🧠:<think>Alright, so I have this problem about modeling social interactions using graph theory and differential equations. It's split into two parts. Let me try to tackle them one by one.Starting with part 1: We have a group of n individuals, and their interactions are represented by a directed graph G with adjacency matrix A. The entry a_ij represents the influence that individual j has on individual i. The student is looking at how the psychological state x_i(t) of each individual changes over time, and the dynamics are given by the differential equation:dx_i(t)/dt = -α x_i(t) + sum_{j=1}^n a_ij x_j(t)where α is a positive constant for natural decay. The question is about determining the conditions on the eigenvalues of A that ensure all x_i(t) converge to zero as t approaches infinity.Hmm, okay. So this is a system of linear differential equations. I remember that for linear systems, the stability is determined by the eigenvalues of the system matrix. In this case, the system can be written in matrix form as:dx/dt = (A - α I) xWhere I is the identity matrix. So the matrix governing the system is (A - α I). For the solutions to converge to zero, we need all the eigenvalues of (A - α I) to have negative real parts. That is, the system must be asymptotically stable.So, if λ is an eigenvalue of A, then the corresponding eigenvalue of (A - α I) is (λ - α). For the real part of (λ - α) to be negative, we need Re(λ) - α < 0, which implies Re(λ) < α.Therefore, the condition is that all eigenvalues of A must have real parts less than α. So, if the real part of every eigenvalue of A is less than α, then the system will converge to zero as t approaches infinity.Wait, but in graph theory, especially for directed graphs, the adjacency matrix can have complex eigenvalues. So, it's not just about the magnitude but the real part of each eigenvalue. So, if all eigenvalues of A have real parts less than α, then the system is stable.But hold on, is there a more specific condition? For example, if A is a Metzler matrix, which is common in systems where interactions are non-negative, then the stability can be determined by the dominant eigenvalue. But in this case, A is an adjacency matrix, which typically has non-negative entries, but it's directed, so it's a general matrix, not necessarily Metzler.Wait, no, actually, adjacency matrices can have negative entries if we consider negative influences, but in the standard case, they are non-negative. But in this problem, the student just says a_ij represents influence, which could be positive or negative? Hmm, the problem statement doesn't specify, so maybe we can assume it's a general matrix.But regardless, the key point is that for the system dx/dt = (A - α I)x to be asymptotically stable, all eigenvalues of (A - α I) must have negative real parts. Which translates to all eigenvalues of A having real parts less than α.So, the condition is that the real part of every eigenvalue of A is less than α.Moving on to part 2: The group is divided into two subgroups with their own adjacency matrices A1 and A2, and cross-group interactions represented by B. The overall adjacency matrix is:A = [A1 B; B^T A2]We need to analyze the stability of the psychological states in this structure, specifically how the eigenvalues of A1, A2, and B affect the overall stability.So, similar to part 1, the system is dx/dt = (A - α I)x, where now A is the block matrix as given. So, the eigenvalues of (A - α I) will determine stability.But since A is a block matrix, maybe we can analyze its eigenvalues in terms of A1, A2, and B.I recall that for block matrices, especially symmetric ones, the eigenvalues can sometimes be related to the eigenvalues of the blocks, but it's not straightforward. Since A is symmetric (because B is transposed in the lower block), it's a symmetric matrix, so all its eigenvalues are real.Therefore, the eigenvalues of A are real, and for stability, we need all eigenvalues of A to be less than α.But how do the eigenvalues of A1, A2, and B affect the eigenvalues of the block matrix A?Hmm, this is