c e^(-αt) + CThen,μ(t) = μ_c + C e^(αt)Applying μ(0) = μ₀:μ₀ = μ_c + C => C = μ₀ - μ_cSo,μ(t) = μ_c + (μ₀ - μ_c) e^(αt)Wait, but if α is positive, then as t increases, the term (μ₀ - μ_c) e^(αt) will dominate, which would mean μ(t) moves away from μ_c if μ₀ ≠ μ_c. That contradicts the intuition that the mean should approach μ_c over time.Ah, I think I messed up the sign in the differential equation. Let me check the original equation: dμ/dt = α(μ - μ_c). So if μ > μ_c, then dμ/dt is positive, meaning μ increases further away. That doesn't make sense for a mean shifting towards μ_c. Maybe the correct equation should be dμ/dt = -α(μ - μ_c). That way, if μ > μ_c, dμ/dt is negative, pulling μ back towards μ_c.But the problem statement says dμ/dt = α(μ - μ_c). Hmm. Maybe α is negative? Or perhaps I need to reconsider.Wait, perhaps α is a rate constant, which could be positive or negative. If α is positive, then the equation dμ/dt = α(μ - μ_c) implies that μ moves away from μ_c if μ > μ_c, which is not a convergence but divergence. That doesn't make sense for a central tendency.Alternatively, if α is negative, then the equation would effectively be dμ/dt = -|α|(μ - μ_c), which would cause μ to approach μ_c. Maybe the problem assumes α is negative? Or perhaps I misapplied the integrating factor.Wait, no, integrating factor was correct. Let me see the solution again:μ(t) = μ_c + (μ₀ - μ_c) e^(αt)If α is positive, then the term (μ₀ - μ_c) e^(αt) grows, so μ(t) moves away from μ_c. If α is negative, then e^(αt) decays, so μ(t) approaches μ_c. So perhaps in the problem, α is a negative constant, representing a decay rate towards μ_c.Alternatively, maybe the equation should be dμ/dt = -α(μ - μ_c). Let me check the problem statement again.It says: "the mean of the distribution changes according to the differential equation dμ/dt = α(μ - μ_c), where α is a constant rate of change and μ_c is a central or average preference toward which the district tends to shift."So, it says μ tends to shift toward μ_c. So, if μ > μ_c, then dμ/dt should be negative, right? Because it's moving towards μ_c. So, if dμ/dt = α(μ - μ_c), then for μ > μ_c, dμ/dt is positive, which would move μ further away. That contradicts the intended behavior.Therefore, perhaps the correct equation is dμ/dt = -α(μ - μ_c). That way, if μ > μ_c, dμ/dt is negative, pulling μ down towards μ_c.But the problem states it as dμ/dt = α(μ - μ_c). Maybe α is negative? Or perhaps the model is set up differently.Alternatively, maybe the problem is correct, and the solution is as I found, with the mean moving away from μ_c if α is positive. But that contradicts the intended behavior.Wait, perhaps I made a mistake in the integrating factor. Let me go through the steps again.Given dμ/dt = α(μ - μ_c)Rewrite as dμ/dt - α μ = -α μ_cIntegrating factor is e^(∫-α dt) = e^(-α t)Multiply both sides:e^(-α t) dμ/dt - α e^(-α t) μ = -α μ_c e^(-α t)Left side is d/dt [μ e^(-α t)]Integrate both sides:μ e^(-α t) = ∫ -α μ_c e^(-α t) dt + CCompute integral:∫ -α μ_c e^(-α t) dt = μ_c ∫ -α e^(-α t) dt = μ_c [e^(-α t)] + CSo,μ e^(-α t) = μ_c e^(-α t) + CMultiply both sides by e^(α t):μ(t) = μ_c + C e^(α t)Apply μ(0) = μ₀:μ₀ = μ_c + C => C = μ₀ - μ_cThus,μ(t) = μ_c + (μ₀ - μ_c) e^(α t)So, unless α is negative, this would mean that μ(t) moves away from μ_c. But the problem says it tends to shift