answer：Okay, so I'm trying to figure out if I can arrange the digits 2, 0, 1, and 9 in a 3x3 grid. Each digit has to be in a separate cell, and every row, column, and both diagonals need to have three different digits. Hmm, that sounds a bit tricky, but let's see.First, I know that a 3x3 grid has 9 cells, and I have only four different digits to work with. That means some digits will have to repeat, right? But wait, the problem says each row, column, and diagonal must have three different digits. So, does that mean each row, column, and diagonal can't have any repeating digits? I think that's what it's saying.So, if I have to fill 9 cells with only four different digits, and each row, column, and diagonal must have all different digits, that seems impossible. Because with only four digits, I can't have nine unique digits in the grid. But maybe I'm misunderstanding something.Let me think again. The problem says "one digit per cell," so each cell has one digit, but it doesn't say that each digit has to be used exactly once. So, maybe digits can repeat, but in such a way that in every row, column, and diagonal, the three digits are different. That might be possible.Okay, so I need to arrange these four digits in the grid, allowing repeats, but ensuring that in every row, column, and diagonal, there are no duplicates. Let's try to visualize the grid.Let's label the grid cells from 1 to 9:1 2 34 5 67 8 9Now, I need to assign digits 2, 0, 1, and 9 to these cells, possibly repeating, but ensuring that in every row (1-2-3, 4-5-6, 7-8-9), every column (1-4-7, 2-5-8, 3-6-9), and both diagonals (1-5-9, 3-5-7), there are no repeating digits.This seems challenging because with only four digits, it's hard to avoid repetition in such a constrained grid. Maybe I can start by placing the digits in a way that spreads them out.Let's try placing 2 in cell 1. Then, in the first row, I need two more different digits. Let's put 0 in cell 2 and 1 in cell 3. So, first row: 2, 0, 1.Now, moving to the second row. Cell 4 is in the first column, which already has 2. So, cell 4 can't be 2. Let's try 9 in cell 4. Then, cell 5 is in the second column, which has 0, so cell 5 can't be 0. Let's put 2 in cell 5. Now, cell 6 is in the third column, which has 1, so cell 6 can't be 1. Let's put 9 in cell 6. So, second row: 9, 2, 9.Wait, but in the second row, I have two 9s. That's not allowed because each row must have three different digits. So, this arrangement doesn't work. Maybe I should try a different approach.Let's try placing 2 in cell 1 again. First row: 2, 0, 1.Second row: Let's try 1 in cell 4. Then, cell 5 can't be 0 or 1, so it has to be 9. Then, cell 6 can't be 1 or 9, so it has to be 2. So, second row: 1, 9, 2.Now, third row: Cell 7 is in the first column, which has 2 and 1, so it can't be 2 or 1. It has to be 0 or 9. Let's put 0 in cell 7. Then, cell 8 is in the second column, which has 0 and 9, so it can't be 0 or 9. It has to be 2 or 1. Let's put 2 in cell 8. Then, cell 9 is in the third column, which has 1 and 2, so it can't be 1 or 2. It has to be 0 or 9. Let's put 9 in cell 9. So, third row: 0, 2, 9.Now, let's check the columns:First column: 2, 1, 0 – all different.Second column: 0, 9, 2 – all different.Third column: 1, 2, 9 – all different.Diagonals:1-5-9: 2, 9, 9 – wait, two 9s. That's not allowed.So, this arrangement also doesn't work. Maybe I need to adjust the third row.Let's try putting 9 in cell 7 instead of 0. Then, cell 8 would have to be 0 or 1. Let's put 0 in cell 8. Then, cell 9 would have to be 1 or 2. Let's put 1 in cell 9. So, third row: 9, 0, 1.Now, check the columns:First column: 2, 1, 9 – all different.Second column: 0, 9, 0 – two 0s. Not allowed.Hmm, that doesn't work either. Maybe I need to try a different starting point.Let's try placing 0 in cell 1. First row: 0, 2, 1.Second row: Cell 4 can't be 0, so let's put 9. Cell 5 can't be 2 