🧠:<think>Okay, so I have this problem about modeling the impact of open access policies on medical journals. It's divided into two sub-problems. Let me take them one by one.Sub-problem 1: The differential equation given is (frac{dC}{dt} = kC(1 - frac{C}{M})). This looks familiar—it's the logistic growth model, right? So, it models population growth with a carrying capacity. In this case, it's modeling the number of citations (C) over time (t).Given:- Initial condition: (C(0) = 100)- Carrying capacity (M = 1000)- Growth rate (k = 0.1)I need to solve this differential equation to find (C(t)).First, I remember that the logistic equation can be solved using separation of variables. Let me write it out:[frac{dC}{dt} = kCleft(1 - frac{C}{M}right)]Rewriting this, we get:[frac{dC}{Cleft(1 - frac{C}{M}right)} = k , dt]To integrate both sides, I can use partial fractions on the left side. Let me set up the integral:[int frac{1}{Cleft(1 - frac{C}{M}right)} dC = int k , dt]Let me simplify the integrand on the left. Let me make a substitution to make it easier. Let (u = frac{C}{M}), so (C = Mu) and (dC = M du). Substituting, the integral becomes:[int frac{1}{Mu(1 - u)} cdot M du = int frac{1}{u(1 - u)} du]That simplifies nicely. Now, I can decompose (frac{1}{u(1 - u)}) into partial fractions:[frac{1}{u(1 - u)} = frac{A}{u} + frac{B}{1 - u}]Multiplying both sides by (u(1 - u)):[1 = A(1 - u) + B u]Let me solve for A and B. Let me set (u = 0):[1 = A(1 - 0) + B(0) implies A = 1]Now, set (u = 1):[1 = A(1 - 1) + B(1) implies B = 1]So, the partial fractions decomposition is:[frac{1}{u(1 - u)} = frac{1}{u} + frac{1}{1 - u}]Therefore, the integral becomes:[int left( frac{1}{u} + frac{1}{1 - u} right) du = int frac{1}{k} dt]Wait, actually, the right side was (int k dt), so it's (kt + C). Let me correct that.So, integrating the left side:[int frac{1}{u} du + int frac{1}{1 - u} du = ln|u| - ln|1 - u| + C]Which simplifies to:[lnleft|frac{u}{1 - u}right| + C]Substituting back (u = frac{C}{M}):[lnleft|frac{frac{C}{M}}{1 - frac{C}{M}}right| = kt + C]Simplify the argument of the logarithm:[lnleft|frac{C}{M - C}right| = kt + C]Exponentiating both sides to eliminate the logarithm:[frac{C}{M - C} = e^{kt + C} = e^{kt} cdot e^{C}]Let me denote (e^{C}) as a constant (K). So,[frac{C}{M - C} = K e^{kt}]Solving for (C):Multiply both sides by (M - C):[C = K e^{kt} (M - C)]Expand the right side:[C = K M e^{kt} - K C e^{kt}]Bring all terms with (C) to the left:[C + K C e^{kt} = K M e^{kt}]Factor out (C):[C (1 + K e^{kt}) = K M e^{kt}]Solve for (C):[C = frac{K M e^{kt}}{1 + K e^{kt}}]Now, let's apply the initial condition (C(0) = 100). At (t = 0):[100 = frac{K M e^{0}}{1 + K e^{0}} = frac{K M}{1 + K}]We know (M = 1000), so:[100 = frac{K cdot 1000}{1 + K}]Multiply both sides by (1 + K):[100 (1 + K) = 1000 K]Expand:[100 + 100 K = 1000 K]Subtract (100 K) from both sides:[100 = 900 K]So,[K = frac{100}{900} = frac{1}{9}]Therefore, the expression for (C(t)) is:[C(t) = frac{frac{1}{9} cdot 1000 cdot e^{0.1 t}}{1 + frac{1}{9} e^{0.1 t}} = frac{frac{1000}{9} e^{0.1 t}}{1 + frac{1}{9} e^{0.1 t}}]Simplify numerator and denominator by multiplying numerator and denominator by 9:[C(t) = frac{1000 e^{0.1 t}}{9 + e^{0.1 t}}]Alternatively, factor out (e^{0.1 t}) in the denominator:[C(t) = frac{1000 e^{0.1 t}}{e^{0.1 t} + 9} = frac{1000}{1 + 9 e^{-0.1 t}}]Either form is acceptable, but the second form might be more intuitive as it shows the approach to the carrying capacity as (t) increases.So, that's the solution for Sub-problem 1.Sub-problem 2: Now, the advisor wants to consider the economic impact. The cost function is given by (E(t) = E_0 e^{rt}), where (E_0 = 10,000) and (r = 0.05). We need to calculate the total cost incurred over the first 10 years.Total cost over a period is typically the integral of the cost function over that time. So, the total cost (TC) from (t = 0) to (t = 10) is:[TC = int_{0}^{10} E(t) , dt = int_{0}^{10} 10,000 e^{0.05 t} , dt]Let me compute this integral.First, factor out the constant:[TC = 10,000 int_{0}^{10} e^{0.05 t} , dt]The integral of (e^{kt}) is (frac{1}{k} e^{kt}), so here (k = 0.05):[int e^{0.05 t} dt = frac{1}{0.05} e^{0.05 t} + C = 20 e^{0.05 t} + C]Therefore, evaluating from 0 to 10:[TC = 10,000 left[ 20 e^{0.05 cdot 10} - 20 e^{0} right] = 10,000 times 20 left( e^{0.5} - 1 right)]Compute (e^{0.5}). I know that (e^{0.5} approx 1.64872).So,[TC = 10,000 times 20 times (1.64872 - 1) = 10,000 times 20 times 0.64872]Calculate step by step:First, 20 * 0.64872 = 12.9744Then, 10,000 * 12.9744 = 129,744So, the total cost incurred over the first 10 years is approximately 129,744.Wait, let me double-check the calculations.Compute (e^{0.5}):Yes, (e^{0.5}) is approximately 1.64872.So, (1.64872 - 1 = 0.64872).Multiply by 20: 0.64872 * 20 = 12.9744.Multiply by 10,000: 12.9744 * 10,000 = 129,744.Yes, that's correct.Alternatively, if I compute it more precisely:(e^{0.5}) is about 1.6487212707.So, 1.6487212707 - 1 = 0.6487212707.Multiply by 20: 0.6487212707 * 20 = 12.974425414.Multiply by 10,000: 12.974425414 * 10,000 = 129,744.25414.So, approximately 129,744.25.Depending on rounding, it's about 129,744.Therefore, the total cost is approximately 129,744 over the first 10 years.Final Answer1. The number of citations as a function of time is (boxed{dfrac{1000}{1 + 9e^{-0.1t}}}).2. The total cost incurred over the first 10 years is (boxed{129744}) dollars.</think>

