🧠:<think>Okay, so I need to figure out the parametric equations for a horse's canter pirouette. Let me start by understanding what a canter pirouette is. From what I remember, it's a dressage movement where the horse makes a circular motion with a specific radius and maintains a uniform angular velocity. The horse starts at a point (0, r) and completes the circle, ending up back at the same point after time T.First, I need to derive the parametric equations for the horse's position without any deviations. Since it's a circular motion, I can use the standard parametric equations for a circle. In Cartesian coordinates, a circle with radius r can be described using sine and cosine functions.The general parametric equations for a circle are:x(t) = r * cos(theta(t))y(t) = r * sin(theta(t))But here, the horse starts at (0, r), which is the top of the circle. In standard position, that corresponds to an angle of 90 degrees or π/2 radians. So, I need to adjust the parametric equations to start at this point.Also, the horse completes one full rotation in time T. That means the angular velocity ω is 2π / T radians per second. So, theta(t) should be a function that starts at π/2 and increases linearly with time, completing 2π radians by time T.Let me write that out. The angle theta(t) as a function of time is:theta(t) = (2π / T) * t + π/2Wait, let me check. At t = 0, theta(0) should be π/2, which is correct. At t = T, theta(T) = (2π / T)*T + π/2 = 2π + π/2. But that's 5π/2, which is more than 2π. Hmm, that's actually a full rotation plus an extra π/2. That would mean the horse ends up at the same point, but rotated an extra π/2. Wait, no, because the circle is periodic with period 2π, so 5π/2 is equivalent to π/2. So, yes, the horse would end up back at (0, r). So that works.So, plugging theta(t) into the parametric equations:x(t) = r * cos(theta(t)) = r * cos((2π / T) * t + π/2)y(t) = r * sin(theta(t)) = r * sin((2π / T) * t + π/2)Alternatively, I can use trigonometric identities to simplify these expressions. Remember that cos(a + b) = cos(a)cos(b) - sin(a)sin(b) and sin(a + b) = sin(a)cos(b) + cos(a)sin(b).Let me apply that to x(t):x(t) = r * [cos((2π / T) * t) * cos(π/2) - sin((2π / T) * t) * sin(π/2)]But cos(π/2) is 0 and sin(π/2) is 1, so:x(t) = r * [0 - sin((2π / T) * t)] = -r * sin((2π / T) * t)Similarly, for y(t):y(t) = r * [sin((2π / T) * t) * cos(π/2) + cos((2π / T) * t) * sin(π/2)]Again, cos(π/2) is 0 and sin(π/2) is 1, so:y(t) = r * [0 + cos((2π / T) * t)] = r * cos((2π / T) * t)So, the parametric equations simplify to:x(t) = -r * sin((2π / T) * t)y(t) = r * cos((2π / T) * t)Let me verify this. At t = 0, x(0) = -r * sin(0) = 0, y(0) = r * cos(0) = r. That's correct, starting at (0, r). At t = T/4, x(T/4) = -r * sin(π/2) = -r, y(T/4) = r * cos(π/2) = 0. So, the horse is at (-r, 0), which is the leftmost point. At t = T/2, x(T/2) = -r * sin(π) = 0, y(T/2) = r * cos(π) = -r. So, the horse is at (0, -r), the bottom. At t = 3T/4, x(3T/4) = -r * sin(3π/2) = r, y(3T/4) = r * cos(3π/2) = 0. So, the horse is at (r, 0), the rightmost point. Finally, at t = T, x(T) = -r * sin(2π) = 0, y(T) = r * cos(2π) = r. So, back to the starting point. Perfect.So, part 1 is done. The parametric equations are:x(t) = -r * sin((2π / T) * t)y(t) = r * cos((2π / T) * t)Now, moving on to part 2. We need to introduce a slight oscillatory deviation in the radial direction. The deviation is given by δr(t) = A sin(2π t / T). So, this is a radial oscillation with amplitude A and frequency 2π / T, which is the same as the angular velocity. So, the radius isn't constant anymore; it's oscillating as the horse moves around the circle.So, the new radius at any time t is r(t) = r + δr(t) = r + A sin(2π t / T). Therefore, the parametric equations need to be adjusted to use this time-varying radius.So, the position equations become:x(t) = -r(t) * sin((2π / T) * t)y(t) = r(t) * cos((2π / T) * t)Substituting r(t):x(t) = -(r + A sin(2π t / T)) * sin((2π / T) * t)y(t) = (r + A sin(2π t / T)) * cos((2π / T) * t)Alternatively, we can expand these expressions if needed, but I think this form is acceptable. However, let me see if I can simplify them further.Let's expand x(t):x(t) = -r sin((2π / T) t) - A sin(2π t / T) sin((2π / T) t)Similarly, y(t):y(t) = r cos((2π / T) t) + A sin(2π t / T) cos((2π / T) t)Now, using trigonometric identities, sin(a) sin(b) = [cos(a - b) - cos(a + b)] / 2 and sin(a) cos(b) = [sin(a + b) + sin(a - b)] / 2.Applying these to x(t):First term: -r sin((2π / T) t) remains as is.Second term: -A sin(2π t / T) sin((2π / T) t) = -A [cos(0) - cos(4π t / T)] / 2 = -A [1 - cos(4π t / T)] / 2Similarly, for y(t):First term: r cos((2π / T) t) remains as is.Second term: A sin(2π t / T) cos((2π / T) t) = A [sin(4π t / T) + sin(0)] / 2 = A sin(4π t / T) / 2So, putting it all together:x(t) = -r sin((2π / T) t) - (A/2) [1 - cos(4π t / T)]y(t) = r cos((2π / T) t) + (A/2) sin(4π t / T)Simplify x(t):x(t) = -r sin((2π / T) t) - A/2 + (A/2) cos(4π t / T)Similarly, y(t) remains:y(t) = r cos((2π / T) t) + (A/2) sin(4π t / T)So, these are the refined parametric equations with the radial oscillation.Let me check if this makes sense. At t = 0:x(0) = -r sin(0) - A/2 + (A/2) cos(0) = 0 - A/2 + A/2 = 0y(0) = r cos(0) + (A/2) sin(0) = r + 0 = rGood, starts at (0, r). At t = T/4:x(T/4) = -r sin(π/2) - A/2 + (A/2) cos(π) = -r - A/2 - A/2 = -r - Ay(T/4) = r cos(π/2) + (A/2) sin(π) = 0 + 0 = 0So, the horse is at (-r - A, 0). That makes sense because the radial deviation is maximum in the negative x-direction at this point.Similarly, at t = T/2:x(T/2) = -r sin(π) - A/2 + (A/2) cos(2π) = 0 - A/2 + A/2 = 0y(T/2) = r cos(π) + (A/2) sin(2π) = -r + 0 = -rSo, at the bottom point, the radial deviation cancels out in the x-direction, and the y-position is -r, which is correct.At t = 3T/4:x(3T/4) = -r sin(3π/2) - A/2 + (A/2) cos(3π) = r - A/2 - A/2 = r - Ay(3T/4) = r cos(3π/2) + (A/2) sin(3π) = 0 + 0 = 0So, the horse is at (r - A, 0). That makes sense because the radial deviation is maximum in the positive x-direction here.Finally, at t = T:x(T) = -r sin(2π) - A/2 + (A/2) cos(4π) = 0 - A/2 + A/2 = 0y(T) = r cos(2π) + (A/2) sin(4π) = r + 0 = rSo, back to the starting point. Perfect.Therefore, the parametric equations with the radial oscillation are:x(t) = -r sin((2π / T) t) - A/2 + (A/2) cos(4π t / T)y(t) = r cos((2π / T) t) + (A/2) sin(4π t / T)Alternatively, we can write them as:x(t) = -r sin(ω t) - A/2 + (A/2) cos(2ω t)y(t) = r cos(ω t) + (A/2) sin(2ω t)Where ω = 2π / T.But the problem didn't specify to express it in terms of ω, so I think the first form is fine.So, to summarize:1. The original parametric equations without deviation are:x(t) = -r sin((2π / T) t)y(t) = r cos((2π / T) t)2. With the radial oscillation, the equations become:x(t) = -r sin((2π / T) t) - A/2 + (A/2) cos(4π t / T)y(t) = r cos((2π / T) t) + (A/2) sin(4π t / T)I think that's the solution.</think>

🧠:<think>Alright, so I'm trying to help this nonprofit director address housing inequality in an urban area. There are two sub-problems here, and I need to figure out how to approach each one. Let me start with Sub-problem 1.Sub-problem 1: Minimizing Inequality with Constant Total Housing UnitsOkay, the director wants to allocate housing units in such a way that the inequality measure ( I ) is minimized. The inequality measure is defined as the square root of the average squared deviation from the mean, which sounds a lot like the standard deviation. So, ( I ) is essentially the standard deviation of the housing units across neighborhoods.The formula given is:[ I = sqrt{frac{1}{n} sum_{i=1}^{n} (H_i - bar{H})^2} ]where ( bar{H} ) is the average number of housing units.The constraint here is that the total number of housing units must remain constant. So, the sum of all ( H_i ) stays the same. Let me denote the total as ( T ), so ( T = sum_{i=1}^{n} H_i ). Therefore, ( bar{H} = frac{T}{n} ).Since ( I ) is the standard deviation, and we know that the standard deviation is minimized when all the data points are equal. That is, when all ( H_i ) are equal to ( bar{H} ). So, if all neighborhoods have the same number of housing units, the inequality measure ( I ) would be zero, which is the minimum possible.But wait, the problem says "allocate funding to reduce inequality by improving the housing conditions." So, does that mean they can only add housing units, or can they also redistribute existing units? The constraint is that the total number must remain constant, so I think it's about redistributing, not adding or removing.Therefore, to minimize ( I ), the optimal allocation is to make all ( H_i ) equal. So, each neighborhood should have ( bar{H} = frac{T}{n} ) housing units. That way, the variance is zero, and so is the inequality measure.Let me verify this. If all ( H_i ) are equal, then each ( H_i - bar{H} = 0 ), so the sum of squares is zero, and hence ( I = 0 ). That makes sense. So, the minimal inequality is achieved when all neighborhoods have the same number of housing units.Therefore, the optimal allocation is to redistribute the housing units so that each neighborhood has exactly ( bar{H} ) units. Since the total is fixed, this is the only way to minimize the standard deviation.Sub-problem 2: Maximizing Reduction in Inequality with a Fixed BudgetNow, moving on to Sub-problem 2. Here, the director has a fixed budget ( B ) and each improvement of a housing unit costs ( C ). So, the total number of housing units that can be improved is ( frac{B}{C} ). Let me denote this as ( k = frac{B}{C} ). So, we can improve ( k ) housing units in total.The goal is to determine the maximum reduction in ( I ) that can be achieved with this budget. So, we need to figure out how to allocate these ( k ) improvements across the neighborhoods to reduce the inequality