or 9, so it has to be 1. Cell 6 can't be 1 or 9, so it has to be 2. Second row: 9, 1, 2.Third row: Cell 7 can't be 0 or 9, so it has to be 1 or 2. Let's put 1 in cell 7. Cell 8 can't be 1 or 2, so it has to be 0 or 9. Let's put 0 in cell 8. Cell 9 can't be 1 or 0, so it has to be 2 or 9. Let's put 2 in cell 9. Third row: 1, 0, 2.Now, check the columns:First column: 0, 9, 1 – all different.Second column: 2, 1, 0 – all different.Third column: 1, 2, 2 – two 2s. Not allowed.This is frustrating. Maybe I need to try a different arrangement altogether.Let's try placing 1 in cell 1. First row: 1, 2, 0.Second row: Cell 4 can't be 1, so let's put 9. Cell 5 can't be 2 or 9, so it has to be 0. Cell 6 can't be 0 or 9, so it has to be 2. Second row: 9, 0, 2.Third row: Cell 7 can't be 1 or 9, so it has to be 0 or 2. Let's put 0 in cell 7. Cell 8 can't be 0 or 2, so it has to be 1 or 9. Let's put 1 in cell 8. Cell 9 can't be 0 or 1, so it has to be 2 or 9. Let's put 2 in cell 9. Third row: 0, 1, 2.Now, check the columns:First column: 1, 9, 0 – all different.Second column: 2, 0, 1 – all different.Third column: 0, 2, 2 – two 2s. Not allowed.This is not working. Maybe I need to consider that it's impossible.Wait, maybe I should think about the number of digits. I have four digits and nine cells. Each digit has to appear multiple times. But the constraints are that in every row, column, and diagonal, there are no duplicates. That seems very restrictive.In a 3x3 grid, each digit would need to appear at least twice, but with four digits, that's eight cells, leaving one cell to have a third repetition. But with the constraints, it's hard to place them without violating the no-duplicates rule.Maybe it's impossible. I think after trying several arrangements and always ending up with duplicates in rows, columns, or diagonals, it's likely that such an arrangement isn't possible.

answer：First, I need to determine the side length of the equilateral triangle. Since the perimeter of the equilateral triangle is 45, I can divide this by 3 to find the length of one side.Next, I'll consider the isosceles triangle. It has two equal sides and a base. I know that at least one of the sides of the isosceles triangle is equal to the side of the equilateral triangle, which I've already calculated.Using the perimeter of the isosceles triangle, which is 40, I can set up an equation to solve for the base. By substituting the known side length into the equation, I can find the value of the base.Finally, I'll solve the equation to determine the base of the isosceles triangle.

answer：Okay, so I have this geometry problem here, and I need to figure out how to prove that point P is the midpoint of EF and that DP is perpendicular to EF. Let me try to unpack the problem step by step.First, let me visualize the given information. There are two isosceles triangles, ABC and DBC, both with base BC. That means in triangle ABC, sides AB and AC are equal, and in triangle DBC, sides DB and DC are equal. Also, we know that angle ABD is a right angle, which is π/2 radians or 90 degrees. M is the midpoint of BC, so BM equals MC.Now, points E, F, and P are chosen with specific conditions:- E is on AB,- P is on MC,- C is on AF,- And the angles BDE, ADP, and CDF are all equal.I need to prove two things:1. P is the midpoint of EF.2. DP is perpendicular to EF.Alright, let's start by trying to draw this figure. I imagine triangle ABC with base BC. Since ABC is isosceles with base BC, AB equals AC. Then, triangle DBC is also isosceles with base BC, so DB equals DC. Given that angle ABD is 90 degrees, point D must be positioned such that when connected to B, it forms a right angle with AB.Since M is the midpoint of BC, BM equals MC. Points E, F, and P are defined with E on AB, P on MC, and C on AF. The angles BDE, ADP, and CDF are all equal. Let's denote this common angle as θ.I think it might help to