more complex. Maybe we can consider the eigenvalues of the blocks and how they interact.One approach is to consider the eigenvalues of A1 and A2. If both A1 and A2 have eigenvalues with real parts less than α, but also considering the cross-influence B, which can potentially introduce eigenvalues that might be larger.Alternatively, perhaps we can use the concept of the spectrum of block matrices. For a block matrix like this, the eigenvalues are not simply the union of the eigenvalues of A1 and A2 because of the cross terms B and B^T.I remember that for a block matrix:[ A1 B ][ B^T A2 ]the eigenvalues can be found by solving the characteristic equation det(A - λ I) = 0, which expands to det( (A1 - λ I)(A2 - λ I) - B B^T ) = 0.Wait, no, actually, the determinant of a block matrix isn't just the product minus something. It's more complicated. For a 2x2 block matrix, the determinant can sometimes be expressed in terms of the blocks, but it's not straightforward.Alternatively, maybe we can use the fact that for symmetric block matrices, the eigenvalues are bounded by the eigenvalues of the diagonal blocks and the off-diagonal blocks.I think there's a theorem that says that the eigenvalues of the block matrix are bounded by the eigenvalues of A1 and A2, adjusted by the norm of B.Wait, more specifically, I recall that for a block matrix like this, the eigenvalues are bounded by the eigenvalues of A1 and A2 plus or minus the spectral norm of B.Is that correct? Let me think.Yes, I think the eigenvalues of the block matrix [A1 B; B^T A2] are bounded by the eigenvalues of A1 and A2 plus or minus the spectral radius of B.But actually, it's more precise to say that the eigenvalues of the block matrix lie within the union of the intervals [λ_min(A1) - ||B||, λ_max(A1) + ||B||] and [λ_min(A2) - ||B||, λ_max(A2) + ||B||], where ||B|| is the spectral norm of B.So, if we want all eigenvalues of A to be less than α, we need that the maximum eigenvalue of A is less than α. Therefore, the maximum eigenvalue of A is less than or equal to the maximum of (λ_max(A1) + ||B||, λ_max(A2) + ||B||).Therefore, to ensure that all eigenvalues of A are less than α, we need:λ_max(A1) + ||B|| < α and λ_max(A2) + ||B|| < α.Similarly, the minimum eigenvalues would be greater than λ_min(A1) - ||B|| and λ_min(A2) - ||B||, but since we are concerned with the maximum eigenvalues for stability, the key condition is on the upper bounds.So, the conditions are:1. The maximum eigenvalue of A1 plus the spectral norm of B must be less than α.2. The maximum eigenvalue of A2 plus the spectral norm of B must be less than α.This ensures that the maximum eigenvalue of the entire block matrix A is less than α, leading to all eigenvalues of (A - α I) having negative real parts, hence asymptotic stability.Alternatively, if we think in terms of the cross-influence matrix B, if the cross-influence is too strong (i.e., ||B|| is too large), it can push the eigenvalues of A above α, causing instability.Therefore, the stability depends on both the internal influence structures (eigenvalues of A1 and A2) and the strength of cross-group interactions (spectral norm of B). If either subgroup has a high maximum eigenvalue or the cross-influence is too strong, it can destabilize the system.So, to sum up, for the divided group structure, the overall stability is ensured if the maximum eigenvalues of both A1 and A2, when added to the spectral norm of B, are still less than α.I think that's the gist of it. Let me just check if I missed anything.In part 1, the key was that all eigenvalues of A must have real parts less than α. In part 2, since A is symmetric, all eigenvalues are real, so we just need the maximum eigenvalue of A to be less than α. And the maximum eigenvalue of A is bounded by the maximum eigenvalues of A1 and A2 plus the spectral norm of B. So, if both A1 and A2 have their maximum eigenvalues plus the cross-influence norm less than α, then the whole system is stable.Yeah, that makes sense. So, I think I have a good grasp on both parts now.Final Answer1. All eigenvalues of ( A ) must have real parts less than ( alpha ). Thus, the condition is that the real part of every eigenvalue of ( A ) is less than ( alpha ). The final answer is (boxed{text{All eigenvalues of } A text{ have real parts less than } alpha}).2. The overall stability is ensured if the maximum eigenvalues of ( A_1 ) and ( A_2 ), when added to the spectral norm of ( B ), are less than ( alpha ). Therefore, the conditions are (boxed{lambda_{text{max}}(A_1) + |B| < alpha text{ and } lambda_{text{max}}(A_2) + |B| < alpha}).</think>