toward μ_c. Therefore, perhaps α is negative. Let me assume that α is a negative constant, so that e^(α t) decays over time.Alternatively, perhaps the problem intended the equation to be dμ/dt = -α(μ - μ_c), which would make more sense for convergence. But since the problem states it as dμ/dt = α(μ - μ_c), I have to proceed with that.So, the solution is μ(t) = μ_c + (μ₀ - μ_c) e^(α t). If α is positive, this diverges; if α is negative, it converges. Since the problem mentions the district tends to shift toward μ_c, perhaps α is negative. So, I can write the solution as:μ(t) = μ_c + (μ₀ - μ_c) e^(α t)But to make it clear that it converges, perhaps express it with a negative exponent. Alternatively, write it as μ(t) = μ_c + (μ₀ - μ_c) e^{-|α| t}, assuming α is positive but the negative exponent ensures convergence.But since the problem didn't specify, I'll stick with the solution as derived.Moving on to part 2: The polarization index P(t) is defined as the standard deviation of the means μ₁(t), μ₂(t), ..., μ_N(t) across all districts at time t. Each μ_i evolves independently according to the differential equation in part 1. I need to derive an expression for P(t) in terms of the initial means μ₁(0), μ₂(0), ..., μ_N(0), the rate α, and the central preference μ_c.First, recall that the standard deviation is the square root of the variance. So, P(t) = sqrt[ (1/N) Σ (μ_i(t) - μ_avg(t))² ], where μ_avg(t) is the average of the μ_i(t).But since each μ_i(t) is given by μ_c + (μ_i(0) - μ_c) e^(α t), let's compute μ_avg(t):μ_avg(t) = (1/N) Σ μ_i(t) = (1/N) Σ [μ_c + (μ_i(0) - μ_c) e^(α t)]= μ_c + e^(α t) (1/N) Σ (μ_i(0) - μ_c)= μ_c + e^(α t) [ (1/N Σ μ_i(0)) - μ_c ]Let me denote μ_avg(0) = (1/N Σ μ_i(0)). So,μ_avg(t) = μ_c + e^(α t) (μ_avg(0) - μ_c)Now, compute the variance of the μ_i(t):Var(t) = (1/N) Σ [μ_i(t) - μ_avg(t)]²Substitute μ_i(t) and μ_avg(t):= (1/N) Σ [ μ_c + (μ_i(0) - μ_c) e^(α t) - (μ_c + e^(α t)(μ_avg(0) - μ_c)) ]²Simplify inside the brackets:= (1/N) Σ [ (μ_i(0) - μ_c) e^(α t) - e^(α t)(μ_avg(0) - μ_c) ]²Factor out e^(α t):= (1/N) Σ [ e^(α t) (μ_i(0) - μ_c - μ_avg(0) + μ_c) ]²Simplify inside:= (1/N) Σ [ e^(α t) (μ_i(0) - μ_avg(0)) ]²= e^(2α t) (1/N) Σ (μ_i(0) - μ_avg(0))²Notice that (1/N) Σ (μ_i(0) - μ_avg(0))² is the initial variance of the means, let's call it Var(0). So,Var(t) = e^(2α t) Var(0)Therefore, the standard deviation P(t) is sqrt(Var(t)) = e^(α t) sqrt(Var(0)).But sqrt(Var(0)) is the initial polarization index P(0). So,P(t) = P(0) e^(α t)But wait, let me express it in terms of the initial means without referring to P(0). Since Var(0) = (1/N) Σ (μ_i(0) - μ_avg(0))², then sqrt(Var(0)) is P(0). So,P(t) = P(0) e^(α t)Alternatively, expressing it directly:P(t) = e^(α t) * sqrt( (1/N) Σ (μ_i(0) - μ_avg(0))² )But since μ_avg(0) is (1/N Σ μ_i(0)), we can write:P(t) = e^(α t) * sqrt( (1/N) Σ (μ_i(0) - (1/N Σ μ_j(0)) )² )So, that's the expression for P(t).But let me check if this makes sense. If α is positive, then P(t) grows exponentially, meaning polarization increases. If α is negative, P(t) decays, meaning polarization decreases. Given that in part 1, if α is positive, μ(t) moves away from μ_c, leading to more spread out means, hence higher polarization. Conversely, if α is negative, means converge towards μ_c, reducing polarization. So this aligns with the result.Therefore, the polarization index P(t) is the initial polarization multiplied by e^(α t).So, summarizing:1. μ(t) = μ_c + (μ₀ - μ_c) e^(α t)2. P(t) = P(0) e^(α t), where P(0) is the initial standard deviation of the means.But to express P(t) in terms of the initial means without referring to P(0), it's:P(t) = e^(α t) * sqrt( (1/N) Σ_{i=1}^N (μ_i(0) - (1/N Σ_{j=1}^N μ_j(0)) )² )Alternatively, factor out the e^(α t):P(t) = e^(α t) * sqrt( (1/N) Σ (μ_i(0) - μ_avg(0))² )So, that's the expression.</think>