assign coordinates to the points to make this more concrete. Let me place point B at (0, 0) and point C at (2a, 0), so that M, the midpoint, is at (a, 0). Since ABC is isosceles with base BC, point A must be somewhere above BC. Let's say A is at (a, b). Similarly, since DBC is isosceles with base BC, point D must be somewhere else above or below BC. Given that angle ABD is 90 degrees, point D must be such that line BD is perpendicular to AB.Wait, AB goes from (0,0) to (a, b). The slope of AB is (b - 0)/(a - 0) = b/a. Therefore, the slope of BD, which is perpendicular to AB, should be -a/b. Since BD starts at B (0,0), the equation of BD is y = (-a/b)x.Now, since D is part of triangle DBC, which is isosceles with DB = DC. So, point D must lie somewhere such that its distance from B equals its distance from C. Let me denote D as (d_x, d_y). Then, the distance from D to B is sqrt((d_x)^2 + (d_y)^2), and the distance from D to C is sqrt((d_x - 2a)^2 + (d_y)^2). Setting these equal:sqrt((d_x)^2 + (d_y)^2) = sqrt((d_x - 2a)^2 + (d_y)^2)Squaring both sides:d_x^2 + d_y^2 = (d_x - 2a)^2 + d_y^2Simplify:d_x^2 = d_x^2 - 4a d_x + 4a^2Cancel d_x^2:0 = -4a d_x + 4a^2Divide both sides by 4a (assuming a ≠ 0):0 = -d_x + aSo, d_x = a.So, point D has an x-coordinate of a. From earlier, we had the equation of BD as y = (-a/b)x. Since D is on BD, its coordinates are (a, (-a/b)*a) = (a, -a^2/b).Wait, but if D is at (a, -a^2/b), is that consistent with triangle DBC being isosceles? Let me check the distances.Distance from D to B: sqrt((a - 0)^2 + (-a^2/b - 0)^2) = sqrt(a^2 + (a^4)/(b^2)).Distance from D to C: sqrt((a - 2a)^2 + (-a^2/b - 0)^2) = sqrt(a^2 + (a^4)/(b^2)).Yes, they are equal, so that's consistent.So, point D is at (a, -a^2/b). Interesting, it's below the base BC.Now, let's note the coordinates of all points:- B: (0, 0)- C: (2a, 0)- M: (a, 0)- A: (a, b)- D: (a, -a^2/b)Now, points E, F, P are defined as follows:- E is on AB,- P is on MC,- C is on AF,- And angles BDE, ADP, and CDF are equal.Let me try to parameterize points E, F, and P.First, let me find the equations of lines AB, AF, and MC.Line AB goes from (0,0) to (a, b). Its parametric equations can be written as:x = ta, y = tb, where t ranges from 0 to 1.Similarly, line MC goes from M(a, 0) to C(2a, 0). Wait, that's just the line segment from (a, 0) to (2a, 0). So, any point P on MC can be written as (a + s, 0), where s ranges from 0 to a.Wait, but in the problem statement, P is on MC, but in the original problem, P is on MC, but in the problem statement, it's written as "P ∈ (MC)", which might mean the line segment MC. Similarly, E is on AB, so E ∈ (AB).Also, C is on AF, meaning that point C lies on the line AF. So, AF is a line from A(a, b) to F, and point C(2a, 0) lies on AF. Therefore, AF is the line connecting A(a, b) and C(2a, 0). Let me find the equation of AF.The slope of AF is (0 - b)/(2a - a) = (-b)/a. So, the equation is y - b = (-b/a)(x - a). Simplifying, y = (-b/a)x + 2b.So, point F is somewhere on AF beyond C, but actually, since C is on AF, and F is another point on AF. Wait, but the problem says "C ∈ (AF)", which might mean that C is on AF, but F is another point on AF. So, AF is the line from A to F, passing through C. So, F is a point on AF beyond C.Wait, but in the problem statement, it's written as "C ∈ (AF)", which might mean that C is on AF, but F is a different point. So, AF is a line passing through A and F, with C lying on it. So, AF is the line from A to F, passing through C. So, F is a point on AF beyond C.But in our coordinate system, AF is the line from A(a, b) through C(2a, 0). So, F must be another point on this line beyond C. Let me parameterize point F.Let me use a parameter k such that when k=0, we are at A(a, b), and when k=1, we are at C(2a, 0). So, the parametric equations for AF can be written as:x = a + k(a), y = b - k(b), where k ≥ 0.So, point F is at (a + k*a, b - k*b) for some k > 1, since it's beyond C.Similarly, point E is on AB. Let me parameterize E with a parameter t, where t=0 is at B(0,0) and t=1 is at A(a, b). So, E is at (ta, tb) for some t between 0 and 1.Point P is on MC, which is the line segment from M(a, 0) to C(2a, 0). So, P can be parameterized as (a + s, 0) where s ranges from 0 to a.Now, the problem states that angles BDE, ADP, and CDF are equal. Let's denote this common angle as θ.So, angle BDE = angle ADP = angle CDF = θ.Let me try to express these angles in terms of coordinates.First, angle BDE is the angle at point D between points B, D, and E.Similarly, angle ADP is the angle at point D between points A, D, and P.Angle CDF is the angle at point D between points C, D, and F.Since all these angles are equal, we can set up equations based on their slopes or use vectors to find the relationships.Let me compute the vectors involved in each angle.For angle BDE:- Points: B(0,0), D(a, -a²/b), E(ta, tb)- Vectors: DB = B - D = (0 - a, 0 - (-a²/b)) = (-a, a²/b) DE = E - D = (ta - a, tb - (-a²/b)) = (a(t - 1), tb + a²/b)For angle ADP:- Points: A(a, b), D(a, -a²/b), P(a + s, 0)- Vectors: DA = A - D = (a - a, b - (-a²/b)) = (0, b + a²/b) DP = P - D = (a + s - a, 0 - (-a²/b)) = (s, a²/b)For angle CDF:- Points: C(2a, 0), D(a, -a²/b), F(a + ka, b - kb)- Vectors: DC = C - D = (2a - a, 0 - (-a²/b)) = (a, a²/b) DF = F - D = (a + ka - a, b - kb - (-a²/b)) = (ka, b - kb + a²/b)Since all these angles are equal, the tangent of these angles should be equal. The tangent of an angle between two vectors can be found using the formula:tanθ = |(v × w)| / (v · w)Where v and w are the vectors, × denotes the cross product (in 2D, it's the scalar magnitude), and · denotes the dot product.Let me compute tanθ for each angle.First, for angle BDE:Vectors DB = (-a, a²/b) and DE = (a(t - 1), tb + a²/b)Cross product (magnitude): (-a)(tb + a²/b) - (a²/b)(a(t - 1)) = -a tb - a³/b - a³(t - 1)/b = -a tb - a³/b - a³ t /b + a³/b = -a tb - a³ t /bDot product: (-a)(a(t - 1)) + (a²/b)(tb + a²/b) = -a²(t - 1) + (a² tb)/b + (a^4)/(b²) = -a² t + a² + a² t + a^4 / b² = a² + a^4 / b²So, tanθ_BDE = | -a tb - a³ t /b | / (a² + a^4 / b² )Simplify numerator: | -a t (b + a² / b ) | = a t (b + a² / b )Denominator: a² + a^4 / b² = a²(1 + a² / b² )So, tanθ_BDE = [a t (b + a² / b ) ] / [a²(1 + a² / b² ) ] = [ t (b + a² / b ) ] / [a (1 + a² / b² ) ]Simplify numerator and denominator:Numerator: t ( (b² + a²)/b )Denominator: a ( (b² + a²)/b² )So, tanθ_BDE = [ t ( (b² + a²)/b ) ] / [ a ( (b² + a²)/b² ) ] = [ t / b ] / [ a / b² ] = t b / aSo, tanθ_BDE = (t b)/aNow, let's compute tanθ for angle ADP.Vectors DA = (0, b + a²/b ) and DP = (s, a²/b )Cross product (magnitude): 0*(a²/b) - (b + a²/b)*s = -s (b + a²/b )Dot product: 0*s + (b + a²/b)(a²/b ) = (b + a²/b)(a²/b ) = a² b / b + a^4 / b² = a² + a^4 / b²So, tanθ_ADP = | -s (b + a²/b ) | / (a² + a^4 / b² ) = s (b + a²/b ) / (a² + a^4 / b² )Similarly, as before:Numerator: s ( (b² + a²)/b )Denominator: a²(1 + a² / b² ) = a²( (b² + a²)/b² )So, tanθ_ADP = [ s ( (b² + a²)/b ) ] / [ a² ( (b² + a²)/b² ) ] = [ s / b ] / [ a² / b² ] = s b / a²So, tanθ_ADP = (s b)/a²Now, for angle CDF.Vectors DC = (a, a²/b ) and DF = (ka, b - kb + a²/b )Cross product (magnitude): a*(b - kb + a²/b ) - (a²/b)*(ka ) = a(b - kb + a²/b ) - a² k a / b = a b - a k b + a³ / b - a³ k / bDot product: a*(ka ) + (a²/b)*(b - kb + a²/b ) = a² k + (a²/b)(b - kb + a²/b ) = a² k + a² - a² k + a^4 / b² = a² + a^4 / b²So, tanθ_CDF = | a b - a k b + a³ / b - a³ k / b | / (a² + a^4 / b² )Factor numerator:= | a b (1 - k ) + a³ / b (1 - k ) | = | (1 - k )(a b + a³ / b ) | = | (1 - k ) a (b + a² / b ) |Denominator: a² + a^4 / b² = a²(1 + a² / b² ) = a²( (b² + a²)/b² )So, tanθ_CDF = | (1 - k ) a (b + a² / b ) | / [ a² ( (b² + a²)/b² ) ] = | (1 - k ) (b + a² / b ) | / [ a ( (b² + a²)/b² ) ]Simplify numerator and denominator:Numerator: | (1 - k ) ( (b² + a²)/b ) |Denominator: a ( (b² + a²)/b² )So, tanθ_CDF = | (1 - k ) / b | / [ a / b² ] = | (1 - k ) b | / aSince angles are equal, their tangents are equal. So, tanθ_BDE = tanθ_ADP = tanθ_CDF.So, we have:(t b)/a = (s b)/a² = | (1 - k ) b | / aLet me write these equalities:1. (t b)/a = (s b)/a²2. (t b)/a = | (1 - k ) b | / aFrom equation 1:(t b)/a = (s b)/a²Divide both sides by b/a:t = s / aSo, s = a tFrom equation 2:(t b)/a = | (1 - k ) b | / aMultiply both sides by a/b:t = |1 - k|Assuming that k > 1 (since F is beyond C on AF), 1 - k is negative, so |1 - k| = k - 1.Thus, t = k - 1So, k = t + 1Now, we have s = a t and k = t + 1.So, points E, P, F can be expressed in terms of t.Point E is at (ta, tb)Point P is at (a + s, 0) = (a + a t, 0) = (a(1 + t), 0)Point F is at (a + ka, b - kb) = (a + a(t + 1), b - b(t + 1)) = (a(2 + t), b(1 - t - 1)) = (a(2 + t), -b t)So, F is at (a(2 + t), -b t)Now, we need to find the coordinates of E, F, and P in terms of t, and then show that P is the midpoint of EF and DP is perpendicular to EF.First, let's find the coordinates of E, F, and P:E: (ta, tb)F: (a(2 + t), -b t)P: (a(1 + t), 0)Now, let's check if P is the midpoint of EF.Midpoint of EF would be:x-coordinate: (ta + a(2 + t))/2 = (ta + 2a + a t)/2 = (2a + 2a t)/2 = a(1 + t)y-coordinate: (tb + (-b t))/2 = (tb - b t)/2 = 0So, the midpoint of EF is (a(1 + t), 0), which is exactly point P.Therefore, P is the midpoint of EF.Now, we need to show that DP is perpendicular to EF.First, let's find the coordinates of D, P, and the direction vector of EF.Point D: (a, -a²/b)Point P: (a(1 + t), 0)Vector DP: P - D = (a(1 + t) - a, 0 - (-a²/b)) = (a t, a²/b)Now, let's find the direction vector of EF.Point E: (ta, tb)Point F: (a(2 + t), -b t)Vector EF: F - E = (a(2 + t) - ta, -b t - tb) = (2a + a t - a t, -b t - b t) = (2a, -2b t)So, vector EF is (2a, -2b t)Vector DP is (a t, a²/b)To check if DP is perpendicular to EF, their dot product should be zero.Dot product: (a t)(2a) + (a²/b)(-2b t) = 2a² t - 2a² t = 0Yes, the dot product is zero, so DP is perpendicular to EF.Therefore, we've shown that P is the midpoint of EF and DP is perpendicular to EF.I think this completes the proof.

answer：Okay, so I have this problem about the normal distribution, and I need to tackle three parts: (a), (b), and (c). Let me start by understanding what each part is asking.Starting with part (a), it says that given independent random variables ξ₁, ξ₂, ..., ξₙ, each following a normal distribution with mean a and variance σ₀², I need to show that the sample mean, which is the average of these ξ's, is unbiased and has a variance of σ₀²/n.Alright, so the sample mean is defined as:[bar{xi} = frac{1}{n} sum_{i=1}^{n} xi_i]First, to show that it's unbiased, I need to compute the expectation of the sample mean and show it equals a. Since expectation is linear, the expectation of the average should be the average of the expectations. Each ξ_i has expectation a, so adding them up and dividing by n should still give a. That seems straightforward.Next, for the variance. The variance of the sum of independent variables is the sum of their variances. Each ξ_i has variance σ₀², so the sum has variance nσ₀². Then, dividing by n, the variance becomes σ₀²/n. So that should work.Moving on to part (b), it's about showing that the sample mean is an efficient estimator. Efficient in this context means it's unbiased and has the smallest possible variance. The problem mentions the Rao-Cramer inequality, which gives a lower bound on the variance of any unbiased estimator. I need to show that the variance of the sample mean meets this bound, hence proving it's efficient.The Rao-Cramer inequality states that for any unbiased estimator T, the variance is at least the reciprocal of the Fisher information. The Fisher information here is computed from the second derivative of the log-likelihood function. So