🧠:<think>Okay, so I have this problem about a travel photographer testing a new camera in extreme environments. The problem has two parts. Let me try to tackle them one by one.Starting with part 1: I need to calculate the difference in battery life and image quality between a desert environment and a snowy mountain environment. The functions given are for battery life and image quality.First, let's write down the functions again to make sure I have them right.Battery life is modeled by:[ B(T, H) = 100e^{-0.05T} - frac{H}{2} ]where ( T ) is temperature in Celsius and ( H ) is humidity percentage.Image quality is given by:[ Q(T, L) = frac{200}{1 + e^{0.1(T - L)}} ]where ( L ) is luminosity in lumens.Alright, so for the desert, the conditions are:- Temperature, ( T = 40^circ )C- Humidity, ( H = 20% )- Luminosity, ( L = 100,000 ) lumensAnd for the snowy mountain:- Temperature, ( T = -10^circ )C- Humidity, ( H = 80% )- Luminosity, ( L = 5,000 ) lumensSo, I need to compute ( B ) and ( Q ) for both environments and then find the differences.Let me start with the desert.Desert Battery Life:[ B(40, 20) = 100e^{-0.05 times 40} - frac{20}{2} ]Calculating the exponent first:[ -0.05 times 40 = -2 ]So, ( e^{-2} ) is approximately ( e^{-2} approx 0.1353 )Multiply by 100:[ 100 times 0.1353 = 13.53 ]Then subtract ( frac{20}{2} = 10 ):[ 13.53 - 10 = 3.53 ]So, battery life in the desert is approximately 3.53 units.Desert Image Quality:[ Q(40, 100000) = frac{200}{1 + e^{0.1(40 - 100000)}} ]First, compute the exponent:[ 0.1 times (40 - 100000) = 0.1 times (-99960) = -9996 ]So, ( e^{-9996} ) is practically zero because the exponent is a large negative number. Therefore, the denominator becomes ( 1 + 0 = 1 ).Thus, image quality is:[ frac{200}{1} = 200 ]Wait, that seems too high. Let me double-check. The formula is ( Q(T, L) = frac{200}{1 + e^{0.1(T - L)}} ). So, when ( T ) is much less than ( L ), the exponent becomes negative, making ( e^{0.1(T - L)} ) very small. So, the denominator is approximately 1, and ( Q ) approaches 200. That makes sense because if luminosity is very high, image quality is excellent.But in the desert, luminosity is 100,000, which is very high, so image quality is near maximum.Snowy Mountain Battery Life:[ B(-10, 80) = 100e^{-0.05 times (-10)} - frac{80}{2} ]Calculating the exponent:[ -0.05 times (-10) = 0.5 ]So, ( e^{0.5} approx 1.6487 )Multiply by 100:[ 100 times 1.6487 = 164.87 ]Subtract ( frac{80}{2} = 40 ):[ 164.87 - 40 = 124.87 ]So, battery life in the snowy mountain is approximately 124.87 units.Snowy Mountain Image Quality:[ Q(-10, 5000) = frac{200}{1 + e^{0.1(-10 - 5000)}} ]Compute the exponent:[ 0.1 times (-10 - 5000) = 0.1 times (-5010) = -501 ]Again, ( e^{-501} ) is practically zero, so denominator is 1, and image quality is 200.Wait, that's the same as the desert. But in the mountain, luminosity is much lower, only 5,000 lumens. Hmm, maybe my initial thought was wrong.Wait, let me think again. The formula is ( Q(T, L) = frac{200}{1 + e^{0.1(T - L)}} ). So, when ( T ) is much less than ( L ), the exponent is negative, so ( e^{0.1(T - L)} ) is small, so ( Q ) is near 200. But in the mountain, ( T = -10 ), ( L = 5000 ). So, ( T - L = -5010 ), so exponent is -501, which is still a very small number. So, ( Q ) is 200.But in the desert, ( T = 40 ), ( L = 100,000 ). So, ( T - L = -99,960 ), exponent is -9996, which is even smaller. So, both image qualities are 200? That seems odd because in the mountain, luminosity is low, but the formula gives the same image quality as the desert.Wait, maybe I