🧠:<think>Alright, so I've got these two problems to tackle. Let me start with the first one.Problem 1 is about finding a subgraph H in a given graph G that maximizes the total interaction strength. Each edge has a weight, and I need to formulate an optimization problem for this. Hmm, okay. So, I think this is similar to finding a maximum weight subgraph. But wait, what exactly defines a subgraph here? A subgraph can include any subset of nodes and edges, right? But in this case, since the nodes are drug compounds and edges are interactions, maybe the subgraph H should include all nodes but only a subset of edges? Or perhaps it's about selecting nodes such that the sum of the edges between them is maximized? That might be more like a clique problem, but with weights.Wait, the question says H is a subgraph, so it can include any subset of nodes and edges. But to maximize the total interaction strength, which is the sum of the weights of the edges in H. So, essentially, I need to select a subset of edges E' ⊆ E such that the sum of their weights is maximized. But that's trivial because you can just take all edges with positive weights. But maybe there's a constraint, like the subgraph has to be connected? Or perhaps it's about selecting a subset of nodes and all edges between them, which would be a clique. But the problem doesn't specify any constraints, so maybe it's just selecting the edges with the highest weights. But that seems too simple.Wait, maybe I'm misunderstanding. If H is a subgraph, it can be any subset of nodes and edges, but perhaps the problem is to find a connected subgraph with maximum total edge weight. That would make more sense because otherwise, it's just selecting all positive edges. So, perhaps the problem is to find a connected subgraph H where the sum of the edge weights is maximized. That sounds more challenging.Alternatively, maybe it's about selecting a subset of nodes such that the sum of the weights of the edges within that subset is maximized. That would be like a maximum weight clique problem, but in a general graph, which is also NP-hard. Hmm.Wait, the question says "subgraph H ⊆ G", which usually means H can be any subset of nodes and edges, but perhaps the problem is to select a subset of edges such that the sum is maximized. But without any constraints, that's just selecting all edges with positive weights. So maybe the problem is more about selecting a subset of nodes and including all edges between them, which would be a clique. But in that case, the problem is equivalent to the maximum clique problem, which is NP-hard.Alternatively, maybe it's about finding a spanning subgraph, which includes all nodes but a subset of edges. But again, without constraints, it's just selecting all positive edges.Wait, perhaps the problem is to find a connected subgraph with maximum total edge weight. That would make sense because in a disconnected subgraph, you could have multiple components, but maybe the problem wants a single connected component. So, the optimization problem would be: find a connected subgraph H such that the sum of the weights of its edges is maximized.Yes, that seems plausible. So, the formulation would be:Maximize Σ w(e) for all e in E'Subject to:- E' is a subset of E- The subgraph (V', E') is connected, where V' is the set of nodes incident to E'But wait, actually, V' can be any subset of V, and E' is a subset of E where all edges in E' have both endpoints in V'. So, the problem is to choose V' ⊆ V and E' ⊆ E such that E' only contains edges between nodes in V', and the subgraph (V', E') is connected, and the sum of w(e) for e in E' is maximized.Alternatively, if we don't require connectivity, it's trivial, so I think the problem is assuming connectivity.So, the optimization problem is to find a connected subgraph H with maximum total edge weight.Now, to prove that this problem is NP-hard. I know that the maximum weight clique problem is NP-hard. If I can reduce the maximum weight clique problem to this problem, then it would show that this problem is also NP-hard.Wait, but in the maximum weight clique problem, we're looking for a subset of nodes where every pair is connected by an edge (i.e., a clique) and the sum of the weights is maximized. In our case, we're looking for a connected subgraph, which is a more general concept because a connected subgraph doesn't have to be a clique; it just needs to be connected.But perhaps I can transform an instance of the maximum weight clique problem into an instance of our problem. Let me think.Suppose I have a graph G where I want to find a maximum weight clique. I can create a new graph G' where each node is the same, but I assign weights to the edges such that edges not present in G have a very low weight (like negative infinity or something). Then, finding a connected subgraph with maximum total weight in G' would be equivalent to finding a clique in G with maximum total edge weight, because non-clique edges would have such low weights that they wouldn't be included. But wait, in our problem, the subgraph must be connected, but in the clique problem, the clique is automatically connected. So, if I can show that solving our problem for G' would give me the maximum weight clique in G, then since the clique problem is NP-hard, our problem is also NP-hard.Alternatively, another approach is to note that the problem of finding a connected subgraph with maximum total edge weight is equivalent to the problem of finding a maximum weight tree, but that's not exactly the same because a tree is minimally connected, whereas our subgraph can have cycles.Wait, actually, the maximum weight connected subgraph problem is known to be NP-hard. I think it's a standard result. So, perhaps I can just cite that.But to be thorough, let me try to construct a reduction. Let's take the maximum weight clique problem. Given a graph G with edge weights, we can transform it into an instance of our problem by keeping the same graph and weights. Then, any clique in G is a connected subgraph in our problem, and the maximum weight clique would correspond to the maximum weight connected subgraph. But wait, is that true? Because in our problem, the connected subgraph can include more edges than just those in the clique, but those edges would have to be within the subset of nodes. However, if the clique has the maximum total edge weight, then adding any other edges would either not increase the total (if their weights are non-positive) or could potentially increase it, but in the clique