I need to compute the log-likelihood of a single observation, take its second derivative with respect to a, take the expectation, and then plug it into the inequality.Given the density function:[p_{(a, sigma_0^2)}(x) = frac{1}{sqrt{2pisigma_0^2}} e^{-frac{(x - a)^2}{2sigma_0^2}}]The log-likelihood is:[ln p = -frac{1}{2} ln(2pisigma_0^2) - frac{(x - a)^2}{2sigma_0^2}]Taking the first derivative with respect to a:[frac{partial ln p}{partial a} = frac{(x - a)}{sigma_0^2}]Taking the second derivative:[frac{partial^2 ln p}{partial a^2} = -frac{1}{sigma_0^2}]So the expectation of the second derivative is just -1/σ₀². Plugging this into the Rao-Cramer inequality:[mathrm{Var}(T) geq frac{sigma_0^2}{n}]Since the variance of the sample mean is exactly σ₀²/n, it meets the lower bound, proving it's efficient.Now, part (c) is about constructing a confidence interval for a. It involves showing that the standardized sample mean follows a standard normal distribution and then using this to form a confidence interval.Given that:[sqrt{n} frac{bar{xi} - a}{sigma_0} sim mathscr{N}(0,1)]This is a direct consequence of the Central Limit Theorem, since the sample mean is normally distributed with mean a and variance σ₀²/n. Standardizing it gives a standard normal variable.Then, to find the confidence interval, we need to find a value z_α such that the probability that the standardized variable lies between -z_α and z_α is 1 - α. This z_α is the critical value from the standard normal distribution corresponding to the desired confidence level.So, the confidence interval for a is:[left(bar{x} - frac{sigma_0}{sqrt{n}} z_{alpha}, bar{x} + frac{sigma_0}{sqrt{n}} z_{alpha}right)]And the coverage probability is 1 - α, meaning that with probability 1 - α, the interval contains the true mean a.I think I have a good grasp of the steps now. Let me try to formalize these thoughts into proper proofs.For part (a), I'll compute the expectation and variance of the sample mean step by step. For part (b), I'll derive the Fisher information and apply the Rao-Cramer inequality. For part (c), I'll use the properties of the normal distribution and the definition of confidence intervals.I might need to recall some properties of expectations and variances, as well as the Central Limit Theorem and the definition of efficient estimators. Also, understanding the log-likelihood function and its derivatives is crucial for part (b).I should also make sure to clearly define each step and justify each part of the proofs to ensure clarity and correctness.Alright, let's get started with part (a).**Step-by-Step Explanation and Proof****Part (a): Unbiasedness and Variance of the Sample Mean**Given independent random variables ξ₁, ξ₂, ..., ξₙ, each following a normal distribution with mean a and variance σ₀², we define the sample mean as:[bar{xi} = frac{1}{n} sum_{i=1}^{n} xi_i]**1. Unbiasedness:**To show that the sample mean is unbiased, we compute its expectation:[mathrm{E}[bar{xi}] = mathrm{E}left[frac{1}{n} sum_{i=1}^{n} xi_iright]]Using the linearity of expectation:[mathrm{E}[bar{xi}] = frac{1}{n} sum_{i=1}^{n} mathrm{E}[xi_i]]Since each ξ_i is normally distributed with mean a:[mathrm{E}[xi_i] = a quad text{for all } i]Therefore:[mathrm{E}[bar{xi}] = frac{1}{n} sum_{i=1}^{n} a = frac{1}{n} cdot n cdot a = a]This shows that the sample mean is unbiased:[mathrm{E}[bar{xi}] = a]**2. Variance of the Sample Mean:**Next, we compute the variance of the sample mean:[mathrm{Var}(bar{xi}) = mathrm{Var}left(frac{1}{n} sum_{i=1}^{n} xi_iright)]Since the ξ_i are independent, the variance of the sum is the sum of the variances:[mathrm{Var}left(frac{1}{n} sum_{i=1}^{n} xi_iright) = frac{1}{n^2} sum_{i=1}^{n} mathrm{Var}(xi_i)]Each ξ_i has variance σ₀²:[mathrm{Var}(xi_i) = sigma_0^2 quad text{for all } i]Thus:[mathrm{Var}(bar{xi}) = frac{1}{n^2} sum_{i=1}^{n} sigma_0^2 = frac{1}{n^2} cdot n cdot sigma_0^2 = frac{sigma_0^2}{n}]Therefore, the variance of the sample mean is:[mathrm{Var}(bar{xi}) = frac{sigma_0^2}{n}]**Conclusion for Part (a):**The sample mean (bar{xi}) is an unbiased estimator of a with variance (frac{sigma_0^2}{n}).