🧠:<think>Okay, so I have this problem about a fantasy tennis league, and I need to figure out two things. First, I need to calculate the probability that player 1 beats player 2 given their skill ratings. Second, I have to simulate the expected number of wins for each player in a round-robin tournament with four players. Let me take it step by step.Starting with the first part: calculating the probability ( P_{12} ) that player 1 beats player 2. The formula given is the logistic function:[P_{ij} = frac{1}{1 + e^{-(s_i - s_j)}}]So, for player 1 and player 2, their skill ratings are ( s_1 = 950 ) and ( s_2 = 850 ). Plugging these into the formula, I get:[P_{12} = frac{1}{1 + e^{-(950 - 850)}}]Simplifying the exponent:[950 - 850 = 100]So now the equation becomes:[P_{12} = frac{1}{1 + e^{-100}}]Hmm, ( e^{-100} ) is a very small number because the exponent is negative and large in magnitude. I remember that ( e^{-x} ) approaches zero as ( x ) becomes large. So, ( e^{-100} ) is practically zero. Therefore, the denominator is approximately ( 1 + 0 = 1 ), which means ( P_{12} ) is approximately 1. But wait, let me verify that. Maybe I should compute ( e^{-100} ) more precisely. I know that ( e^{-100} ) is equal to ( 1 / e^{100} ). Calculating ( e^{100} ) is a massive number. Let me see, ( e ) is approximately 2.71828, so ( e^{100} ) is about ( 2.71828^{100} ). That's way beyond any calculator I have, but I can recall that ( e^{100} ) is approximately ( 2.688117 times 10^{43} ). So, ( e^{-100} ) is roughly ( 3.72 times 10^{-44} ). That's an extremely small number, almost zero.Therefore, ( P_{12} = 1 / (1 + 3.72 times 10^{-44}) approx 1 ). So, the probability that player 1 beats player 2 is almost 100%. That makes sense because player 1 has a much higher skill rating than player 2.Moving on to the second part: calculating the expected number of wins ( E[W_i] ) for each player in a round-robin tournament with four players. The players are ranked with skill ratings ( s_1 = 950 ), ( s_2 = 850 ), ( s_3 = 700 ), and ( s_4 = 600 ).In a round-robin tournament, each player plays every other player exactly once. So, each player will have 3 matches. The expected number of wins for each player is the sum of the probabilities that they win each of their matches.So, for each player ( i ), ( E[W_i] = sum_{j neq i} P_{ij} ).Therefore, I need to calculate the probability of each player beating each of the other three players and sum those probabilities.Let me list out all the players:- Player 1: ( s_1 = 950 )- Player 2: ( s_2 = 850 )- Player 3: ( s_3 = 700 )- Player 4: ( s_4 = 600 )So, for each player, I need to compute their probability of winning against the other three.Starting with Player 1:Player 1 plays against Player 2, Player 3, and Player 4.Compute ( P_{12} ), ( P_{13} ), and ( P_{14} ).We already calculated ( P_{12} ) earlier, which is approximately 1. But