problem, we're restricted to cliques.Hmm, maybe this reduction isn't straightforward. Alternatively, perhaps I can use the fact that the problem is a generalization of the maximum weight clique problem, and since it's more general, it's at least as hard.Alternatively, consider the problem where all edge weights are positive. Then, the maximum weight connected subgraph would be the entire graph, which is trivial. But if we have both positive and negative edge weights, then it's non-trivial. So, perhaps the problem is only interesting when edge weights can be negative, which complicates things.Wait, but the original problem doesn't specify whether the weights are positive or can be negative. It just says each edge has an associated weight. So, perhaps the problem is to find a connected subgraph where the sum of the weights is maximized, regardless of whether individual edges are positive or negative.In that case, it's similar to the problem of finding a maximum weight connected subgraph, which is indeed NP-hard. I think this is a known result. For example, in the case where all edge weights are positive, the maximum weight connected subgraph is the entire graph, but when some edges have negative weights, you have to be careful.But to formally prove it's NP-hard, I can reduce from the maximum weight clique problem. Suppose I have a graph G with edge weights. I can create a new graph G' where each node is connected to every other node, but the edge weights are set as follows: for edges that exist in G, keep their weights, and for edges that don't exist in G, assign a very large negative weight. Then, finding a maximum weight connected subgraph in G' would force the subgraph to be a clique in G, because otherwise, including non-edges would introduce large negative weights, which would reduce the total. Therefore, the maximum weight connected subgraph in G' corresponds to the maximum weight clique in G. Since the maximum weight clique problem is NP-hard, our problem is also NP-hard.Okay, that seems like a valid reduction. So, the optimization problem is to find a connected subgraph H with maximum total edge weight, and it's NP-hard.Now, moving on to Problem 2. The differential equation is:dC(t)/dt = -kC(t) + Σ_{i=1}^n a_i sin(b_i t + c_i)We need to find the steady-state concentration C(t) if it exists and describe when it doesn't.A steady-state solution is a particular solution that the system approaches as t approaches infinity. For linear differential equations with periodic inputs, the steady-state solution is typically a combination of sinusoids with the same frequencies as the inputs.So, let's consider the homogeneous solution first. The homogeneous equation is dC/dt = -kC, which has the solution C_h(t) = C_0 e^{-kt}, where C_0 is the initial concentration.For the particular solution, since the input is a sum of sinusoids, we can find a particular solution for each term and sum them up. Each term is of the form a_i sin(b_i t + c_i). The particular solution for each term will be of the form A_i sin(b_i t + c_i) + B_i cos(b_i t + c_i).Wait, actually, since the forcing function is sin(b_i t + c_i), the particular solution can be written as A_i sin(b_i t + c_i) + B_i cos(b_i t + c_i). Alternatively, it can be written as M_i sin(b_i t + c_i + φ_i), where M_i and φ_i are constants to be determined.Let me proceed step by step.The differential equation is linear and nonhomogeneous. The general solution is the sum of the homogeneous solution and a particular solution.First, find the homogeneous solution:dC_h/dt = -k C_hSolution: C_h(t) = C_h0 e^{-kt}Now, find a particular solution C_p(t) for each term in the sum. Let's consider one term a_i sin(b_i t + c_i). We can write the particular solution for this term as:C_p,i(t) = A_i sin(b_i t + c_i) + B_i cos(b_i t + c_i)Now, compute the derivative:dC_p,i/dt = A_i b_i cos(b_i t + c_i) - B_i b_i sin(b_i t + c_i)Substitute into the differential equation:A_i b_i cos(b_i t + c_i) - B_i b_i sin(b_i t + c_i) = -k (A_i sin(b_i t + c_i) + B_i cos(b_i t + c_i)) + a_i sin(b_i t + c_i)Now, collect like terms:Left side: A_i b_i cos(...) - B_i b_i sin(...)Right side: -k A_i sin(...) - k B_i cos(...) + a_i sin(...)Equate coefficients of sin and cos:For sin(b_i t + c_i):- B_i b_i = -k A_i + a_iFor cos(b_i t + c_i):A_i b_i = -k B_iSo, we have a system of two equations:1. -B_i b_i = -k A_i + a_i2. A_i b_i = -k B_iFrom equation 2: A_i = (-k B_i)/b_iSubstitute into equation 1:- B_i b_i = -k (-k B_i)/b_i + a_iSimplify:- B_i b_i = (k^2 B_i)/b_i + a_iMultiply both sides by b_i:- B_i b_i^2 = k^2 B_i + a_i b_iBring all terms to one side:- B_i b_i^2 - k^2 B_i - a_i b_i = 0Factor out B_i:B_i (-b_i^2 - k^2) - a_i b_i = 0Solve for B_i:B_i = (a_i b_i) / (b_i^2 + k^2)Then, from equation 2:A_i = (-k B_i)/b_i = (-k (a_i b_i)/(b_i^2 + k^2))/b_i = -k a_i / (b_i^2 + k^2)So, the particular solution for each term is:C_p,i(t) = A_i sin(b_i t + c_i) + B_i cos(b_i t + c_i)= (-k a_i / (b_i^2 + k^2)) sin(b_i t + c_i) + (a_i b_i / (b_i^2 + k^2)) cos(b_i t + c_i)We can write this as:C_p,i(t) = (a_i / (b_i^2 + k^2)) [ -k sin(b_i t + c_i) + b_i cos(b_i t + c_i) ]This can be further simplified using the amplitude-phase form:Let’s factor out a_i / (b_i^2 + k^2):C_p,i(t) = (a_i / (b_i^2 + k^2)) [ b_i cos(b_i t + c_i) - k sin(b_i t + c_i) ]Notice that b_i cos(...) - k sin(...) can be written as M_i cos(b_i t + c_i + φ_i), where M_i = sqrt(b_i^2 + k^2) and tan φ_i = k / b_i.But since we already have the expression in terms of sin and cos, we can proceed.Therefore, the particular solution for each term is as above, and the total particular solution is the sum over all i:C_p(t) = Σ_{i=1}^n (a_i / (b_i^2 + k^2)) [ b_i cos(b_i t + c_i) - k sin(b_i t + c_i) ]So, the general solution is:C(t) = C_h(t) + C_p(t) = C_0 e^{-kt} + Σ_{i=1}^n (a_i / (b_i^2 + k^2)) [ b_i cos(b_i t + c_i) - k sin(b_i t + c_i) ]As t approaches infinity, the homogeneous solution C_h(t) tends to zero because e^{-kt} decays exponentially. Therefore, the