made a mistake. Let me check the formula again.The image quality function is ( Q(T, L) = frac{200}{1 + e^{0.1(T - L)}} ). So, when ( T ) is less than ( L ), the exponent is negative, so ( e^{0.1(T - L)} ) is small, so ( Q ) is near 200. When ( T ) is greater than ( L ), the exponent is positive, so ( e^{0.1(T - L)} ) is large, making ( Q ) near 0.So, in both cases, since ( T ) is much less than ( L ), ( Q ) is near 200. So, both image qualities are 200. That's interesting.But wait, in the mountain, luminosity is 5,000, which is still quite high, but ( T ) is -10, so ( T - L = -5010 ). So, the exponent is -501, which is a very small number, so ( e^{-501} ) is practically zero. So, yes, image quality is 200.In the desert, luminosity is 100,000, so ( T - L = -99,960 ), which is even more negative, so ( e^{-9996} ) is also practically zero, so image quality is also 200.So, both image qualities are 200. Therefore, the difference in image quality is zero.But that seems counterintuitive because in the mountain, luminosity is lower, but the formula doesn't account for that? Wait, no, the formula does account for luminosity through the exponent. But in both cases, luminosity is much higher than temperature, so the exponent is very negative, making image quality max out at 200.So, in both environments, image quality is 200, so the difference is zero.But let me think again. Maybe I should compute the exact value, but it's so small that it's negligible.Alternatively, maybe the formula is intended to have image quality decrease when luminosity is too high or too low? Or perhaps the formula is designed such that image quality peaks when ( T = L ). Let me see.If ( T = L ), then the exponent is 0, so ( e^{0} = 1 ), so ( Q = frac{200}{1 + 1} = 100 ). So, image quality is 100 when ( T = L ). So, when ( T ) is much less than ( L ), image quality approaches 200, and when ( T ) is much greater than ( L ), image quality approaches 0.So, in both desert and mountain, since ( T < L ), image quality is near 200, but in the mountain, ( T ) is much less than ( L ) as well, so it's also near 200. So, the difference is indeed zero.Hmm, interesting. So, the image quality is the same in both environments.Now, for battery life, in the desert, it's approximately 3.53, and in the mountain, it's approximately 124.87. So, the difference is 124.87 - 3.53 = 121.34.So, the battery life is much better in the mountain, which makes sense because the temperature is lower, and the battery life function has an exponential term that increases with lower temperatures, and humidity is higher, which subtracts from the battery life, but in the mountain, the exponential term dominates.Wait, in the desert, temperature is high, so ( e^{-0.05T} ) is small, so battery life is low. In the mountain, temperature is low, so ( e^{-0.05T} ) is larger, so battery life is higher, despite the higher humidity.So, the difference in battery life is about 121.34 units, and the difference in image quality is 0.Wait, but let me make sure I didn't make a calculation error.For the desert battery life:[ B(40, 20) = 100e^{-2} - 10 ][ e^{-2} approx 0.1353 ][ 100 * 0.1353 = 13.53 ][ 13.53 - 10 = 3.53 ]Yes, that's correct.For the mountain:[ B(-10, 80) = 100e^{0.5} - 40 ][ e^{0.5} approx 1.6487 ][ 100 * 1.6487 = 164.87 ][ 164.87 - 40 = 124.87 ]Yes, that's correct.So, the difference in battery life is 124.87 - 3.53 = 121.34.And image quality is 200 in both, so difference is 0.Okay, so that's part 1.Now, part 2: The photographer wants to determine the optimal temperature ( T ) and luminosity ( L ) for achieving the highest possible image quality score ( Q(T, L) ), given that the camera operates best within a temperature range of (-20^circ)C to (50^circ)C and a luminosity range of 1,000 to 100,000 lumens.So, we need to formulate the optimization problem and find the critical points that potentially maximize image quality.First, let's write down the image quality function again:[ Q(T, L) = frac{200}{1 + e^{0.1(T - L)}} ]We need to maximize ( Q(T, L) ) with respect to ( T ) and ( L ), subject to:[ -20 leq