🧠:<think>Okay, so I have these two problems about solar farms and their energy production and storage. Let me try to figure them out step by step.Starting with the first one, Solar Farm A. It's near the equator, so I guess that means it gets a lot of sunlight, which is why the irradiance is 5.5 kWh/m²/day. The efficiency is 18%, and the total area is 50,000 m². I need to find the total annual energy production in MWh.Hmm, okay. So, first, I think I need to calculate the daily energy production and then multiply by the number of days in a year. Let me write that down.Daily production would be the irradiance multiplied by the area and then multiplied by the efficiency. So, that's 5.5 kWh/m²/day * 50,000 m² * 18%. Let me compute that.First, 5.5 * 50,000. Let me do 5 * 50,000 = 250,000 and 0.5 * 50,000 = 25,000. So, 250,000 + 25,000 = 275,000. So, 5.5 * 50,000 = 275,000 kWh/m²/day? Wait, no, that's just the total without considering efficiency yet.Wait, no, actually, 5.5 kWh/m²/day times 50,000 m² gives the total energy per day before efficiency. So, 5.5 * 50,000 = 275,000 kWh per day. Then, we have to apply the efficiency of 18%. So, 275,000 * 0.18.Let me calculate that. 275,000 * 0.1 = 27,500. 275,000 * 0.08 = 22,000. So, 27,500 + 22,000 = 49,500 kWh per day.Now, to get the annual production, I need to multiply this daily amount by the number of days in a year. Assuming 365 days, so 49,500 * 365.Let me compute that. 49,500 * 300 = 14,850,000. 49,500 * 65 = let's see, 49,500 * 60 = 2,970,000 and 49,500 * 5 = 247,500. So, 2,970,000 + 247,500 = 3,217,500. Adding that to 14,850,000 gives 18,067,500 kWh per year.But the question asks for MWh, so I need to convert kWh to MWh by dividing by 1,000. So, 18,067,500 / 1,000 = 18,067.5 MWh per year.Wait, let me double-check my calculations. 5.5 * 50,000 is indeed 275,000. 275,000 * 0.18 is 49,500. 49,500 * 365: 49,500 * 300 is 14,850,000, 49,500 * 65 is 3,217,500, so total 18,067,500 kWh, which is 18,067.5 MWh. That seems correct.Moving on to the second problem. Solar Farms B and C both generate and store the same amount of energy daily. Farm B uses a battery storage with 10 MWh capacity and 85% round-trip efficiency. Farm C uses hydrogen storage with 40% efficiency each way, so total round-trip is 40%? Or is it 40% each way, meaning 40% * 40% = 16%? Hmm, the question says "conversion efficiency from electricity to hydrogen and back to electricity is 40%." So, I think that means the overall round-trip efficiency is 40%. So, not 16%, but 40%.Wait, no, actually, sometimes when they say conversion efficiency each way, it's multiplied. So, if you convert electricity to hydrogen at 40% efficiency, and then hydrogen back to electricity at 40% efficiency, the total round-trip efficiency is 0.4 * 0.4 = 0.16, which is 16%. So, I need to clarify that.The question says: "the conversion efficiency from electricity to hydrogen and back to electricity is 40%." Hmm, that wording is a bit ambiguous. It could mean that each step is 40%, so total is 16%, or it could mean that the overall round-trip is 40%. I think it's more likely that each step is 40%, so the total is 16%. Because if it was the overall, they might have specified it as round-trip efficiency.But let me check the problem statement again: "conversion efficiency from electricity to hydrogen and back to electricity is 40%." So, it's 40% for both processes combined. So, that would mean the round-trip efficiency is 40%. So, if they generate X amount of energy, they can store it and retrieve 40% of X.Wait, but in the case of battery storage, the round-trip efficiency is 85%, meaning that if you put in X, you get back 0.85X. For hydrogen, it's 40%, so you get back 0.4X.But the problem says both farms generate and store the same amount of energy daily. So, let me denote the amount of energy generated and stored as E.For Farm B, the usable energy retrieved is E * 0.85.For Farm C, the usable energy retrieved is E * 0.4.But wait, the problem says they generate and store the same