measure as much as possible.First, let's recall that ( I ) is the standard deviation, which is sensitive to how spread out the housing units are. To reduce ( I ), we need to make the distribution of ( H_i ) more equal. That is, we should focus on increasing the number of housing units in neighborhoods with fewer units, thereby reducing the variance.So, the strategy here is likely to allocate the improvements to the neighborhoods with the least number of housing units. By doing so, we can bring up the lower values of ( H_i ) closer to the mean, which should decrease the overall variance.But let me think more formally. Let's denote the initial number of housing units as ( H_1, H_2, ldots, H_n ), with ( sum H_i = T ). The initial mean is ( bar{H} = frac{T}{n} ). The initial inequality measure is ( I_{text{initial}} = sqrt{frac{1}{n} sum (H_i - bar{H})^2} ).After allocating ( k ) improvements, each costing ( C ), we have a new set of housing units ( H'_1, H'_2, ldots, H'_n ), where ( sum H'_i = T + k ) (since each improvement adds one unit, I assume). Wait, hold on. Is each improvement adding a unit or just improving an existing one? The problem says "improving one housing unit in any neighborhood costs ( C ) units of currency." So, does that mean each improvement increases the number of housing units by one? Or does it mean that each unit is being improved, but the number of units remains the same?This is a crucial point. If improving a housing unit doesn't change the number of units, but just their quality, then the total number of housing units remains ( T ). However, if improving a unit means adding a new unit, then the total increases by ( k ).Looking back at the problem statement: "improving one housing unit in any neighborhood costs ( C ) units of currency." It doesn't specify whether this increases the number of units or just improves the existing ones. Since the first sub-problem was about allocation with a fixed total, this one might be about improving existing units without changing the total. But the wording is a bit ambiguous.Wait, the first sub-problem was about allocation, so maybe the second is about improving, which could mean upgrading existing units, not adding new ones. So, perhaps the total number of housing units remains ( T ), but their quality is improved. However, the inequality measure ( I ) is based on the number of housing units, not their quality. So, if we can only improve the quality without changing the count, then ( I ) remains the same. That can't be right because the problem asks for a reduction in ( I ).Alternatively, maybe improving a housing unit allows us to convert it into more units, effectively increasing the number. So, each improvement adds a unit. So, the total number becomes ( T + k ).But this is unclear. Let me check the problem statement again."improving one housing unit in any neighborhood costs ( C ) units of currency."Hmm. It doesn't specify whether it's adding a unit or just improving an existing one. Since the inequality measure is based on the number of units, if we can only improve the quality without changing the count, then ( I ) can't be reduced. Therefore, it must be that improving a unit allows us to increase the number of units, i.e., each improvement adds a new unit.Therefore, the total number of housing units becomes ( T + k ), and the new mean is ( bar{H}' = frac{T + k}{n} ).But wait, if we can only add units, then the optimal way to reduce inequality is to add as many units as possible to the neighborhoods with the least number of units. Because adding units to the lower ones will bring them closer to the new mean, thereby reducing the variance.Alternatively, if we can only improve existing units without changing the count, then we can't affect ( I ). So, I think the correct interpretation is that each improvement adds a unit, so the total increases.Therefore, with a budget ( B ), we can add ( k = frac{B}{C} ) units. Our goal is to distribute these ( k ) units across the neighborhoods to minimize the new inequality measure ( I' ).To minimize ( I' ), we should distribute the additional units in a way that equalizes the neighborhoods as much as possible. That is, we should add units to the neighborhoods with the fewest housing units until all are as equal as possible.This is similar to the concept of the "minimum variance" allocation. The minimal variance occurs when all ( H'_i ) are as equal as possible. So, we need to distribute ( k ) units to the neighborhoods to make their counts as equal as possible.Let me formalize this. Let the initial counts be ( H_1, H_2, ldots, H_n ), sorted in non-decreasing order. So, ( H_1 leq H_2 leq ldots leq H_n ).We have ( k ) units to distribute. To minimize the variance, we should add as many units as possible to the neighborhoods with the lowest ( H_i ).The process would be:1. Calculate how many units each neighborhood needs to