steady-state concentration is the particular solution:C_ss(t) = Σ_{i=1}^n (a_i / (b_i^2 + k^2)) [ b_i cos(b_i t + c_i) - k sin(b_i t + c_i) ]This is a periodic function with the same frequencies as the inputs. So, the steady-state concentration exists and is periodic.However, if the system is not stable, meaning if k ≤ 0, then the homogeneous solution might not decay. Wait, in the original equation, k is a decay constant, so it should be positive. If k were zero or negative, the homogeneous solution would not decay, and the steady-state might not exist or be unbounded.So, the conditions for the steady-state to exist are that k > 0, ensuring that the transient solution decays to zero, leaving the periodic particular solution as the steady-state.Therefore, the steady-state concentration exists if k > 0 and is given by the sum above. If k ≤ 0, the homogeneous solution does not decay, and the concentration might not settle to a steady-state, potentially growing without bound or oscillating indefinitely without damping.Wait, but if k = 0, the equation becomes dC/dt = Σ a_i sin(b_i t + c_i), which is a linear differential equation with no decay. The solution would be the integral of the sum of sinusoids, which would be a sum of cosines, but without any decay, so the concentration would grow without bound if the inputs are persistent. Wait, no, integrating sin functions gives cos functions, but the integral over time would accumulate, leading to unbounded growth. So, if k = 0, the solution would not have a steady-state; it would keep increasing or oscillating with increasing amplitude.If k < 0, the homogeneous solution would grow exponentially, so the concentration would blow up unless the particular solution cancels it out, but that's not possible because the particular solution is bounded.Therefore, the steady-state exists if and only if k > 0. If k ≤ 0, the steady-state does not exist because the concentration either grows without bound or oscillates indefinitely without settling.So, summarizing:The steady-state concentration is the particular solution, which is a combination of sinusoids with the same frequencies as the inputs, scaled by factors involving the decay constant k. It exists provided that k > 0. If k ≤ 0, the steady-state does not exist because the homogeneous solution either remains constant or grows without bound, preventing the system from reaching a steady-state.I think that's the conclusion.</think>

🧠:<think>Alright, so Alex is organizing this charity event, and he's got two main problems to solve. Let me try to figure them out step by step.Starting with the first problem: calculating the total area of the banner. The banner is a large right-angled triangle made up of three smaller right-angled triangles. Each smaller triangle has legs of 3 feet, 4 feet, and 5 feet respectively. Also, each smaller triangle's hypotenuse becomes one leg of the next larger triangle. Hmm, okay, so it's like a fractal or something where each hypotenuse becomes a leg for the next triangle.First, I need to visualize this. There's a big triangle, and inside it, there are three smaller right-angled triangles arranged in such a way that each hypotenuse is a leg for the next. So, starting with the smallest triangle, which has legs 3 and 4 feet. Let me calculate its hypotenuse first.Using the Pythagorean theorem: c = sqrt(a² + b²). So for the first triangle, c = sqrt(3² + 4²) = sqrt(9 + 16) = sqrt(25) = 5 feet. Okay, so the hypotenuse is 5 feet, which becomes one leg of the next triangle. The next triangle has legs 4 feet and 5 feet? Wait, no, the problem says the legs are 3, 4, and 5 feet respectively. So maybe each subsequent triangle uses the hypotenuse as one leg and the next given leg as the other.Wait, let me read that again: "the legs of the smaller triangles have lengths 3 feet, 4 feet, and 5 feet, respectively. Each smaller triangle's hypotenuse forms one leg of the next larger triangle." So, the first triangle has legs 3 and 4, hypotenuse 5. The next triangle uses that hypotenuse (5 feet) as one leg and the next given leg, which is 5 feet? Wait, no, the legs are 3, 4, 5. So maybe the first triangle is 3 and 4, hypotenuse 5. Then the next triangle is 5 and 5? Because the next leg is 5? Or is the next leg 4? Wait, the legs are 3, 4, 5 respectively.Wait, maybe the three smaller triangles each have legs 3, 4, 5. So each is a right-angled triangle with legs 3, 4, 5? But that can't be because 3,4,5 is a Pythagorean triple, so each triangle is a 3-4-5 triangle. But then each hypotenuse is 5, which becomes a leg of the next triangle. So the next triangle would have legs 5 and something. But the legs are given as 3,4,5. So maybe it's arranged so that each subsequent triangle uses the hypotenuse as one leg and the next given leg as the other.Wait, maybe I need to consider that the three smaller triangles are arranged such that each hypotenuse is a leg for the next. So starting with 3 and 4, hypotenuse 5. Then the next triangle has legs 5 and 5? Because the next given leg is 5? Or is it 5 and 4? Hmm, this is confusing.Alternatively, perhaps the three smaller triangles each have legs 3, 4, and 5, but arranged in a way that their hypotenuses form the legs of the larger triangle. So the large triangle's legs would be the hypotenuses of the smaller triangles.Wait, the problem says: "the legs of the smaller triangles have lengths 3 feet, 4 feet, and 5 feet, respectively. Each smaller triangle's hypotenuse forms one leg of the next larger triangle." So, the first triangle has legs 3 and 4, hypotenuse 5. Then the next triangle has legs 5 and 5? Because the next given leg is 5. Then the hypotenuse of that triangle would be sqrt(5² + 5²) = sqrt(50) ≈ 7.07 feet. Then the next triangle would have legs 7.07 and 5? Wait, but the legs are given as 3,4,5. Maybe I'm overcomplicating.Wait, perhaps the three smaller triangles each have legs 3, 4, and 5, and each hypotenuse becomes a leg for the next triangle. So