T leq 50 ][ 1000 leq L leq 100000 ]To find the maximum, we can take partial derivatives and set them equal to zero.First, let's compute the partial derivative of ( Q ) with respect to ( T ).Let me denote ( Q = frac{200}{1 + e^{0.1(T - L)}} )Let me rewrite it as:[ Q = 200 times left(1 + e^{0.1(T - L)}right)^{-1} ]So, partial derivative with respect to ( T ):[ frac{partial Q}{partial T} = 200 times (-1) times left(1 + e^{0.1(T - L)}right)^{-2} times 0.1 e^{0.1(T - L)} ]Simplify:[ frac{partial Q}{partial T} = -20 times frac{e^{0.1(T - L)}}{left(1 + e^{0.1(T - L)}right)^2} ]Similarly, partial derivative with respect to ( L ):[ frac{partial Q}{partial L} = 200 times (-1) times left(1 + e^{0.1(T - L)}right)^{-2} times (-0.1) e^{0.1(T - L)} ]Simplify:[ frac{partial Q}{partial L} = 20 times frac{e^{0.1(T - L)}}{left(1 + e^{0.1(T - L)}right)^2} ]To find critical points, we set both partial derivatives equal to zero.So, set ( frac{partial Q}{partial T} = 0 ):[ -20 times frac{e^{0.1(T - L)}}{left(1 + e^{0.1(T - L)}right)^2} = 0 ]The denominator is always positive, and ( e^{0.1(T - L)} ) is always positive, so the only way this is zero is if the numerator is zero, but the numerator is ( -20 times ) positive, which can't be zero. So, no solution here.Similarly, set ( frac{partial Q}{partial L} = 0 ):[ 20 times frac{e^{0.1(T - L)}}{left(1 + e^{0.1(T - L)}right)^2} = 0 ]Again, the numerator is positive, so this can't be zero. So, no critical points inside the domain.Therefore, the maximum must occur on the boundary of the domain.So, we need to check the boundaries of ( T ) and ( L ).But before that, let's analyze the function ( Q(T, L) ).We can rewrite ( Q(T, L) ) as:[ Q(T, L) = frac{200}{1 + e^{0.1(T - L)}} ]Let me consider ( x = T - L ). Then,[ Q = frac{200}{1 + e^{0.1x}} ]We can analyze ( Q ) as a function of ( x ).When ( x ) is very large (positive), ( e^{0.1x} ) is very large, so ( Q ) approaches 0.When ( x ) is very large (negative), ( e^{0.1x} ) approaches 0, so ( Q ) approaches 200.Therefore, ( Q ) is maximized when ( x ) is as negative as possible, i.e., when ( T - L ) is as negative as possible, meaning ( T ) is as small as possible and ( L ) is as large as possible.Given the constraints:- ( T ) can be as low as -20°C- ( L ) can be as high as 100,000 lumensTherefore, to maximize ( Q ), we should set ( T ) to its minimum value (-20°C) and ( L ) to its maximum value (100,000 lumens).So, the optimal point is ( T = -20 ), ( L = 100,000 ).But let's verify this by checking the boundaries.The function ( Q(T, L) ) depends on ( T - L ). To maximize ( Q ), we need to minimize ( T - L ), which is equivalent to maximizing ( L - T ).So, to minimize ( T - L ), we need to set ( T ) as small as possible and ( L ) as large as possible.Therefore, the maximum image quality occurs at ( T = -20 ) and ( L = 100,000 ).Let me compute ( Q(-20, 100000) ):[ Q(-20, 100000) = frac{200}{1 + e^{0.1(-20 - 100000)}} ][ = frac{200}{1 + e^{-10002}} ]Since ( e^{-10002} ) is practically zero, ( Q ) is approximately 200.Similarly, if we set ( T = -20 ) and ( L = 100000 ), we get the maximum image quality.Alternatively, if we set ( T ) to -20 and ( L ) to 100,000, we get the highest possible ( Q ).Therefore, the optimal temperature is -20°C and optimal luminosity is 100,000 lumens.But let me check if there are other boundary points that could give a higher ( Q ).For example, if ( T ) is at its minimum (-20) and ( L ) is at its maximum (100,000), that's the best case.If ( T ) is higher, say 50°C, and ( L ) is 100,000, then ( T - L = -99,950 ), so ( Q ) is still approximately 200.Wait, but if ( T ) is 50 and ( L ) is 100,000, ( T - L = -99,950 ), so ( e^{0.1(-99,950)} = e^{-9995} ), which is practically zero, so ( Q ) is still 200.Similarly, if ( T ) is -20 and ( L ) is 1000, then ( T - L = -1020 ), so ( e^{-102} ) is practically zero, so ( Q ) is 200.Wait, so actually, as long as ( T ) is much less than ( L ), ( Q ) is 200. So, in the entire domain where ( T leq L ), ( Q ) is 200. Wait, no, because when ( T ) approaches ( L ), ( Q ) approaches 100.Wait, let me think