amount of energy daily. So, they both store E, but retrieve different amounts.But the question is asking for the amount of usable energy that can be retrieved from each storage system per day. So, if they store the same E, then retrieved energy is E * efficiency.But the problem says "generate and store the same amount of energy daily." So, does that mean they both store the same E, or they both generate the same E and store it?I think it's the latter: both farms generate the same amount of energy daily and store it. So, E is the same for both, but the storage systems have different efficiencies.But the problem says "they generate and store the same amount of energy daily." So, perhaps they both store the same E, but given their storage capacities, which are different.Wait, no. Solar Farm B uses a battery storage system that can store up to 10 MWh. So, the maximum storage is 10 MWh. But if they generate and store the same amount, does that mean they store 10 MWh each day? Or they store the same amount, but the storage capacities are different.Wait, the problem says: "Solar Farm B uses a battery storage system that can store up to 10 MWh of energy with a round-trip efficiency of 85%. Solar Farm C uses a hydrogen storage system where the conversion efficiency from electricity to hydrogen and back to electricity is 40%. If both farms generate and store the same amount of energy daily, calculate the amount of usable energy (in MWh) that can be retrieved from each storage system per day."So, both farms generate and store the same amount of energy daily. Let's denote that amount as E. So, they both store E MWh each day.But Farm B's storage can store up to 10 MWh. So, if E is the amount stored, E must be less than or equal to 10 MWh.But the problem doesn't specify E, it just says they store the same amount. So, perhaps E is 10 MWh? Because Farm B can store up to 10 MWh, so if they store the same amount, maybe E is 10 MWh.Wait, but the problem says "generate and store the same amount of energy daily." So, they generate E and store E. So, for Farm B, the storage capacity is 10 MWh, so E can be up to 10 MWh. For Farm C, the storage system's capacity isn't given, but the efficiency is 40%.But the problem doesn't specify the storage capacity of Farm C, only the efficiency. So, maybe we don't need to worry about the storage capacity, just the efficiency.So, if both farms generate and store E MWh daily, then the usable energy retrieved is E * efficiency.But the problem says "they generate and store the same amount of energy daily." So, E is the same for both.But the question is asking for the amount of usable energy retrieved from each storage system per day. So, for Farm B, it's E * 0.85, and for Farm C, it's E * 0.4.But we don't know E. Wait, unless E is the maximum storage capacity for Farm B, which is 10 MWh. So, if they store the same amount, E would be 10 MWh for both. So, Farm B stores 10 MWh, retrieves 10 * 0.85 = 8.5 MWh. Farm C stores 10 MWh, retrieves 10 * 0.4 = 4 MWh.But the problem says "they generate and store the same amount of energy daily." So, if they generate E and store E, then E must be the same for both. But Farm B's storage is limited to 10 MWh, so E can't exceed that. So, perhaps E is 10 MWh for both.Alternatively, maybe E is the amount generated, which is stored, but for Farm B, the storage is 10 MWh, so if they generate more than 10 MWh, they can only store 10 MWh. But the problem says they generate and store the same amount, so perhaps they generate E and store E, with E being the same for both, but E can't exceed Farm B's storage capacity.But since the problem doesn't specify the generation capacity, just that they store the same amount, I think we can assume that E is 10 MWh for both, as that's the maximum for Farm B.So, for Farm B: 10 MWh stored, retrieved energy is 10 * 0.85 = 8.5 MWh.For Farm C: 10 MWh stored, retrieved energy is 10 * 0.4 = 4 MWh.But wait, the problem says "conversion efficiency from electricity to hydrogen and back to