the first triangle is 3-4-5, hypotenuse 5. Then the next triangle has legs 5 and 4? Because the next given leg is 4? Wait, no, the legs are 3,4,5 respectively. So maybe the first triangle is 3-4-5, hypotenuse 5. Then the next triangle has legs 5 and 5? Because the next given leg is 5. Then the hypotenuse of that is sqrt(5² +5²)=5√2. Then the third triangle has legs 5√2 and 5? Wait, but the legs are given as 3,4,5. Maybe I'm misunderstanding.Wait, perhaps the three smaller triangles each have legs 3, 4, and 5, but arranged in a sequence where each hypotenuse is a leg for the next. So first triangle: 3 and 4, hypotenuse 5. Second triangle: 5 and 5, hypotenuse 5√2. Third triangle: 5√2 and 5, hypotenuse sqrt((5√2)^2 +5^2)=sqrt(50 +25)=sqrt(75)=5√3. Then the large triangle would have legs 5√3 and something? Wait, no, the large triangle is composed of these three smaller triangles. Maybe the large triangle's legs are the sum of the legs? Or maybe it's a right triangle where each smaller triangle is attached to form the larger one.Alternatively, perhaps the large triangle is made by arranging the three smaller triangles such that their hypotenuses form the legs. So the large triangle's legs are the hypotenuses of the smaller triangles. So first triangle: 3-4-5, hypotenuse 5. Second triangle: 4-5- something? Wait, no, the legs are 3,4,5 respectively. Maybe each triangle is built on the hypotenuse of the previous one.Wait, maybe it's better to think of it as a geometric progression. Each triangle's hypotenuse becomes a leg for the next, so the legs of the triangles are 3,4,5, then 5, something, something. But the problem says the legs of the smaller triangles are 3,4,5. So each triangle has legs 3,4,5? But that can't be because 3,4,5 is a right triangle, so each triangle is 3-4-5. Then how are they arranged to form a larger triangle?Wait, maybe the three smaller triangles are arranged such that their hypotenuses form the legs of the larger triangle. So the large triangle has legs equal to the hypotenuses of the smaller triangles. But each smaller triangle has hypotenuse 5, so the large triangle would have legs 5 and 5, making it an isosceles right triangle with area (5*5)/2=12.5. But that seems too small because the smaller triangles each have area (3*4)/2=6, so three of them would be 18, which is more than 12.5. So that can't be.Wait, maybe the large triangle is composed of the three smaller triangles. So the area of the large triangle is the sum of the areas of the three smaller triangles. So first triangle: 3-4-5, area 6. Second triangle: legs 4 and 5, area (4*5)/2=10. Third triangle: legs 5 and something? Wait, no, the legs are 3,4,5. Maybe each subsequent triangle uses the hypotenuse as one leg and the next given leg as the other.Wait, let me try again. The first triangle has legs 3 and 4, hypotenuse 5. The second triangle uses that hypotenuse (5) as one leg and the next given leg, which is 5, as the other leg. So second triangle: legs 5 and 5, hypotenuse 5√2. Third triangle uses that hypotenuse (5√2) as one leg and the next given leg, which is... but we only have three legs: 3,4,5. So maybe the third triangle uses 5√2 and 5? Wait, but the legs are given as 3,4,5. Maybe I'm overcomplicating.Alternatively, perhaps the three smaller triangles are arranged such that their hypotenuses form the legs of the larger triangle. So the large triangle's legs are the hypotenuses of the smaller triangles. So first triangle: 3-4-5, hypotenuse 5. Second triangle: 4-5- something? Wait, no, the legs are 3,4,5. Maybe the large triangle's legs are 5 and 5√2, but I'm not sure.Wait, maybe the total area is just the sum of the areas of the three smaller triangles. So first triangle: 3*4/2=6. Second triangle: 4*5/2=10. Third triangle: 5* something? Wait, no, the legs are 3,4,5. Maybe each triangle is 3-4-5, so each has area 6. So three triangles would be 18. But the large triangle's area would be more than that, right? Because it's composed of them.Wait, perhaps the large triangle is a right triangle whose legs are the sum of the legs of the smaller triangles. So if the smaller triangles have legs 3,4,5, then the large triangle's legs would be 3+4=7 and 4+5=9? No, that doesn't make sense because the hypotenuse would then be sqrt(7² +9²)=sqrt(49+81)=sqrt(130)≈11.4. But then the area would be (7*9)/2=31.5. But the sum of the smaller areas is 6+10+ something? Wait, no, the smaller triangles are each 3-4-5, so each has area 6, so three would be 18. But 18 is less than 31.5, so that can't be.Wait, maybe the large triangle is made by arranging the three smaller triangles such that each hypotenuse is a leg. So first triangle: 3-4-5. Then the second triangle has legs 5 and 4, hypotenuse sqrt(5² +4²)=sqrt(41). Then the third triangle has legs sqrt(41) and 5, hypotenuse sqrt(41 +25)=sqrt(66). Then the large triangle would have legs sqrt(66) and something? I'm getting confused.Alternatively, maybe the large triangle is a right triangle where each of its legs is composed of the hypotenuses of the smaller triangles. So if the smaller triangles have hypotenuses 5, sqrt(41), and sqrt(66), then the large triangle's legs would be 5 + sqrt(41) + sqrt(66). But that seems too complicated.Wait, maybe I'm overcomplicating. Let me think differently. The problem says the banner is a large right-angled triangle composed of three smaller right-angled triangles. Each smaller triangle's hypotenuse forms one leg of the next larger triangle. So the first triangle has legs 3 and 4, hypotenuse 5. Then the second triangle has legs 5 and 4, hypotenuse sqrt(5² +4²)=sqrt(41). Then the third triangle has legs sqrt(41) and 5, hypotenuse sqrt(41 +25)=sqrt(66). Then the large triangle would have legs sqrt(66) and something? Wait, no, the large triangle is the combination of the three smaller ones. Maybe the large triangle's legs are the sum of the legs of the smaller triangles? So 3+4+5=12 for one leg and 4+5+ something? I'm not sure.Wait, maybe the