again. If ( T = L ), then ( Q = 100 ). If ( T < L ), ( Q ) is greater than 100, approaching 200 as ( T ) becomes much less than ( L ). If ( T > L ), ( Q ) is less than 100, approaching 0 as ( T ) becomes much greater than ( L ).So, in the domain ( T in [-20, 50] ) and ( L in [1000, 100000] ), the maximum ( Q ) is 200, achieved when ( T ) is as small as possible and ( L ) is as large as possible, but also when ( T ) is small and ( L ) is large enough such that ( T - L ) is very negative.But actually, in the entire domain, as long as ( T ) is much less than ( L ), ( Q ) is 200. So, the maximum is 200, achieved when ( T ) is minimized and ( L ) is maximized.Therefore, the optimal point is ( T = -20 ), ( L = 100,000 ).But let me check if there are other points where ( Q ) could be higher. For example, if ( T ) is -20 and ( L ) is 100,000, ( Q ) is 200. If ( T ) is -20 and ( L ) is 100,000, same result. If ( T ) is -20 and ( L ) is 100,000, same.Wait, actually, any point where ( T ) is much less than ( L ) will give ( Q = 200 ). So, the entire boundary where ( T ) is at its minimum and ( L ) is at its maximum, but also any point where ( T ) is low enough and ( L ) is high enough such that ( T - L ) is very negative.But in terms of optimization, the maximum value of ( Q ) is 200, and it's achieved when ( T - L ) is minimized, i.e., ( T ) is as small as possible and ( L ) is as large as possible.Therefore, the critical point is at ( T = -20 ), ( L = 100,000 ).But wait, earlier when I computed the partial derivatives, I found that there are no critical points inside the domain because the partial derivatives never equal zero. So, the maximum must be on the boundary.Therefore, the optimal temperature is -20°C and optimal luminosity is 100,000 lumens.So, summarizing:1. Difference in battery life: 124.87 - 3.53 = 121.34 units. Image quality difference: 0.2. The optimal temperature is -20°C and optimal luminosity is 100,000 lumens.I think that's it.</think>

🧠:<think>Alright, so I've got these two math problems to solve, and I need to figure them out step by step. Let me start with the first one about Alex's acting career modeled by a polynomial.Sub-problem 1: Finding the coefficients of the polynomialOkay, so the polynomial is given by ( P(t) = at^3 + bt^2 + ct + d ). We have four points: when t is 1, 2, 3, and 4, the corresponding P(t) values are 10, 12, 18, and 34. So, we can set up four equations based on these points.Let me write them out:1. When t = 1: ( a(1)^3 + b(1)^2 + c(1) + d = 10 ) Simplifies to: ( a + b + c + d = 10 )2. When t = 2: ( a(2)^3 + b(2)^2 + c(2) + d = 12 ) Simplifies to: ( 8a + 4b + 2c + d = 12 )3. When t = 3: ( a(3)^3 + b(3)^2 + c(3) + d = 18 ) Simplifies to: ( 27a + 9b + 3c + d = 18 )4. When t = 4: ( a(4)^3 + b(4)^2 + c(4) + d = 34 ) Simplifies to: ( 64a + 16b + 4c + d = 34 )So now I have a system of four equations:1. ( a + b + c + d = 10 ) -- Equation (1)2. ( 8a + 4b + 2c + d = 12 ) -- Equation (2)3. ( 27a + 9b + 3c + d = 18 ) -- Equation (3)4. ( 64a + 16b + 4c + d = 34 ) -- Equation (4)I need to solve for a, b, c, d. Let me try subtracting equations to eliminate variables step by step.First, subtract Equation (1) from Equation (2):Equation (2) - Equation (1):( (8a - a) + (4b - b) + (2c - c) + (d - d) = 12 - 10 )Which simplifies to:( 7a + 3b + c = 2 ) -- Let's call this Equation (5)Next, subtract Equation (2) from Equation (3):Equation (3) - Equation (2):( (27a - 8a) + (9b - 4b) + (3c - 2c) + (d - d) = 18 - 12 )Simplifies to:( 19a + 5b + c = 6 ) -- Equation (6)Then, subtract Equation (3) from Equation (4):Equation (4) - Equation (3):( (64a - 27a) + (16b - 9b) + (4c - 3c) + (d - d) = 34 - 18 )Simplifies to:( 37a + 7b + c = 16 ) -- Equation (7)Now, I have three new equations: (5), (6), and (7). Let me write them:5. ( 7a + 3b + c = 2 )6. ( 19a + 5b + c = 6 )7. ( 37a + 7b + c = 16 )Now, let's subtract Equation (5) from Equation (6):Equation (6) - Equation (5):( (19a - 7a) + (5b - 3b) + (c - c) = 6 - 2 )Simplifies to:( 12a + 2b = 4 )Divide both sides by 2:( 6a + b = 2 ) -- Equation (8)Similarly, subtract