electricity is 40%." So, does that mean the round-trip efficiency is 40%, or each step is 40%? If each step is 40%, then the round-trip is 0.4 * 0.4 = 0.16, so 16%. But the problem says "conversion efficiency from electricity to hydrogen and back to electricity is 40%." So, it's the overall efficiency, not each step. So, round-trip is 40%.Therefore, for Farm C, the retrieved energy is 10 * 0.4 = 4 MWh.So, the answer would be Farm B retrieves 8.5 MWh, and Farm C retrieves 4 MWh per day.But let me make sure. If the conversion efficiency is 40% each way, then it's 40% to make hydrogen, and 40% to convert back, so total is 16%. But the problem says "conversion efficiency from electricity to hydrogen and back to electricity is 40%." So, it's the overall efficiency, not each step. So, it's 40% round-trip.Therefore, the answer is 8.5 MWh for B and 4 MWh for C.Alternatively, if it was each step, it would be 16%, so 1.6 MWh, but I think it's 40% overall.So, to summarize:1. Solar Farm A: 18,067.5 MWh annually.2. Solar Farm B: 8.5 MWh usable per day.Solar Farm C: 4 MWh usable per day.I think that's it.</think>

🧠:<think>Okay, so I have this problem about analyzing voting patterns in a provincial election. There are two parts, and I need to figure out both. Let me start with the first one.Problem 1: Expected Proportion of Votes for Party AAlright, so we have n districts, each with a different number of voters. In each district i, the total number of voters is V_i. The proportions of votes for Party A, B, and C in district i are p_{Ai}, p_{Bi}, and p_{Ci} respectively, and they add up to 1. The problem states that these proportions follow a Dirichlet distribution with parameters α_A, α_B, and α_C. I remember that the Dirichlet distribution is a multivariate generalization of the Beta distribution and is often used as a prior distribution in Bayesian statistics for categorical data. For a Dirichlet distribution with parameters α_A, α_B, α_C, the expected value of each proportion is given by the ratio of the respective parameter to the sum of all parameters. So, E[p_{Ai}] = α_A / (α_A + α_B + α_C). Since this is true for each district, does that mean the expected proportion for Party A across the entire province is just the same?Wait, but hold on. The province is made up of multiple districts, each with different numbers of voters. So, the overall proportion of votes for Party A in the province isn't just the average of the expected proportions in each district, because districts have different sizes. It's a weighted average, where each district's expected proportion is weighted by its number of voters.So, the total number of voters in the province is the sum of all V_i, which is V = V_1 + V_2 + ... + V_n. The expected number of votes for Party A in district i is V_i * E[p_{Ai}] = V_i * (α_A / (α_A + α_B + α_C)). Therefore, the total expected votes for Party A in the province is the sum over all districts of V_i * (α_A / (α_A + α_B + α_C)).To get the expected proportion, we divide this total by the total number of voters V. So, the expected proportion E[P_A] is:E[P_A] = (Σ V_i * (α_A / (α_A + α_B + α_C))) / VBut since (α_A / (α_A + α_B + α_C)) is a constant for all districts, we can factor it out:E[P_A] = (α_A / (α_A + α_B + α_C)) * (Σ V_i) / VBut Σ V_i is just V, so this simplifies to:E[P_A] = α_A / (α_A + α_B + α_C)Wait, so is it just the same as the expected proportion in a single district? That seems a bit counterintuitive because districts have different numbers of voters. But actually, since each district's proportion is weighted by its size, and the expectation is linear, it still ends up being the same as the expectation for a single district. That makes sense because the Dirichlet distribution parameters are the same across all districts, so the overall expectation is just the same as each individual district's expectation.So, the expected proportion of votes for Party A across the entire province is α_A divided by the sum of α_A, α_B, and α_C.Problem 2: Overall Variance of the Proportion of Votes for Party ANow, moving on to the second part. The correspondent notes that the variance of the proportion of votes for Party A in each district i is σ_i². These variances are independent across districts, and each district has a large number of voters. We need to derive the overall variance of the proportion of votes for Party A across the entire province.Hmm, okay. So, variance is a measure of spread, and since the districts are independent, the total variance should be the sum of variances from each district, appropriately weighted.But wait, the overall proportion is a weighted average of the district proportions. So, the variance of a weighted sum of independent random variables is the sum of the variances multiplied by the square of the weights.Let me recall: If we have a weighted sum Y = w_1 X_1 + w_2 X_2 + ... + w_n X_n, where the X_i are independent, then Var(Y) = w_1² Var(X_1) + w_2² Var(X_2) + ... + w_n² Var(X_n).In this case, the overall proportion P_A is a weighted average of p_{Ai} with weights V_i / V. So, each p_{Ai} is multiplied by V_i / V, and then summed up.Therefore, the variance of P_A should be the sum over all districts of (V_i / V)² * Var(p_{Ai}).Given that Var(p_{Ai}) is σ_i², the overall variance Var(P_A) is:Var(P_A) = Σ [(V_i / V)² * σ_i²]But wait, the problem mentions that the total number of votes in each district is large. Does that affect anything? Well, when the number of voters is large, the variance of the proportion might be approximated differently, but since we are given Var(p_{Ai}) = σ_i², I think we can proceed with that.So, putting it all together, the overall variance is the sum of (V_i squared over V squared) times σ_i squared for each district.Alternatively, we can factor out 1/V squared:Var(P_A) = (1 / V²) * Σ [V_i² * σ_i²]That seems correct. Let me check if I missed anything. The weights are V_i / V, so when squared, they are (V_i²)/(V²). Multiply each by the variance σ_i², and sum over all districts. Yes, that makes sense.So, the overall variance is the sum of (V_i squared times σ_i squared) divided by V squared.Wait a second, but in the first part, the expectation was just α_A / (α_A + α_B + α_C), regardless of the district sizes. But for the variance, it's a weighted sum of variances, weighted by the square of the district sizes relative to the total.Is there another way to think about this? Maybe using the law of total variance or something else? Let me consider.Alternatively, since each district's proportion is independent, the variance of the overall proportion is the sum of variances of each district's contribution.But the overall proportion is the sum of (V_i / V) * p_{Ai}, so the variance is the sum of Var[(V_i / V) * p_{Ai}]. Since the districts are independent, the covariance terms are zero.Each term Var[(V_i / V) * p_{Ai}] = (V_i / V)^2 * Var(p_{Ai}) = (V_i² / V²) * σ_i².So, adding them up gives the same result as before. So, that seems consistent.Therefore, the overall variance is indeed the sum over all districts of (V_i squared times σ_i squared) divided by V squared.So, to summarize:1. The expected proportion of votes for Party A across the entire province is α_A divided by the sum of α_A, α_B, and α_C.2. The overall variance of the proportion of votes for Party A is the sum of (V_i squared times σ_i squared) divided by V squared.I think that's it. Let me just write that down clearly.Final Answer1. The expected proportion of votes for Party A is boxed{dfrac{alpha_A}{alpha_A + alpha_B + alpha_C}}.2. The overall variance of the proportion of votes for Party A is boxed{dfrac{1}{V^2} sum_{i=1}^{n} V_i^2 sigma_i^2}.</think>