reach the current maximum or a more equal distribution.But actually, since we can choose where to add, the optimal is to make all neighborhoods as equal as possible. So, the minimal variance occurs when all ( H'_i ) are either ( lfloor frac{T + k}{n} rfloor ) or ( lceil frac{T + k}{n} rceil ).Therefore, the minimal possible variance is achieved when the housing units are as equal as possible after adding ( k ) units.But wait, the initial distribution might not be equal, so adding units to the lower ones will bring them up.Let me think step by step.Suppose we have neighborhoods ordered by ( H_1 leq H_2 leq ldots leq H_n ).We need to add ( k ) units. To minimize the variance, we should add as many as possible to the neighborhoods with the lowest ( H_i ).So, the strategy is:1. Start with the neighborhood with the smallest ( H_i ). Add units until it reaches the next neighborhood's ( H_i ).2. Then, move to the next neighborhood and repeat until all neighborhoods are as equal as possible or we run out of units.This is similar to the concept of "filling up" the lower neighborhoods to match the higher ones.But let's make it more precise.Let me denote the target as making all neighborhoods have the same number of units, which would be ( bar{H}' = frac{T + k}{n} ).If ( T + k ) is divisible by ( n ), then each neighborhood can have exactly ( bar{H}' ), and the variance would be zero. However, if not, some neighborhoods will have ( lfloor bar{H}' rfloor ) and others ( lceil bar{H}' rceil ).But in reality, the initial distribution might not allow us to reach this perfect equality because we can only add units, not remove them. So, we can only increase the number of units in some neighborhoods, but not decrease them.Therefore, the minimal variance is achieved when we add units to the neighborhoods with the lowest ( H_i ) until we can't add any more without exceeding the budget.Wait, but actually, the minimal variance is achieved when the distribution is as equal as possible, regardless of the order. So, perhaps the optimal way is to calculate how much each neighborhood needs to reach the new mean, and add units accordingly.But since we can only add units, not remove them, we have to ensure that we don't take away from any neighborhood.So, the minimal variance occurs when we add units to the neighborhoods with the lowest ( H_i ) until the distribution is as equal as possible.Let me try to model this.Let the initial counts be ( H_1 leq H_2 leq ldots leq H_n ).We need to add ( k ) units. Let me denote the new counts as ( H'_1, H'_2, ldots, H'_n ), where ( H'_i geq H_i ) for all ( i ), and ( sum H'_i = T + k ).To minimize the variance, we need to make the ( H'_i ) as equal as possible.The minimal variance occurs when the ( H'_i ) are as close as possible to each other. So, we need to distribute the ( k ) units to the neighborhoods starting from the smallest.Let me define the target as ( m = frac{T + k}{n} ). If ( m ) is an integer, then each neighborhood should have ( m ) units. If not, some will have ( lfloor m rfloor ) and some ( lceil m rceil ).But since we can only add units, we need to see how much we need to add to each neighborhood to reach at least ( m ).Wait, no. Actually, we can only add units, so we need to ensure that each neighborhood is at least as large as the target. But if the target is higher than some neighborhoods, we have to add units to those.But actually, the target is the new mean, so we need to make sure that the sum is ( T + k ). So, the minimal variance is achieved when the distribution is as equal as possible, which is when each neighborhood is either ( lfloor m rfloor ) or ( lceil m rceil ).But since we can only add units, we have to add units to the neighborhoods with the smallest ( H_i ) until we reach the desired distribution.Let me think of an example.Suppose we have 3 neighborhoods with ( H = [1, 2, 3] ), so ( T = 6 ). Suppose ( k = 3 ), so ( T + k = 9 ), and ( m = 3 ). So, we need to make each neighborhood have 3 units. So, we add 2 units to the first neighborhood and 1 unit to the second. The new distribution is [3, 3, 3], which has zero variance.Another example: ( H = [1, 1, 4] ), ( T = 6 ), ( k = 3 ), so ( m = 3 ). We need to add 2 units to the first neighborhood and 1 unit to the second. New distribution: [3, 2, 4]. Wait, that's not equal. Wait, no, we can only add to the first two neighborhoods. So, adding 2 to the first and 1 to the second gives [3, 2, 4], which has a variance. Alternatively, maybe we should add differently.Wait, actually, in this case, the target is 3 for each, so we need to add 2 to the first, 2 to the second, but we only have 3 units. So, we can add 2 to the first and 1 to the second, making [3, 2, 4]. But that's not equal. Alternatively, add 1 to the first and 2 to the second, making [2, 3, 4]. That also has a variance.Wait, so in this case, we can't make all equal because we only