[mathrm{E}[bar{xi}] = a quad text{and} quad mathrm{Var}(bar{xi}) = frac{sigma_0^2}{n}]**Part (b): Efficiency of the Sample Mean via Rao-Cramer Inequality**To establish that the sample mean is an efficient estimator, we need to show that it achieves the Cramér-Rao lower bound. The Rao-Cramer inequality states that for any unbiased estimator (T) of parameter (a):[mathrm{Var}(T) geq frac{1}{n I(a)}]where (I(a)) is the Fisher information. For the normal distribution, we compute the Fisher information as follows.**1. Log-Likelihood Function:**The probability density function for a single observation ξ₁ is:[p_{(a, sigma_0^2)}(x) = frac{1}{sqrt{2pisigma_0^2}} e^{-frac{(x - a)^2}{2sigma_0^2}}]The log-likelihood function for a single observation is:[ln p_{(a, sigma_0^2)}(x) = -frac{1}{2} ln(2pisigma_0^2) - frac{(x - a)^2}{2sigma_0^2}]**2. First and Second Derivatives:**First derivative with respect to (a):[frac{partial ln p}{partial a} = frac{(x - a)}{sigma_0^2}]Second derivative with respect to (a):[frac{partial^2 ln p}{partial a^2} = -frac{1}{sigma_0^2}]**3. Fisher Information:**The Fisher information (I(a)) is the expectation of the negative second derivative of the log-likelihood:[I(a) = -mathrm{E}left[frac{partial^2 ln p}{partial a^2}right] = -mathrm{E}left[-frac{1}{sigma_0^2}right] = frac{1}{sigma_0^2}]**4. Applying Rao-Cramer Inequality:**For n independent observations, the Fisher information is (n I(a) = frac{n}{sigma_0^2}). Thus, the Cramér-Rao lower bound is:[mathrm{Var}(T) geq frac{1}{n I(a)} = frac{sigma_0^2}{n}]Since the variance of the sample mean (bar{xi}) is exactly (frac{sigma_0^2}{n}), it achieves the lower bound. Therefore, the sample mean is an efficient estimator.**Conclusion for Part (b):**The sample mean (bar{xi}) is an efficient estimator of (a) because it attains the Cramér-Rao lower bound.[mathrm{Var}(bar{xi}) = frac{sigma_0^2}{n}]**Part (c): Confidence Interval for the Mean**We need to show that:[sqrt{n} frac{bar{xi} - a}{sigma_0} sim mathscr{N}(0,1)]and use this to construct a confidence interval for (a) with confidence level (1 - alpha).**1. Standard Normal Distribution:**Given that (bar{xi}) is normally distributed with mean (a) and variance (frac{sigma_0^2}{n}):[bar{xi} sim mathscr{N}left(a, frac{sigma_0^2}{n}right)]Standardizing (bar{xi}):[Z = frac{bar{xi} - a}{sqrt{frac{sigma_0^2}{n}}} = sqrt{n} frac{bar{xi} - a}{sigma_0}]This standardized variable (Z) follows a standard normal distribution:[Z sim mathscr{N}(0,1)]**2. Constructing the Confidence Interval:**To find a confidence interval for (a) with confidence level (1 - alpha), we need to find a value (z_{alpha}) such that:[mathrm{P}left(-z_{alpha} leq Z leq z_{alpha}right) = 1 - alpha]This translates to:[mathrm{P}left(-z_{alpha} leq sqrt{n} frac{bar{xi} - a}{sigma_0} leq z_{alpha}right) = 1 - alpha]Rewriting this inequality to solve for (a):[mathrm{P}left(bar{xi} - frac{sigma_0}{sqrt{n}} z_{alpha} leq a leq bar{xi} + frac{sigma_0}{sqrt{n}} z_{alpha}right) = 1 - alpha]Thus, the confidence interval for (a) is:[left(bar{x} - frac{sigma_0}{sqrt{n}} z_{alpha}, bar{x} + frac{sigma_0}{sqrt{n}} z_{alpha}right)]where (bar{x}) is the observed sample mean.**3. Coverage Probability:**The coverage probability is the probability that the interval contains the true mean (a):[mathrm{P}_{(a, sigma_0^2)}left(bar{xi} - frac{sigma_0}{sqrt{n}} z_{alpha} leq a leq bar{xi} + frac{sigma_0}{sqrt{n}} z_{alpha}right) = 1 - alpha]This confirms that the interval has a coverage probability of (1 - alpha).**Conclusion for Part (c):**The standardized sample mean follows a standard normal distribution, and the interval:[left(bar{x} - frac{sigma_0}{sqrt{n}} z_{alpha}, bar{x} + frac{sigma_0}{sqrt{n}} z_{alpha}right)]is a confidence interval for (a) with confidence level (1 - alpha).