🧠:<think>Okay, so I have this problem about Australian athletes in the Olympics, and I need to figure out two probabilities. Let me take it step by step.First, the problem is divided into two parts: Performance Analysis and Medal Distribution. I'll tackle them one by one.1. Performance Analysis: Probability of Winning at Least One Gold MedalAlright, so we have three sports: swimming, cycling, and athletics. Each has its own probability of an Australian athlete winning a gold medal. The probabilities are:- Swimming: 0.3- Cycling: 0.2- Athletics: 0.25And these events are independent. I need to find the probability that they win at least one gold medal in these three sports.Hmm, when I see "at least one," I remember that it's often easier to calculate the complement probability and subtract it from 1. The complement of winning at least one medal is winning no medals at all. So, if I can find the probability of not winning any medals, I can subtract that from 1 to get the desired probability.Let me write that down:P(at least one gold) = 1 - P(no golds in swimming, cycling, and athletics)Since the events are independent, the probability of not winning any medals is the product of the probabilities of not winning each individual medal.So, first, I need the probabilities of not winning a gold in each sport.For swimming: P(not gold) = 1 - 0.3 = 0.7For cycling: P(not gold) = 1 - 0.2 = 0.8For athletics: P(not gold) = 1 - 0.25 = 0.75Therefore, the probability of not winning any medals is:P(no golds) = 0.7 * 0.8 * 0.75Let me compute that.First, 0.7 * 0.8 = 0.56Then, 0.56 * 0.75. Hmm, 0.56 * 0.75. Let me think: 0.5 * 0.75 is 0.375, and 0.06 * 0.75 is 0.045. So adding those together, 0.375 + 0.045 = 0.42.So, P(no golds) = 0.42Therefore, P(at least one gold) = 1 - 0.42 = 0.58Wait, that seems straightforward. Let me double-check my calculations.0.7 * 0.8 is indeed 0.56. Then, 0.56 * 0.75. Let me compute it another way: 0.56 * 3/4. 0.56 divided by 4 is 0.14, so 0.14 * 3 = 0.42. Yep, that's correct.So, 1 - 0.42 is 0.58. So, the probability is 0.58, or 58%.That seems reasonable. I don't think I made a mistake there.2. Medal Distribution: Probability of Exactly Two Gold MedalsNow, the second part says that the total number of gold medals won by Australian athletes in these three sports follows a Poisson distribution with an average λ of 1.2 gold medals per Olympics.I need to determine the probability that they win exactly two gold medals.Alright, Poisson distribution formula is:P(X = k) = (λ^k * e^(-λ)) / k!Where:- λ is the average rate (1.2 in this case)- k is the number of occurrences (2 in this case)- e is the base of the natural logarithm, approximately 2.71828So, plugging in the numbers:P(X = 2) = (1.2^2 * e^(-1.2)) / 2!Let me compute each part step by step.First, compute 1.2 squared:1.2^2 = 1.44Next, compute e^(-1.2). I know that e^(-1) is approximately 0.3679, and e^(-0.2) is approximately 0.8187. So, e^(-1.2) = e^(-1) * e^(-0.2) ≈ 0.3679 * 0.8187.Let me calculate that:0.3679 * 0.8187First, 0.3 * 0.8 = 0.240.3 * 0.0187 = ~0.005610.0679 * 0.8 = ~0.054320.0679 * 0.0187 ≈ ~0.00127Adding all these together:0.24 + 0.00561 = 0.245610.24561 + 0.05432 = 0.299930.29993 + 0.00127 ≈ 0.3012So, e^(-1.2) ≈ 0.3012Alternatively, I can use a calculator for a more precise value, but since I don't have one, I'll go with this approximation.Now, 1.44 * 0.3012 ≈ ?Let me compute 1 * 0.3012 = 0.30120.44 * 0.3012 ≈ 0.1325So, adding together: 0.3012 + 0.1325 ≈ 0.4337So, the numerator is approximately 0.4337Now, the denominator is 2! = 2So, P(X = 2) ≈ 0.4337 / 2 ≈ 0.21685So, approximately 0.21685, or 21.685%Wait, is that correct? Let me check my steps again.1.2 squared is 1.44, correct.e^(-1.2) ≈ 0.3012, correct.1.44 * 0.3012 ≈ 0.4337, correct.Divide by 2: 0.4337 / 2 ≈ 0.21685, yes.Alternatively, if I use a calculator for e^(-1.2), it's approximately 0.3011942.So, 1.44 * 0.3011942 ≈ 1.44 * 0.3011942.Let me compute 1 * 0.3011942 = 0.30119420.44 * 0.3011942 ≈ 0.44 * 0.3 = 0.132, 0.44 * 0.0011942 ≈ 0.000525. So total ≈ 0.132525Adding to 0.3011942: 0.3011942 + 0.132525 ≈ 0.4337192Divide by 2: 0.4337192 / 2 ≈ 0.2168596So, approximately 0.21686, which is about 21.69%So, rounding to four decimal places, 0.2169 or 21.69%Alternatively, if I use more precise calculations:Compute 1.2^2 = 1.44Compute e^(-1.2):We can use the Taylor