have 3 units to add, and the third neighborhood is already at 4, which is above the target. So, we can't reduce it, so the minimal variance is achieved by making the first two as high as possible without exceeding the budget.Wait, but actually, the minimal variance occurs when the distribution is as equal as possible. So, in this case, we can make the first two neighborhoods as high as possible, but since the third is already above the target, we can't do much about it.So, perhaps the minimal variance is achieved by adding as much as possible to the lowest neighborhoods until we can't add any more.Therefore, the general approach is:1. Sort the neighborhoods in ascending order of ( H_i ).2. Starting from the smallest, add as many units as possible to each until we either reach the target mean or run out of budget.3. The target mean is ( m = frac{T + k}{n} ).4. For each neighborhood, calculate how much it needs to reach ( m ). If ( H_i < m ), we need to add ( m - H_i ) units. If ( H_i geq m ), we don't add anything.5. Sum up all the required additions for neighborhoods below ( m ). If the total required is less than or equal to ( k ), then we can make all neighborhoods at least ( m ), and the variance will be minimized.6. If the total required is more than ( k ), then we have to distribute the ( k ) units as much as possible to the neighborhoods starting from the smallest.Wait, but in reality, ( m ) might not be an integer, and we might not be able to make all neighborhoods exactly ( m ). So, we have to distribute the ( k ) units to make the neighborhoods as equal as possible.Let me formalize this.Let ( m = frac{T + k}{n} ).For each neighborhood ( i ), the required addition is ( r_i = max(0, m - H_i) ).But since ( m ) might not be an integer, and we can only add whole units, we need to adjust.Alternatively, we can think of it as:The minimal variance is achieved when the distribution is as equal as possible, which is when the neighborhoods are either ( lfloor m rfloor ) or ( lceil m rceil ).But since we can only add units, we have to ensure that we don't reduce any neighborhood's count.So, the process is:1. Calculate ( m = frac{T + k}{n} ).2. For each neighborhood, if ( H_i < lfloor m rfloor ), we need to add ( lfloor m rfloor - H_i ) units.3. If ( H_i ) is between ( lfloor m rfloor ) and ( lceil m rceil ), we might need to add 0 or 1 units depending on the distribution.But this is getting complicated. Maybe a better approach is to consider that the minimal variance is achieved when the neighborhoods are as equal as possible, which is when they are either ( lfloor m rfloor ) or ( lceil m rceil ).Therefore, the number of neighborhoods that will have ( lceil m rceil ) units is ( (T + k) mod n ), and the rest will have ( lfloor m rfloor ).But since we can only add units, we have to make sure that we don't reduce any neighborhood's count. So, we need to add units to the neighborhoods with the lowest ( H_i ) until they reach at least ( lfloor m rfloor ), and then distribute the remaining units to reach ( lceil m rceil ).Wait, perhaps it's better to think in terms of how much we need to add to each neighborhood to reach the target.Let me denote ( m = frac{T + k}{n} ).For each neighborhood ( i ), the deficit is ( d_i = m - H_i ). If ( d_i > 0 ), we need to add ( d_i ) units. If ( d_i leq 0 ), we don't add anything.But since ( m ) might not be an integer, we have to adjust. So, we can calculate the total deficit as ( sum_{i=1}^{n} max(0, m - H_i) ). If this total deficit is less than or equal to ( k ), then we can cover all deficits and have some leftover units to distribute. If it's more than ( k ), we can only cover part of the deficits.Wait, but actually, the total deficit should be exactly ( k ), because ( T + k = n times m ). So, ( sum_{i=1}^{n} (H'_i - H_i) = k ), where ( H'_i geq H_i ).Therefore, the minimal variance occurs when the ( H'_i ) are as equal as possible, which is when they are either ( lfloor m rfloor ) or ( lceil m rceil ).Therefore, the maximum reduction in ( I ) is achieved by making the distribution as equal as possible, which is when the neighborhoods are either ( lfloor m rfloor ) or ( lceil m rceil ).But to calculate the exact reduction, we need to compute the initial variance and the new variance after adding the ( k ) units optimally.However, since the problem asks for the maximum reduction in ( I ), not the exact value, perhaps we can express it in terms of the initial variance and the potential reduction.But maybe a better approach is to realize that the minimal possible ( I ) after adding ( k ) units is when the distribution is as equal as possible, so the new ( I ) is the standard deviation of a distribution where each ( H'_i ) is either ( lfloor m rfloor ) or ( lceil m rceil ).Therefore, the maximum reduction in ( I ) is the difference between the initial ( I ) and this minimal possible ( I' ).But to express