[sqrt{n} frac{bar{xi} - a}{sigma_0} sim mathscr{N}(0,1)]and the confidence interval has coverage probability (1 - alpha).**Final Answer**(a) The sample mean is unbiased with variance (frac{sigma_0^2}{n}):[boxed{mathrm{E}[bar{xi}] = a quad text{and} quad mathrm{Var}(bar{xi}) = frac{sigma_0^2}{n}}](b) The sample mean is efficient, achieving the Cramér-Rao lower bound:[boxed{mathrm{Var}(bar{xi}) = frac{sigma_0^2}{n}}](c) The confidence interval for (a) with level (1 - alpha) is:[boxed{left(bar{x} - frac{sigma_0}{sqrt{n}} z_{alpha}, bar{x} + frac{sigma_0}{sqrt{n}} z_{alpha}right)}]

answer：First, I need to determine the percentage of seats that were vacant. Since 62% of the seats were filled, the remaining percentage must be vacant.Next, I'll calculate the number of vacant seats by finding 38% of the total 600 seats. This involves multiplying 600 by 0.38.Finally, performing the multiplication will give me the exact number of vacant seats.

answer：Okay, so I have this problem where I need to find the distance between two vertices of quadratic equations. The equations given are y = x² - 6x + 13 and y = x² + 2x + 4. The vertices are labeled as C and D respectively, and I need to find the distance between them.First, I remember that the vertex form of a quadratic equation is y = a(x - h)² + k, where (h, k) is the vertex. So, to find the vertices of these equations, I need to convert them into vertex form. That means I'll have to complete the square for both equations.Starting with the first equation: y = x² - 6x + 13. I need to complete the square for the x terms. The coefficient of x² is 1, which is good because it makes things simpler. The coefficient of x is -6, so I take half of that, which is -3, and then square it to get 9. So, I can rewrite the equation as:y = (x² - 6x + 9) + 13 - 9Wait, why did I subtract 9? Oh, right, because I added 9 inside the parentheses to complete the square, so I have to subtract it outside to keep the equation balanced. Simplifying that, I get:y = (x - 3)² + 4So, the vertex C is at (3, 4). That makes sense because in the vertex form, h is 3 and k is 4.Now, moving on to the second equation: y = x² + 2x + 4. Again, I need to complete the square. The coefficient of x is 2, so half of that is 1, and squaring it gives 1. So, I rewrite the equation as:y = (x² + 2x + 1) + 4 - 1Subtracting 1 because I added it inside the parentheses. Simplifying, I get:y = (x + 1)² + 3So, the vertex D is at (-1, 3). That seems correct because h is -1 and k is 3.Now that I have both vertices, C at (3, 4) and D at (-1, 3), I need to find the distance between these two points. I remember the distance formula is derived from the Pythagorean theorem, which is:Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]Let me assign the coordinates: for point C, x₁ is 3 and y₁ is 4. For point D, x₂ is -1 and y₂ is 3. Plugging these into the formula:Distance = √[(-1 - 3)² + (3 - 4)²]Calculating the differences first:x₂ - x₁ = -1 - 3 = -4y₂ - y₁ = 3 - 4 = -1Now, squaring these differences:(-4)² = 16(-1)² = 1Adding them together:16 + 1 = 17Taking the square root of 17:√17So, the distance between the two vertices C and D is √17 units.Wait, let me double-check my calculations to make sure I didn't make any mistakes. For the first equation, completing the square:x² - 6x + 13Take half of -6, which is -3, square it to get 9. So,(x - 3)² = x² - 6x + 9So, 13 - 9 = 4, which gives y = (x - 3)² + 4. That seems right.For the second equation:x² + 2x + 4Half of 2 is 1, square it to get 1. So,(x + 1)² = x² + 2x + 1Then, 4 - 1 = 3, so y = (x + 1)² + 3. That also looks correct.Distance formula:√[(-1 - 3)² + (3 - 4)²] = √[(-4)² + (-1)²] = √[16 + 1] = √17Yes, that all checks out. I think I did everything correctly. So, the distance between C and D is √17.