series expansion for e^x around x = 0:e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...But since x is negative, e^(-1.2) = 1 - 1.2 + (1.2)^2/2! - (1.2)^3/3! + (1.2)^4/4! - ...Let me compute up to, say, the fifth term for better approximation.Compute each term:Term 0: 1Term 1: -1.2Term 2: (1.44)/2 = 0.72Term 3: -(1.728)/6 ≈ -0.288Term 4: (2.0736)/24 ≈ 0.0864Term 5: -(2.48832)/120 ≈ -0.020736Term 6: (2.985984)/720 ≈ 0.004147So, adding these up:1 - 1.2 = -0.2-0.2 + 0.72 = 0.520.52 - 0.288 = 0.2320.232 + 0.0864 = 0.31840.3184 - 0.020736 ≈ 0.2976640.297664 + 0.004147 ≈ 0.301811So, e^(-1.2) ≈ 0.301811, which is very close to the calculator value.So, 1.44 * 0.301811 ≈ 1.44 * 0.301811Let me compute 1 * 0.301811 = 0.3018110.44 * 0.301811 ≈ 0.132796Adding together: 0.301811 + 0.132796 ≈ 0.434607Divide by 2: 0.434607 / 2 ≈ 0.2173035So, approximately 0.2173, which is about 21.73%So, rounding to four decimal places, 0.2173 or 21.73%So, the probability is approximately 21.73%Wait, so depending on the precision, it's about 21.69% or 21.73%. Either way, approximately 21.7%.Alternatively, if I use a calculator for e^(-1.2), it's approximately 0.3011942.So, 1.44 * 0.3011942 = 0.4337192Divide by 2: 0.2168596, which is approximately 0.21686, so 21.686%, which rounds to 21.69%.So, depending on the precision, it's about 21.69% or 21.73%. For the purposes of this problem, I think 21.7% is a reasonable approximation.Alternatively, if I use more precise computation:Compute 1.2^2 = 1.44Compute e^(-1.2) ≈ 0.3011942Multiply: 1.44 * 0.3011942 ≈ 0.4337192Divide by 2: 0.4337192 / 2 = 0.2168596So, 0.2168596 is approximately 0.21686, which is 21.686%, so 21.69%Therefore, the probability is approximately 21.69%So, to summarize:1. The probability of winning at least one gold medal is 0.58 or 58%.2. The probability of winning exactly two gold medals is approximately 0.2169 or 21.69%.Wait, let me just make sure I didn't make any calculation mistakes.For the first part, 0.7 * 0.8 = 0.56, 0.56 * 0.75 = 0.42. So, 1 - 0.42 = 0.58. That seems correct.For the second part, Poisson with λ=1.2, k=2.Yes, the formula is correct. The calculations step by step seem correct. So, I think my answers are accurate.Final Answer1. The probability of winning at least one gold medal is boxed{0.58}.2. The probability of winning exactly two gold medals is approximately boxed{0.2169}.</think>

🧠:<think>Okay, so I have this problem about a medical thriller where doctors are using a new experimental drug. The effectiveness of the drug is modeled by this function E(t) = 100 / (1 + e^{k(t - τ)}). I need to find the constants k and τ given some conditions, and then use those to find when the effectiveness is 50%. Hmm, let me break this down step by step.First, the function E(t) is given as 100 divided by (1 plus e raised to the power of k times (t minus τ)). So, it's a logistic function, right? It starts at a low effectiveness and asymptotically approaches 100% as time increases. That makes sense for a drug's effectiveness over time.The first part says the drug reaches its maximum effectiveness of 90% after 5 days. Wait, maximum effectiveness? But the function approaches 100% as t goes to infinity. So, maybe 90% is the effectiveness at t=5, not the maximum. Hmm, the wording says "reaches its maximum effectiveness of 90% after 5 days." That's a bit confusing because the function should approach 100%. Maybe it's a typo or maybe I'm misunderstanding. Let me check the problem again.It says, "the drug reaches its maximum effectiveness of 90% after 5 days, and its effectiveness was 10% after 1 day." So, maybe 90% is the maximum they've observed so far, but in reality, the function can go up to 100%. Or perhaps the maximum is 90% because of some constraints. Hmm, maybe the function is actually E(t) = 90 / (1 + e^{k(t - τ)}), but no, the problem states it's 100. So perhaps the maximum effectiveness is 90%, meaning that the function never goes beyond 90%. Wait, but the function is 100 / (1 + e^{...