this mathematically, we need to compute both ( I_{text{initial}} ) and ( I' ).However, without knowing the exact initial distribution, it's hard to give a numerical answer. But perhaps we can express the maximum reduction in terms of the initial variance and the potential to equalize the distribution.Alternatively, perhaps the maximum reduction is achieved by making the distribution as equal as possible, which would be when all ( H'_i ) are equal, if possible. If ( T + k ) is divisible by ( n ), then ( I' = 0 ), so the maximum reduction is ( I_{text{initial}} ). If not, ( I' ) is minimal but not zero.But since the problem doesn't specify the initial distribution, I think the answer should be in terms of the initial variance and the potential to equalize.Wait, but maybe we can express the maximum possible reduction in ( I ) as the difference between the initial standard deviation and the minimal possible standard deviation after adding ( k ) units.But without knowing the initial distribution, we can't compute the exact value. Therefore, perhaps the answer is that the maximum reduction is achieved by equalizing the distribution as much as possible, which would require adding units to the neighborhoods with the least housing units until the distribution is as equal as possible.Therefore, the maximum reduction in ( I ) is achieved by making the distribution of housing units as equal as possible, given the budget constraint.But to express this formally, perhaps we can say that the optimal allocation is to add as many units as possible to the neighborhoods with the lowest ( H_i ) until the distribution is as equal as possible, thereby minimizing the variance and thus the inequality measure ( I ).Therefore, the maximum reduction in ( I ) is the difference between the initial ( I ) and the minimal possible ( I' ) after optimally distributing the ( k ) units.But since the problem asks to determine the maximum reduction, perhaps we can express it as:The maximum reduction in ( I ) is equal to the initial ( I ) minus the minimal possible ( I' ), where ( I' ) is the standard deviation of the most equal distribution achievable with the budget ( B ).But without specific values, we can't compute the exact numerical reduction. Therefore, the answer is that the maximum reduction is achieved by equalizing the distribution as much as possible, which involves adding units to the neighborhoods with the least housing units until the distribution is as equal as possible.So, summarizing:For Sub-problem 1, the optimal allocation is to make all neighborhoods have the same number of housing units, which minimizes the inequality measure ( I ).For Sub-problem 2, the maximum reduction in ( I ) is achieved by adding housing units to the neighborhoods with the least number of units until the distribution is as equal as possible, given the budget constraint.</think>

🧠:<think>Alright, so I have this research project where I need to analyze the effectiveness of a new educational software. The software claims to improve students' math test scores by an average of 15% after three months of use. I have two main tasks here: first, to conduct a hypothesis test to see if the software has significantly improved the test scores, and second, to calculate the correlation coefficient between the hours spent using the software and the test score improvement.Starting with the first part. Before using the software, the class of 20 students had an average test score of 70, which was normally distributed with a standard deviation of 10. After using the software for three months, the mean score increased to 80. I need to test if this improvement is statistically significant at a 5% significance level. The population standard deviation is assumed to remain unchanged, so I guess that means it's still 10.Okay, hypothesis testing. I remember that for hypothesis testing, I need to set up the null and alternative hypotheses. The null hypothesis is usually that there's no effect, so in this case, it would be that the software doesn't improve the scores. The alternative hypothesis is that it does improve them. So:- Null hypothesis (H0): μ = 70- Alternative hypothesis (H1): μ > 70Since the population standard deviation is known, and the sample size is 20, which is relatively small, but since the original distribution is normal, I think we can use a z-test. Wait, no, actually, for a z-test, the sample size can be small if the population is normal. So, yeah, z-test is appropriate here.The formula for the z-test statistic is:z = (x̄ - μ) / (σ / sqrt(n))Where:- x̄ is the sample mean after using the software, which is 80.- μ is the population mean before using the software, which is 70.- σ is the population standard deviation, 10.