Equation (6) from Equation (7):Equation (7) - Equation (6):( (37a - 19a) + (7b - 5b) + (c - c) = 16 - 6 )Simplifies to:( 18a + 2b = 10 )Divide both sides by 2:( 9a + b = 5 ) -- Equation (9)Now, we have Equations (8) and (9):8. ( 6a + b = 2 )9. ( 9a + b = 5 )Subtract Equation (8) from Equation (9):( (9a - 6a) + (b - b) = 5 - 2 )Simplifies to:( 3a = 3 )So, ( a = 1 )Now, plug a = 1 into Equation (8):( 6(1) + b = 2 )So, ( 6 + b = 2 )Thus, ( b = 2 - 6 = -4 )Now, we can find c using Equation (5):Equation (5): ( 7a + 3b + c = 2 )Plug in a = 1, b = -4:( 7(1) + 3(-4) + c = 2 )Simplify:( 7 - 12 + c = 2 )( -5 + c = 2 )So, ( c = 2 + 5 = 7 )Finally, find d using Equation (1):Equation (1): ( a + b + c + d = 10 )Plug in a = 1, b = -4, c = 7:( 1 - 4 + 7 + d = 10 )Simplify:( 4 + d = 10 )So, ( d = 10 - 4 = 6 )So, the coefficients are:a = 1, b = -4, c = 7, d = 6Let me double-check these values with all four original equations to make sure.1. ( 1 - 4 + 7 + 6 = 10 ) → 10 = 10 ✔️2. ( 8(1) + 4(-4) + 2(7) + 6 = 8 - 16 + 14 + 6 = 12 ) → 12 = 12 ✔️3. ( 27(1) + 9(-4) + 3(7) + 6 = 27 - 36 + 21 + 6 = 18 ) → 18 = 18 ✔️4. ( 64(1) + 16(-4) + 4(7) + 6 = 64 - 64 + 28 + 6 = 34 ) → 34 = 34 ✔️All equations check out. So, the coefficients are correct.Sub-problem 2: Calculating k and the probabilityAlright, moving on to the second problem about the surgery success rate. The probability density function is given by ( f(x) = k e^{-k x} ), which looks like an exponential distribution. The expected value (mean) is given as 4 hours. We need to find k and then the probability that the surgery is successful within the first 2 hours.First, for an exponential distribution, the expected value (mean) is ( frac{1}{k} ). So, if the mean is 4, then:( frac{1}{k} = 4 ) → ( k = frac{1}{4} )So, k is 1/4.Now, the probability that the surgery is successful within the first 2 hours is the integral of the PDF from 0 to 2. For an exponential distribution, the cumulative distribution function (CDF) is ( F(x) = 1 - e^{-k x} ). So, the probability P(X ≤ 2) is:( F(2) = 1 - e^{-k cdot 2} )Plugging in k = 1/4:( F(2) = 1 - e^{-(1/4) cdot 2} = 1 - e^{-1/2} )Calculating ( e^{-1/2} ) is approximately 0.6065, so:( F(2) ≈ 1 - 0.6065 = 0.3935 )So, approximately 39.35% probability.Let me verify this step by step.First, exponential distribution PDF: ( f(x) = k e^{-k x} ) for x ≥ 0.The CDF is indeed ( F(x) = 1 - e^{-k x} ).Mean (expected value) is ( frac{1}{k} ), so if mean is 4, k = 1/4.Then, P(X ≤ 2) = F(2) = 1 - e^{-2*(1/4)} = 1 - e^{-0.5}.Calculating e^{-0.5} is approximately 0.6065, so 1 - 0.6065 = 0.3935.Yes, that seems correct.So, summarizing:- k = 1/4- Probability within first 2 hours ≈ 0.3935 or 39.35%I think that's all for both sub-problems.Final AnswerSub-problem 1: The coefficients are ( a = boxed{1} ), ( b = boxed{-4} ), ( c = boxed{7} ), and ( d = boxed{6} ).Sub-problem 2: The value of ( k ) is ( boxed{dfrac{1}{4}} ) and the probability is approximately ( boxed{0.3935} ).</think>

🧠:<think>Okay, so I've got these two probability problems to solve. Let me take them one at a time.Problem 1:We need to find the probability that the total number of books published by both celebrities and professional authors in a given year is exactly 25. Given:- Celebrities' books follow a Poisson distribution with λ_c = 5.- Professional authors' books follow a Poisson distribution with λ_a = 20.I remember that the sum of two independent Poisson random variables is also Poisson, with the parameter being the sum of the individual parameters. So, if X ~ Poisson(λ_c) and Y ~ Poisson(λ_a), then X + Y ~ Poisson(λ_c + λ_a).Let me verify that. Yes, the Poisson distribution is additive, so the combined distribution should be Poisson(25). Therefore, the probability that the total is exactly 25 is given by the Poisson probability formula:P(X + Y = 25) = (e^{-25} * 25^{25}) / 25!Wait, hold on. Is that correct? Because λ_c + λ_a is 25, so yes, the combined rate is 25 per year. So, the probability mass function for Poisson(25) evaluated at 25 is indeed (e^{-25} * 25^{25}) / 25!.But just to make sure, let me think about