}), so as t approaches infinity, E(t) approaches 100. So maybe the problem is that the maximum effectiveness observed is 90% at t=5, but the function can go higher? Hmm, maybe I should proceed with the given information.So, given that at t=5, E(t)=90, and at t=1, E(t)=10. So, two equations:1. 90 = 100 / (1 + e^{k(5 - τ)})2. 10 = 100 / (1 + e^{k(1 - τ)})I can write these as:1. 90 = 100 / (1 + e^{k(5 - τ)})2. 10 = 100 / (1 + e^{k(1 - τ)})Let me solve these equations for k and τ.Starting with the first equation:90 = 100 / (1 + e^{k(5 - τ)})Divide both sides by 100:90/100 = 1 / (1 + e^{k(5 - τ)})0.9 = 1 / (1 + e^{k(5 - τ)})Take reciprocals:1/0.9 = 1 + e^{k(5 - τ)}1/0.9 is approximately 1.1111, but let me write it as 10/9.10/9 = 1 + e^{k(5 - τ)}Subtract 1:10/9 - 1 = e^{k(5 - τ)}10/9 - 9/9 = 1/9 = e^{k(5 - τ)}Take natural logarithm:ln(1/9) = k(5 - τ)ln(1) - ln(9) = k(5 - τ)0 - ln(9) = k(5 - τ)So, -ln(9) = k(5 - τ) --- Equation ASimilarly, for the second equation:10 = 100 / (1 + e^{k(1 - τ)})Divide both sides by 100:10/100 = 1 / (1 + e^{k(1 - τ)})0.1 = 1 / (1 + e^{k(1 - τ)})Take reciprocals:1/0.1 = 1 + e^{k(1 - τ)}10 = 1 + e^{k(1 - τ)}Subtract 1:9 = e^{k(1 - τ)}Take natural logarithm:ln(9) = k(1 - τ) --- Equation BNow, I have two equations:Equation A: -ln(9) = k(5 - τ)Equation B: ln(9) = k(1 - τ)Let me write them as:Equation A: k(5 - τ) = -ln(9)Equation B: k(1 - τ) = ln(9)Let me denote ln(9) as L for simplicity. So, L = ln(9). Then:Equation A: k(5 - τ) = -LEquation B: k(1 - τ) = LSo, from Equation A: k(5 - τ) = -LFrom Equation B: k(1 - τ) = LLet me write these as:5k - kτ = -L --- Equation A'k - kτ = L --- Equation B'Now, let me subtract Equation B' from Equation A':(5k - kτ) - (k - kτ) = -L - L5k - kτ - k + kτ = -2LSimplify:(5k - k) + (-kτ + kτ) = -2L4k + 0 = -2LSo, 4k = -2LTherefore, k = (-2L)/4 = (-L)/2Since L = ln(9), then k = (-ln(9))/2But ln(9) is positive, so k is negative. Hmm, is that okay? Let me think.In the function E(t) = 100 / (1 + e^{k(t - τ)}), if k is negative, then as t increases, the exponent becomes more negative, so e^{k(t - τ)} decreases, making the denominator smaller, so E(t) increases. So, that makes sense because the effectiveness increases over time. So, a negative k is acceptable.So, k = (-ln(9))/2.Now, let's find τ.From Equation B: k(1 - τ) = LWe know k = (-ln(9))/2 and L = ln(9). So,(-ln(9)/2)(1 - τ) = ln(9)Multiply both sides by 2:(-ln(9))(1 - τ) = 2 ln(9)Divide both sides by -ln(9):(1 - τ) = (2 ln(9))/(-ln(9)) = -2So, 1 - τ = -2Therefore, τ = 1 + 2 = 3So, τ is 3 days.Let me verify these values.Given k = (-ln(9))/2 ≈ (-2.1972)/2 ≈ -1.0986τ = 3So, let's plug into the first condition: t=5, E=90.E(5) = 100 / (1 + e^{k(5 - τ)}) = 100 / (1 + e^{(-1.0986)(5 - 3)}) = 100 / (1 + e^{-2.1972})e^{-2.1972} ≈ e^{-2.1972} ≈ 0.1111So, 1 + 0.1111 ≈ 1.1111100 / 1.1111 ≈ 90, which matches.Similarly, t=1, E=10.E(1) = 100 / (1 + e^{k(1 - τ)}) = 100 / (1 + e^{(-1.0986)(1 - 3)}) = 100 / (1 + e^{(-1.0986)(-2)}) = 100 / (1 + e^{2.1972})e^{2.1972} ≈ 9So, 1 + 9 = 10100 / 10 = 10, which matches.Great, so k = (-ln(9))/2 and τ = 3.Now, part 2: Using these values, find the time t when E(t) = 50%.So, set E(t) = 50:50 = 100 / (1 + e^{k(t - τ)})Divide both sides by 100:0.5 = 1 / (1 + e^{k(t - τ)})Take reciprocals:2 = 1 + e^{k(t - τ)}Subtract 1:1 = e^{k(t - τ)}Take natural logarithm:ln(1) = k(t - τ)But ln(1) = 0, so:0 = k(t - τ)But k is not zero, so t - τ = 0Therefore, t = τSo, t = 3 days.Wait, that's interesting. So, the effectiveness is 50% at t=3 days. That makes sense because τ is the time when the effectiveness is 50%, as the logistic function crosses 50% at t=τ.Let me verify this.Given τ=3, so at t=3:E(3) = 100 / (1 + e^{k(3 - 3)}) = 100 / (1 + e^{0}) = 100 / (1 + 1) = 50. Yep, that's correct.So, the time when effectiveness is 50% is 3 days.So, summarizing:k = (-ln(9))/2 ≈ -1.0986τ = 3And the time when effectiveness is 50% is t=3.Final Answer1. The constants are ( k = boxed{-frac{ln 9}{2}} ) and ( tau = boxed{3} ).2. The effectiveness reaches 50% at ( t = boxed{3} ) days.</think>