- n is the sample size, 20.Plugging in the numbers:z = (80 - 70) / (10 / sqrt(20)) = 10 / (10 / 4.4721) ≈ 10 / 2.2361 ≈ 4.4721So the z-score is approximately 4.47. Now, I need to compare this to the critical value at a 5% significance level for a one-tailed test. Looking at the z-table, the critical value for a one-tailed test at 5% is about 1.645. My calculated z-score is 4.47, which is way higher than 1.645. Therefore, I can reject the null hypothesis.Alternatively, I could calculate the p-value associated with a z-score of 4.47. The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the one observed, assuming the null hypothesis is true. A z-score of 4.47 is extremely high, so the p-value would be very small, definitely less than 0.05. Therefore, again, we reject the null hypothesis.So, conclusion: The software has significantly improved the test scores at the 5% significance level.Moving on to the second part. I have data on the number of hours each student spent using the software over three months. The test score improvement Y is related to the hours of software use X by the equation Y = 5 + 2X. I need to calculate the correlation coefficient between X and Y and interpret it.First, let's recall that the correlation coefficient, r, measures the strength and direction of the linear relationship between two variables. It ranges from -1 to 1, where 1 is a perfect positive correlation, -1 is a perfect negative correlation, and 0 is no correlation.Given the linear regression equation Y = 5 + 2X, this suggests that for every additional hour of software use, the test score improvement increases by 2 points. The slope is 2, which is positive, indicating a positive correlation.But to find the correlation coefficient, I need more information. Wait, actually, in a simple linear regression, the correlation coefficient can be found using the formula:r = b * (s_x / s_y)Where:- b is the slope of the regression line.- s_x is the standard deviation of X.- s_y is the standard deviation of Y.But wait, do I have the standard deviations of X and Y? The problem doesn't specify them. Hmm. Let me read the problem again.It says: "The data fits a linear regression model where the test score improvement Y is related to the hours of software use X by the equation Y = 5 + 2X." So, it's a perfect linear relationship? Because if Y is exactly equal to 5 + 2X, then every point lies exactly on the regression line, which would mean a perfect correlation. So, the correlation coefficient r would be 1.But wait, in reality, data points don't always lie perfectly on the regression line. Unless it's a deterministic relationship, which in this case, it seems like it is. The equation is given as Y = 5 + 2X, which suggests that Y is a linear function of X without any error term. So, in that case, the correlation coefficient would indeed be 1, because there's a perfect linear relationship.Alternatively, if Y were an estimate based on X with some error, then r would be less than 1. But since the equation is given as exact, r is 1.So, the correlation coefficient is 1, which means there's a perfect positive linear relationship between the number of hours spent using the software and the test score improvement. This implies that as the hours of software use increase, the test score improvement also increases in a perfectly linear fashion.But wait, let me think again. In the context of the problem, the software claims to improve scores by 15% on average. The mean score went from 70 to 80, which is an improvement of 10 points. So, 10 points improvement over 70 is approximately 14.29%, which is close to 15%. So, the software seems to have delivered on its claim.But in the regression model, the improvement Y is 5 + 2X. So, if X is the number of hours, then the improvement is 5 plus twice the hours. So, for example, if a student used the software for 10 hours, their improvement would be 5 + 20 = 25 points. That seems quite high. Wait, but the overall improvement was 10 points on average. So, maybe the average hours used was 2.5 hours, because 5 + 2*2.5 = 10. So, the average hours used was 2.5 hours.But regardless, the correlation coefficient is 1 because Y is a perfect linear function of X.Wait, but in reality, can we have a perfect correlation? Because in real data, there's usually some variability. But since the problem states that the data fits the linear regression model exactly, I think we can assume that r is 1.So, in conclusion, the correlation coefficient is 1, indicating a perfect positive linear relationship. This means that the more hours students spent using the software, the higher their test score improvement, in a perfectly predictable manner.But just to be thorough, let me recall that in a simple linear regression, the correlation coefficient can also be calculated using the formula:r = sqrt(R^2)Where R^2 is the coefficient of determination. In this case, since the regression model perfectly predicts Y, R^2 would be 1, so r would be 1 as well.Alternatively, if I consider that in the regression equation Y = 5 + 2X, the slope b is 2, and if I had the standard deviations of X and Y, I could compute r as b*(s_x/s_y). But since Y is a perfect linear function of X, the standard deviation of Y would be 2 times the standard deviation of X. Because Y = 5 + 2X, so the variability in Y is twice the variability in X. Therefore, s_y = 2*s_x. Then, r = b*(s_x/s_y) = 2*(s_x/(2*s_x)) = 1. So, again, r = 1.Therefore, the correlation coefficient is indeed 1.So, summarizing both parts:1. The hypothesis test shows that the software significantly improved test scores at the 5% significance level.2. The correlation coefficient between hours of software use and test score improvement is 1, indicating a perfect positive linear relationship.</think>