large triangle's legs are the hypotenuses of the smaller triangles. So the first triangle's hypotenuse is 5, the second is sqrt(41), the third is sqrt(66). So the large triangle's legs would be 5 and sqrt(41), making its area (5*sqrt(41))/2. But that doesn't seem right because the sum of the smaller areas would be 6 + (5*4)/2=6+10=16, plus the third triangle's area which is (sqrt(41)*5)/2≈(6.4*5)/2≈16. So total area would be 6+10+16≈32, but the large triangle's area would be (5*sqrt(41))/2≈(5*6.4)/2≈16, which is less than the sum. So that can't be.Wait, maybe the large triangle's legs are the sum of the legs of the smaller triangles. So one leg is 3+5=8, and the other is 4+5=9. Then the area would be (8*9)/2=36. The sum of the smaller areas is 6 (from 3-4-5) + 10 (from 5-4-?) + something. Wait, no, the second triangle would have legs 5 and 4, area 10, and the third triangle would have legs 5 and 5, area 12.5. So total area would be 6+10+12.5=28.5, which is less than 36. Hmm, not sure.Wait, maybe the large triangle is a right triangle where each of its legs is the sum of the legs of the smaller triangles. So one leg is 3+4=7, and the other is 4+5=9. Then the area would be (7*9)/2=31.5. The sum of the smaller areas is 6 (3-4-5) + 10 (4-5-?) + (5-5-?)=6+10+12.5=28.5. So 31.5 is the area of the large triangle, but the sum of the smaller ones is 28.5. That doesn't add up because the large triangle should encompass the smaller ones, so its area should be equal to the sum of the smaller ones. So 31.5 vs 28.5, which is a discrepancy. Maybe I'm missing something.Wait, perhaps the large triangle is made by arranging the three smaller triangles such that their hypotenuses form the legs. So the large triangle's legs are the hypotenuses of the smaller triangles. So first triangle: 3-4-5, hypotenuse 5. Second triangle: 4-5- something, hypotenuse sqrt(41). Third triangle: 5- something- something, hypotenuse sqrt(66). So the large triangle's legs are 5 and sqrt(41), making its area (5*sqrt(41))/2≈16. Then the sum of the smaller areas is 6 (3-4-5) + 10 (4-5-sqrt(41)) + (5*sqrt(41))/2≈6+10+16=32. But the large triangle's area is only 16, which is half of that. So that can't be.Wait, maybe the large triangle is a right triangle where each of its legs is the hypotenuse of the smaller triangles. So the first triangle's hypotenuse is 5, the second is sqrt(41), the third is sqrt(66). So the large triangle's legs are 5 and sqrt(41), making its area (5*sqrt(41))/2≈16. But the sum of the smaller areas is 6+10+ (5*sqrt(41))/2≈6+10+16=32. So the large triangle's area is 16, which is half of 32. That suggests that the large triangle is half the area of the sum of the smaller ones, which doesn't make sense because the large triangle should encompass them.Wait, maybe I'm approaching this wrong. Let me think of it as a geometric series. Each triangle's hypotenuse becomes a leg for the next, so the legs are 3,4,5, then 5, something, something. But the problem says the legs of the smaller triangles are 3,4,5. So each triangle has legs 3,4,5. So each is a 3-4-5 triangle. Then how are they arranged? Maybe they are arranged such that their hypotenuses form the legs of the larger triangle. So the large triangle's legs are 5 and 5, making it an isosceles right triangle with area (5*5)/2=12.5. But the sum of the smaller areas is 3*(3*4/2)=18, which is more than 12.5. So that can't be.Wait, maybe the large triangle is made by arranging the three smaller triangles in a way that their hypotenuses form the legs. So the large triangle's legs are 5 and 5√2, making its area (5*5√2)/2≈17.7. The sum of the smaller areas is 6 (3-4-5) + 10 (5-4-?) + (5*5√2)/2≈6+10+17.7=33.7. But the large triangle's area is 17.7, which is less than the sum. So that doesn't make sense.Wait, maybe the large triangle is a right triangle where each of its legs is the sum of the legs of the smaller triangles. So one leg is 3+4=7, the other is 4+5=9. Then the area is (7*9)/2=31.5. The sum of the smaller areas is 6 (3-4-5) + 10 (4-5-?) + (5*5)/2=6+10+12.5=28.5. So 31.5 is the area of the large triangle, but the sum of the smaller ones is 28.5. That suggests that the large triangle is slightly larger, which makes sense because the smaller triangles are arranged within it. So maybe the total area is 31.5 square feet.But wait, let me think again. The problem says the banner is composed of three smaller right-angled triangles arranged to form a larger right-angled triangle. Each smaller triangle's hypotenuse forms one leg of the next larger triangle. So starting with the first triangle: legs 3 and 4, hypotenuse 5. Then the next triangle has legs 5 and 4, hypotenuse sqrt(5² +4²)=sqrt(41). Then the next triangle has legs sqrt(41) and 5, hypotenuse sqrt(41 +25)=sqrt(66). So the large triangle's legs are sqrt(66) and something? Wait, no, the large triangle is the combination of these three. So the large triangle's legs would be the sum of the legs of the smaller triangles? Or maybe the large triangle's legs are the hypotenuses of the smaller triangles.Wait, maybe the large triangle's legs are the hypotenuses of the smaller triangles. So first triangle: hypotenuse 5. Second triangle: hypotenuse sqrt(41). Third triangle: hypotenuse sqrt(66). So the large triangle's legs are 5 and sqrt(41), making its area (5*sqrt(41))/2≈16. But the sum of the smaller areas is 6 (3-4-5) + 10 (5-4-sqrt(41)) + (sqrt(41)*5)/2≈6+10+16=32. So the large triangle's area is 16, which is half of 32. That suggests that the large triangle is half the area of the sum of the smaller ones, which doesn't make sense because the large triangle should encompass them.Wait, maybe the large triangle is made by arranging the three smaller triangles such that each hypotenuse is a leg, and the large triangle's legs are the sum of the legs of the smaller triangles. So one leg is 3+5=8, the other is 4+5=9. Then the area is (8*9)/2=36. The sum of the smaller areas is 6 (3-4-5) + 10 (5-4-sqrt(41)) + (sqrt(41)*5)/2≈6+10+16=32. So 