it another way. Suppose we model the total as the sum of two independent Poisson variables. The probability that X = k and Y = 25 - k for k from 0 to 25. So, the total probability is the sum over k=0 to 25 of [P(X=k) * P(Y=25 - k)]. Which would be sum_{k=0}^{25} [ (e^{-5} * 5^k / k!) * (e^{-20} * 20^{25 - k} / (25 - k)! ) ].But since 5 + 20 = 25, this sum is equal to e^{-25} * 25^{25} / 25! because of the properties of Poisson distributions. So, yes, that's correct.Therefore, the probability is (e^{-25} * 25^{25}) / 25!.I can compute this numerically if needed, but since the question just asks for the probability, expressing it in terms of the Poisson PMF is sufficient.Problem 2:Now, this is a Bayesian probability problem. We need to find the probability that a book with a critical rating of exactly 6 was written by a professional author.Given:- Ratings for celebrities: Normal(μ_c = 4, σ_c² = 1.5)- Ratings for professional authors: Normal(μ_a = 7, σ_a² = 2)- The number of books by each group follows Poisson distributions as in Problem 1: λ_c = 5, λ_a = 20.So, we need to compute P(Author is Professional | Rating = 6).Using Bayes' theorem:P(A|X) = [P(X|A) * P(A)] / P(X)Where:- A is the event that the author is professional.- X is the event that the rating is 6.First, we need to find P(A), the prior probability that a randomly selected book is by a professional author. Since the number of books is Poisson distributed, the expected number of books by professionals is 20, and by celebrities is 5. So, the total expected number is 25. Therefore, P(A) = 20 / 25 = 4/5 = 0.8.Similarly, P(not A) = 5 / 25 = 1/5 = 0.2.Next, we need P(X|A) and P(X|not A). Since the ratings are continuous variables, the probability that the rating is exactly 6 is technically zero. But I think the question is referring to the probability density at 6.So, P(X=6|A) is the probability density of the normal distribution N(7, 2) at 6.Similarly, P(X=6|not A) is the probability density of N(4, 1.5) at 6.Let me compute these densities.For the professional author's rating:- μ = 7, σ² = 2, so σ = sqrt(2) ≈ 1.4142.- The density at 6 is (1 / (σ * sqrt(2π))) * e^{ - (6 - 7)^2 / (2 * 2) } = (1 / (sqrt(2) * sqrt(2π))) * e^{-1/4}.Simplify:= (1 / (sqrt(4π))) * e^{-0.25}≈ (1 / (2.5066)) * 0.7788≈ 0.3989 * 0.7788 ≈ 0.311.Wait, let me compute it step by step.First, the standard normal density at (6 - 7)/sqrt(2) = (-1)/1.4142 ≈ -0.7071.But actually, the density is:f(6) = (1 / (sqrt(2π) * sqrt(2))) * e^{ - (6 - 7)^2 / (2 * 2) }= (1 / (sqrt(4π))) * e^{-1/4}= (1 / (2 * sqrt(π))) * e^{-0.25}≈ (1 / (2 * 1.77245)) * 0.7788≈ (0.28209) * 0.7788 ≈ 0.2197.Wait, that seems different. Let me recalculate.Wait, the formula is:f(x) = (1 / (σ sqrt(2π))) e^{ - (x - μ)^2 / (2 σ²) }So, for professional authors:σ = sqrt(2) ≈ 1.4142f(6) = (1 / (1.4142 * sqrt(2π))) * e^{ - (6 - 7)^2 / (2 * 2) }= (1 / (1.4142 * 2.5066)) * e^{-1/4}≈ (1 / 3.5355) * 0.7788≈ 0.2829 * 0.7788 ≈ 0.2197.Similarly, for celebrities:μ = 4, σ² = 1.5, so σ ≈ 1.2247.f(6) = (1 / (1.2247 * sqrt(2π))) * e^{ - (6 - 4)^2 / (2 * 1.5) }= (1 / (1.2247 * 2.5066)) * e^{-4 / 3}≈ (1 / 3.070) * e^{-1.3333}≈ 0.3257 * 0.2636 ≈ 0.0858.So, P(X=6|A) ≈ 0.2197P(X=6|not A) ≈ 0.0858Now, applying Bayes' theorem:P(A|X) = [P(X|A) * P(A)] / [P(X|A) * P(A) + P(X|not A) * P(not A)]Plugging in the numbers:= [0.2197 * 0.8] / [0.2197 * 0.8 + 0.0858 * 0.2]= [0.1758] / [0.1758 + 0.01716]= 0.1758 / 0.19296 ≈ 0.911.So, approximately 91.1% probability that the book was written by a professional author.Wait, let me double-check the calculations.Compute numerator: 0.2197 * 0.8 = 0.17576Compute denominator: 0.17576 + (0.0858 * 0.2) = 0.17576 + 0.01716 = 0.19292So, 0.17576 / 0.19292 ≈ 0.911.Yes, that seems correct.Alternatively, using exact expressions:P(A|X) = [f(6|A) * P(A)] / [f(6|A) * P(A) + f(6|not A) * P(not A)]Where f(6|A) is the normal density as above, same for f(6|not A).So, the final probability is approximately 0.911 or 91.1%.Final Answer1. The probability is boxed{dfrac{e^{-25} cdot 25^{25}}{25!}}.2. The probability is approximately boxed{0.911}.</think>