🧠:<think>Okay, so I have two problems to solve here. Let me tackle them one by one.Starting with the first problem: The food blogger is using a set ( S ) to model the variety of ingredients. Each element ( s_i ) is a unique ingredient. The cardinality of the set ( S ) is ( n ). She mentions that the total number of ways to choose subsets of ingredients such that no subset is empty is ( 2^n - 1 ). We are given that ( 2^n - 1 = 1023 ), and we need to find ( n ).Hmm, okay. So, subsets of a set with ( n ) elements are ( 2^n ) in total, right? That includes the empty set. So, if we subtract 1, we get all the non-empty subsets. So, ( 2^n - 1 = 1023 ). So, we need to solve for ( n ) here.Let me write that equation down:( 2^n - 1 = 1023 )So, adding 1 to both sides:( 2^n = 1024 )Now, 1024 is a power of 2. I remember that ( 2^{10} = 1024 ). Let me check that:( 2^1 = 2 )( 2^2 = 4 )( 2^3 = 8 )( 2^4 = 16 )( 2^5 = 32 )( 2^6 = 64 )( 2^7 = 128 )( 2^8 = 256 )( 2^9 = 512 )( 2^{10} = 1024 )Yes, that's correct. So, ( n = 10 ).Wait, that seems straightforward. So, the value of ( n ) is 10.Alright, moving on to the second problem. The blogger models customer satisfaction ratings using a continuous random variable ( X ) which follows a normal distribution with an unknown mean ( mu ) and a standard deviation ( sigma ) of 5. She collects a sample of 25 customer ratings and calculates a sample mean ( bar{X} ) of 80. We need to construct a 95% confidence interval for the true mean ( mu ).Okay, so this is a confidence interval estimation problem. Since the population standard deviation ( sigma ) is known (which is 5), and the sample size is 25, which is relatively small, but since the population is normally distributed, we can use the z-distribution.Wait, hold on. The sample size is 25, which is greater than 30, so some people might argue that the Central Limit Theorem applies, but actually, since the original distribution is normal, even a small sample size is okay to use the z-distribution.But just to confirm, when the population is normal and the standard deviation is known, regardless of sample size, we can use the z-interval. So, in this case, since ( sigma ) is known and the population is normal, we use the z-score.So, the formula for the confidence interval is:( bar{X} pm z_{alpha/2} times frac{sigma}{sqrt{n}} )Where ( bar{X} ) is the sample mean, ( z_{alpha/2} ) is the critical value from the standard normal distribution corresponding to the desired confidence level, ( sigma ) is the population standard deviation, and ( n ) is the sample size.Given that the confidence level is 95%, so ( alpha = 1 - 0.95 = 0.05 ), so ( alpha/2 = 0.025 ). The critical value ( z_{0.025} ) is the value such that the area to the right of it is 0.025. From the standard normal distribution table, this value is 1.96.So, plugging in the numbers:( bar{X} = 80 )( z_{0.025} = 1.96 )( sigma = 5 )( n = 25 )So, the standard error ( SE = frac{sigma}{sqrt{n}} = frac{5}{sqrt{25}} = frac{5}{5} = 1 )Therefore, the margin of error ( E = z_{0.025} times SE = 1.96 times 1 = 1.96 )So, the confidence interval is:( 80 pm 1.96 )Which gives us:Lower bound: ( 80 - 1.96 = 78.04 )Upper bound: ( 80 + 1.96 = 81.96 )So, the 95% confidence interval for the true mean ( mu ) is (78.04, 81.96).Wait, let me double-check my calculations.First, ( sqrt{25} = 5 ), so ( 5 / 5 = 1 ). That's correct.Then, 1.96 * 1 = 1.96. So, adding and subtracting 1.96 from 80 gives 78.04 and 81.96. That seems right.Alternatively, sometimes people might use more decimal places for the z-score, but 1.96 is standard for 95% confidence.So, I think that's correct.So, to recap:1. The value of ( n ) is 10.2. The 95% confidence interval for ( mu ) is (78.04, 81.96).I think that's all. Let me just make sure I didn't make any calculation errors.For the first problem, ( 2^{10} = 1024 ), so subtracting 1 gives 1023. Correct.For the second problem, sample mean 80, z-score 1.96, standard error 1, so interval is 80 ± 1.96. Yep, that's 78.04 to 81.96.So, I think that's solid.Final Answer1. The value of ( n ) is boxed{10}.2. The 95% confidence interval for the true mean customer satisfaction rating is boxed{(78.04, 81.96)}.</think>