36 is the area of the large triangle, which is slightly larger than the sum of the smaller ones. That seems plausible.But I'm not entirely sure. Maybe I should calculate the area of each smaller triangle and then see how they fit into the large one.First triangle: 3-4-5, area=6.Second triangle: legs 5 and 4, area=10.Third triangle: legs 5 and 5, area=12.5.Total area of smaller triangles: 6+10+12.5=28.5.If the large triangle's area is 36, then 28.5 is less than 36, which makes sense because the large triangle encompasses them. But how is the large triangle formed? It's a right triangle with legs 8 and 9, area 36. But how does that relate to the smaller triangles?Alternatively, maybe the large triangle's legs are 5 and 5√2, making its area (5*5√2)/2≈17.7. But that's less than the sum of the smaller areas.Wait, maybe the large triangle is a right triangle where each of its legs is the hypotenuse of the smaller triangles. So the first triangle's hypotenuse is 5, the second is sqrt(41), the third is sqrt(66). So the large triangle's legs are 5 and sqrt(41), making its area (5*sqrt(41))/2≈16. But the sum of the smaller areas is 6+10+ (sqrt(41)*5)/2≈6+10+16=32. So the large triangle's area is 16, which is half of 32. That suggests that the large triangle is half the area of the sum of the smaller ones, which doesn't make sense because the large triangle should encompass them.Wait, maybe I'm approaching this wrong. Let me think of it as a geometric series where each triangle's hypotenuse is a leg for the next. So starting with 3 and 4, hypotenuse 5. Then the next triangle has legs 5 and 4, hypotenuse sqrt(41). Then the next triangle has legs sqrt(41) and 5, hypotenuse sqrt(41 +25)=sqrt(66). So the large triangle's legs are sqrt(66) and something? Wait, no, the large triangle is the combination of these three. So the large triangle's legs would be the sum of the legs of the smaller triangles? Or maybe the large triangle's legs are the hypotenuses of the smaller triangles.Wait, maybe the large triangle's legs are the sum of the legs of the smaller triangles. So one leg is 3+5=8, the other is 4+5=9. Then the area is (8*9)/2=36. The sum of the smaller areas is 6 (3-4-5) + 10 (5-4-sqrt(41)) + (sqrt(41)*5)/2≈6+10+16=32. So 36 is the area of the large triangle, which is slightly larger than the sum of the smaller ones. That seems plausible.But I'm still not entirely confident. Maybe I should calculate the area of the large triangle based on the arrangement. If the large triangle is made by attaching the three smaller triangles such that each hypotenuse is a leg, then the large triangle's legs would be the sum of the legs of the smaller triangles. So one leg is 3+5=8, the other is 4+5=9. Then the area is (8*9)/2=36. That seems logical.So, for the first problem, the total area of the banner is 36 square feet.Now, moving on to the second problem: the sound system's power supply. The performance lasts 1 hour and 20 minutes, which is 1.333... hours. The sound system consumes power at 2 kilowatts per hour. The battery capacity is 3.5 kilowatt-hours. We need to determine if the battery will last through the entire performance.First, calculate the total power consumption. Power consumption = rate * time = 2 kW * (4/3) hours = 8/3 ≈2.6667 kWh.The battery capacity is 3.5 kWh, which is more than 2.6667 kWh. So the battery will last through the entire performance.But wait, let me double-check. 1 hour and 20 minutes is 1 + 20/60 = 1.333... hours. 2 kW * 1.333... hours = 2.666... kWh. The battery is 3.5 kWh, which is more than 2.666... So yes, it will last.Alternatively, if we calculate the remaining capacity: 3.5 - 2.666... ≈0.833... kWh left, which is about 50 minutes of additional power at 2 kW.So, the battery will last through the entire performance.</think>

🧠:<think>Okay, so I have this problem about a Markov Chain for a portfolio of Swiss stocks. The portfolio can be in one of three states: High (H), Medium (M), or Low (L). The transition matrix is given, and I need to solve two parts. Let me take them one by one.Starting with part 1: If the portfolio is currently in the Medium (M) state, compute the probability that it will be in the High (H) state after two months.Hmm, okay. So, Markov Chains. I remember that to find the probability of being in a certain state after a number of steps, we can use the transition matrix raised to the power of that number. Since we're looking at two months, we need to compute P squared, where P is the transition matrix.First, let me write down the transition matrix P:[ P = begin{pmatrix}0.6 & 0.3 & 0.1 0.2 & 0.5 & 0.3 0.1 & 0.3 & 0.6end{pmatrix} ]So, each row represents the current state, and each column represents the next state. The entry P(i,j) is the probability of moving from state i to state j.Since we're starting in state M, which is the second state, the initial state vector would be [0, 1, 0]. After one month, we can multiply this vector by P to get the distribution after one month. Then, to get after two months, we can either multiply by P again or compute P squared first and then multiply.Let me try computing P squared. To do that, I need to perform matrix multiplication of P with itself.Let me denote P squared as P^2. So, each entry (i,j) in P^2 is the sum over k of P(i,k) * P(k,j).Let me compute each entry step by step.First row of P squared:- Entry (1,1): 0.6*0.6 + 0.3*0.2 + 0.1*0.1= 0.36 + 0.06 + 0.01= 0.43- Entry (1,2): 0.6*0.3 + 0.3*0.5 + 0.1*0.3= 0.18 + 0.15 + 0.03= 0.36- Entry (1,3): 0.6*0.1 + 0.3*0.3 + 0.1*0.6= 0.06 + 0.09 + 0.06= 0.21Second row of P squared:- Entry (2,1): 0.2*0.6 + 0.5*0.2 + 0.3*0.1= 0.12 + 0.10 + 0.03= 0.25- Entry (2,2): 0.2*0.3 + 0.5*0.5 + 0.3*0.3= 0.06 + 0.25 + 0.09= 0.40- Entry (2,3): 0.2*0.1 + 0.5*0.3 + 0.3*0.6= 0.02 + 0.15 + 0.18= 0.35Third row of P squared:- Entry (3,1): 0.1*0.6 + 0.3*0.2 + 0.6*0.1= 0.06 + 0.06 + 0.06= 0.18- Entry (3,2): 0.1*0.3 + 0.3*0.5 + 0.6*0.3= 0.03 + 0.15 + 0.18= 0.36- Entry (3,3): 0.1*0.1 + 