let me compute it more accurately.Using the formula:[P_{12} = frac{1}{1 + e^{-(950 - 850)}} = frac{1}{1 + e^{-100}} approx 1]Similarly, ( P_{13} ):[P_{13} = frac{1}{1 + e^{-(950 - 700)}} = frac{1}{1 + e^{-250}}]Again, ( e^{-250} ) is an extremely small number, so ( P_{13} approx 1 ).Same with ( P_{14} ):[P_{14} = frac{1}{1 + e^{-(950 - 600)}} = frac{1}{1 + e^{-350}} approx 1]So, Player 1 is expected to win all three of their matches, so ( E[W_1] = 3 ).Moving on to Player 2:Player 2 plays against Player 1, Player 3, and Player 4.Compute ( P_{21} ), ( P_{23} ), and ( P_{24} ).First, ( P_{21} ):[P_{21} = frac{1}{1 + e^{-(850 - 950)}} = frac{1}{1 + e^{-100}} approx 1 / (1 + 3.72 times 10^{-44}) approx 1]Wait, that can't be right. Wait, no. Wait, ( P_{21} ) is the probability that Player 2 beats Player 1. Since Player 1 has a higher skill rating, this probability should be very low, not almost 1.Wait, let me recast the formula correctly. The formula is:[P_{ij} = frac{1}{1 + e^{-(s_i - s_j)}}]So, for ( P_{21} ), it's Player 2 vs Player 1, so ( s_i = 850 ), ( s_j = 950 ). Therefore:[P_{21} = frac{1}{1 + e^{-(850 - 950)}} = frac{1}{1 + e^{-(-100)}} = frac{1}{1 + e^{100}}]Ah, that's different. So, ( e^{100} ) is a huge number, so ( P_{21} ) is approximately ( 1 / (1 + e^{100}) approx 0 ). So, Player 2 has almost no chance of beating Player 1.Similarly, ( P_{23} ):[P_{23} = frac{1}{1 + e^{-(850 - 700)}} = frac{1}{1 + e^{-150}} approx 1]Because ( e^{-150} ) is very small.And ( P_{24} ):[P_{24} = frac{1}{1 + e^{-(850 - 600)}} = frac{1}{1 + e^{-250}} approx 1]So, Player 2 is expected to lose to Player 1 but win against Players 3 and 4. Therefore, ( E[W_2] = 0 + 1 + 1 = 2 ).Wait, but let me double-check the calculation for ( P_{21} ). It's 1 / (1 + e^{100}), which is a very small number, effectively 0. So, Player 2 has almost 0 probability of beating Player 1. So, their expected wins are 2.Now, moving on to Player 3:Player 3 plays against Player 1, Player 2, and Player 4.Compute ( P_{31} ), ( P_{32} ), and ( P_{34} ).First, ( P_{31} ):[P_{31} = frac{1}{1 + e^{-(700 - 950)}} = frac{1}{1 + e^{-(-250)}} = frac{1}{1 + e^{250}} approx 0]So, Player 3 has almost no chance against Player 1.Next, ( P_{32} ):[P_{32} = frac{1}{1 + e^{-(700 - 850)}} = frac{1}{1 + e^{-(-150)}} = frac{1}{1 + e^{150}} approx 0]Again, Player 3 has almost no chance against Player 2.Finally, ( P_{34} ):[P_{34} = frac{1}{1 + e^{-(700 - 600)}} = frac{1}{1 + e^{-100}} approx 1]So, Player 3 is expected to lose to Players 1 and 2 but win against Player 4. Therefore, ( E[W_3] = 0 + 0 + 1 = 1 ).Lastly, Player 4:Player 4 plays against Player 1, Player 2, and Player 3.Compute ( P_{41} ), ( P_{42} ), and ( P_{43} ).First, ( P_{41} ):[P_{41} = frac{1}{1 + e^{-(600 - 950)}} = frac{1}{1 + e^{-(-350)}} = frac{1}{1 + e^{350}} approx 0]Player 4 has almost no chance against Player 1.Next, ( P_{42} ):[P_{42} = frac{1}{1 + e^{-(600 - 850)}} = frac{1}{1 + e^{-(-250)}} = frac{1}{1 + e^{250}} approx 0]Similarly, Player 4 has almost no chance against Player 2.Finally, ( P_{43} ):[P_{43} = frac{1}{1 + e^{-(600 - 700)}} = frac{1}{1 + e^{-(-100)}} = frac{1}{1 + e^{100}} approx 0]So, Player 4 is expected to lose all three matches. Therefore, ( E[W_4] = 0 + 0 + 0 = 0 ).Wait, but hold on. Let me make sure I didn't make a mistake here. For Player 4, when calculating ( P_{43} ), it's the probability that Player 4 beats Player 3. Since Player 3 has a higher skill rating (700 vs 600), the probability should be low, which is what I got. So, yes, Player 4 is expected to lose all matches.But let me think again about the probabilities. Is it possible that the expected number of wins for Player 2 is 2, Player 3 is 1, and Player 4 is 0? That seems consistent with the skill ratings, as Player 1 is the strongest, followed by Player 2, then Player 3, then Player 4.But just to be thorough, let me compute the exact probabilities for some of these to see if they're indeed negligible or if they contribute a tiny bit.For example, let's compute ( P_{21} ):[P_{21} = frac{1}{1 + e^{100}} approx frac{1}{1 + 2.688117 times 10^{43}} approx 3.72 times 10^{-44}]That's an extremely small number, practically zero.Similarly, ( P_{31} = frac{1}{1 + e^{250}} approx 1 / (1 + e^{250}) ). ( e^{250} ) is an astronomically large number, so this is practically zero.Same with ( P_{32} = frac{1}{1 + e^{150}} approx 0 ), and ( P_{41} = frac{1}{1 + e^{350}} approx 0 ), ( P_{42} = frac{1}{1 + e^{250}} approx 0 ), ( P_{43} = frac{1}{1 + e^{100}} approx 0 ).So, indeed, all these probabilities are effectively zero. Therefore, the expected number of wins are as I calculated.So, summarizing:- Player 1: 3 wins- Player 2: 2 wins- Player 3: 1 win- Player 4: 0 winsTherefore, the expected number of wins for each player is 3, 2, 1, and 0 respectively.But wait, let me think again. Is it possible that in reality, the probabilities aren't exactly zero, so the expected wins might be slightly more than 2 for Player 2, or slightly less than 3 for Player 1? But given the massive skill differences, the probabilities are so small that they don't significantly affect the expectation. For example, even if ( P_{21} ) is ( 3.72 times 10^{-44} ), multiplying that by 1 (the number of matches) gives an expected win of ( 3.72 times 10^{-44} ), which is negligible. So, for all practical purposes, the expected number of wins is 3, 2, 1, 0.Therefore, my final answers are:1. The probability ( P_{12} ) is approximately 1.2. The expected number of wins for each player is 3, 2, 1, and 0 for Players 1, 2, 3, and 4 respectively.But wait, let me make sure I didn't misinterpret the problem. The question says "the league has 4 players with skill ratings ( s_1 = 950 ), ( s_2 = 850 ), ( s_3 = 700 ), and ( s_4 = 600 )". So, the order is 1, 2, 3, 4 with decreasing skill. So, my calculations are correct.Just to be thorough, let me compute one of the non-negligible probabilities, like ( P_{12} ). Wait, earlier I thought it was approximately 1, but let's compute it more precisely.Given ( s_1 = 950 ), ( s_2 = 850 ):[P_{12} = frac{1}{1 + e^{-(950 - 850)}} = frac{1}{1 + e^{-100}} approx frac{1}{1 + 3.72 times 10^{-44}} approx 1 - 3.72 times 10^{-44}]So, it's practically 1, but just slightly less. However, for the purposes of expected wins, since each match is independent, and the expectation is linear, we can sum these probabilities. But since the other probabilities are so small, they don't affect the expectation significantly.Therefore, my conclusion remains the same.Final Answer1. The probability ( P_{12} ) is boxed{1}.2. The expected number of wins ( E[W_i] ) for each player is: - Player 1: boxed{3} - Player 2: boxed{2} - Player 3: boxed{1} - Player 4: boxed{0}</think>