0.3*0.3 + 0.6*0.6= 0.01 + 0.09 + 0.36= 0.46So, putting it all together, P squared is:[ P^2 = begin{pmatrix}0.43 & 0.36 & 0.21 0.25 & 0.40 & 0.35 0.18 & 0.36 & 0.46end{pmatrix} ]Now, since we start in state M, which is the second state, our initial vector is [0, 1, 0]. To find the distribution after two months, we multiply this vector by P squared.So, the multiplication would be:[0, 1, 0] * P^2 = [0*0.43 + 1*0.25 + 0*0.18, 0*0.36 + 1*0.40 + 0*0.36, 0*0.21 + 1*0.35 + 0*0.46] = [0.25, 0.40, 0.35]Therefore, the probability of being in state H after two months is 0.25.Wait, let me double-check that. Alternatively, maybe I should compute it step by step.Starting from M, after one month, the distribution is the second row of P:[0.2, 0.5, 0.3]Then, after the second month, we multiply this by P again.So, let's compute that:- Probability of H: 0.2*0.6 + 0.5*0.2 + 0.3*0.1= 0.12 + 0.10 + 0.03= 0.25- Probability of M: 0.2*0.3 + 0.5*0.5 + 0.3*0.3= 0.06 + 0.25 + 0.09= 0.40- Probability of L: 0.2*0.1 + 0.5*0.3 + 0.3*0.6= 0.02 + 0.15 + 0.18= 0.35Yes, same result. So, the probability of being in H after two months is 0.25.Okay, that seems solid.Moving on to part 2: Given that the portfolio starts in the High (H) state, determine the long-term steady-state distribution of the portfolio's value across the three states.Hmm, steady-state distribution. I remember that the steady-state distribution is a probability vector π such that π = πP. So, it's an eigenvector of P corresponding to eigenvalue 1, with the additional constraint that the sum of the components is 1.So, we need to solve πP = π, where π is a row vector [π_H, π_M, π_L], and π_H + π_M + π_L = 1.Let me write down the equations.From πP = π, we have:π_H = π_H * 0.6 + π_M * 0.2 + π_L * 0.1π_M = π_H * 0.3 + π_M * 0.5 + π_L * 0.3π_L = π_H * 0.1 + π_M * 0.3 + π_L * 0.6And π_H + π_M + π_L = 1So, we have four equations:1. π_H = 0.6π_H + 0.2π_M + 0.1π_L2. π_M = 0.3π_H + 0.5π_M + 0.3π_L3. π_L = 0.1π_H + 0.3π_M + 0.6π_L4. π_H + π_M + π_L = 1Let me rearrange equations 1, 2, 3 to bring all terms to one side.Equation 1:π_H - 0.6π_H - 0.2π_M - 0.1π_L = 00.4π_H - 0.2π_M - 0.1π_L = 0Equation 2:π_M - 0.3π_H - 0.5π_M - 0.3π_L = 0-0.3π_H + 0.5π_M - 0.3π_L = 0Equation 3:π_L - 0.1π_H - 0.3π_M - 0.6π_L = 0-0.1π_H - 0.3π_M + 0.4π_L = 0So, now we have three equations:1. 0.4π_H - 0.2π_M - 0.1π_L = 02. -0.3π_H + 0.5π_M - 0.3π_L = 03. -0.1π_H - 0.3π_M + 0.4π_L = 0And equation 4: π_H + π_M + π_L = 1So, we can set up a system of equations. Let me write them as:0.4π_H - 0.2π_M - 0.1π_L = 0 --> Equation A-0.3π_H + 0.5π_M - 0.3π_L = 0 --> Equation B-0.1π_H - 0.3π_M + 0.4π_L = 0 --> Equation CAnd π_H + π_M + π_L = 1 --> Equation DLet me try to solve this system.First, let's express Equations A, B, C.Equation A: 0.4π_H = 0.2π_M + 0.1π_L --> π_H = (0.2π_M + 0.1π_L)/0.4 = 0.5π_M + 0.25π_LEquation B: -0.3π_H + 0.5π_M - 0.3π_L = 0Equation C: -0.1π_H - 0.3π_M + 0.4π_L = 0Let me substitute π_H from Equation A into Equations B and C.Substituting into Equation B:-0.3*(0.5π_M + 0.25π_L) + 0.5π_M - 0.3π_L = 0Compute:-0.15π_M - 0.075π_L + 0.5π_M - 0.3π_L = 0Combine like terms:(-0.15 + 0.5)π_M + (-0.075 - 0.3)π_L = 00.35π_M - 0.375π_L = 0 --> Let's call this Equation ESimilarly, substitute into Equation C:-0.1*(0.5π_M + 0.25π_L) - 0.3π_M + 0.4π_L = 0Compute:-0.05π_M - 0.025π_L - 0.3π_M + 0.4π_L = 0Combine like terms:(-0.05 - 0.3)π_M + (-0.025 + 0.4)π_L = 0-0.35π_M + 0.375π_L = 0 --> Let's call this Equation FNow, Equations E and F:Equation E: 0.35π_M - 0.375π_L = 0Equation F: -0.35π_M + 0.375π_L = 0Wait, that's interesting. Equations E and F are negatives of each other.Equation E: 0.35π_M - 0.375π_L = 0Equation F: -0.35π_M + 0.375π_L = 0So, if we add them together, we get 0 = 0. So, they are dependent.Therefore, we only have one independent equation from E and F, which is 0.35π_M = 0.375π_LSimplify this ratio:π_M / π_L = 0.375 / 0.35 = (375/1000)/(35/100) = (3/8)/(7/20) = (3/8)*(20/7) = (60/56) = 15/14So, π_M = (15/14)π_LSo, π_M = (15/14)π_LNow, let's go back to Equation A:π_H = 0.5π_M + 0.25π_LSubstitute π_M = (15/14)π_L:π_H = 0.5*(15/14)π_L + 0.25π_L = (15/28)π_L + (7/28)π_L = (22/28)π_L = (11/14)π_LSo, π_H = (11/14)π_LSo, now, we have π_H = (11/14)π_L and π_M = (15/14)π_LNow, let's use Equation D: π_H + π_M + π_L = 1Substitute:(11/14)π_L + (15/14)π_L + π_L = 1Compute:(11/14 + 15/14 + 14/14)π_L = 1(40/14)π_L = 1Simplify:(20/7)π_L = 1 --> π_L = 7/20So, π_L = 7/20Then, π_M = (15/14)*(7/20) = (15/14)*(7/20) = (15*7)/(14*20) = (105)/(280) = 3/8Similarly, π_H = (11/14)*(7/20) = (11*7)/(14*20) = (77)/(280) = 11/40So, π_H = 11/40, π_M = 3/8, π_L = 7/20Let me check if these add up to 1:11/40 + 3/8 + 7/20Convert to 40 denominator:11/40 + 15/40 + 14/40 = (11 + 15 + 14)/40 = 40/40 = 1. Okay, that's good.Let me also verify if πP = π.Compute πP:π = [11/40, 3/8, 7/20]Multiply by P:First entry: (11/40)*0.6 + (3/8)*0.2 + (7/20)*0.1= (11/40)*(3/5) + (3/8)*(1/5) + (7/20)*(1/10)= (33/200) + (3/40) + (7/200)Convert to 200 denominator:33/200 + 15/200 + 7/200 = 55/200 = 11/40. That's correct.Second entry: (11/40)*0.3 + (3/8)*0.5 + (7/20)*0.3= (11/40)*(3/10) + (3/8)*(1/2) + (7/20)*(3/10)= (33/400) + (3/16) + (21/200)Convert to 400 denominator:33/400 + 75/400 + 42/400 = 150/400 = 3/8. Correct.Third entry: (11/40)*0.1 + (3/8)*0.3 + (7/20)*0.6= (11/40)*(1/10) + (3/8)*(3/10) + (7/20)*(3/5)= (11/400) + (9/80) + (21/100)Convert to 400 denominator:11/400 + 45/400 + 84/400 = 140/400 = 7/20. Correct.So, yes, πP = π. So, the steady-state distribution is [11/40, 3/8, 7/20].Alternatively, in decimal form, that's 0.275, 0.375, 0.35.But the question didn't specify the form, so fractions are probably better.So, summarizing:1. The probability of being in H after two months starting from M is 0.25.2. The steady-state distribution is π_H = 11/40, π_M = 3/8, π_L = 7/20.I think that's it.Final Answer1. The probability is boxed{0.25}.2. The steady-state distribution is boxed{left[ frac{11}{40}, frac{3}{8}, frac{7}{20} right]}.</think>