answer:<think>Alright, so I have this problem where Dr. Evelyn is analyzing the historical accuracy of costumes in a film set in the 18th century. There are two sub-problems here, and I need to tackle each one step by step. Let me start with Sub-problem 1.Sub-problem 1: Testing the Proportion of Lace in CostumesOkay, so Dr. Evelyn observed that out of 300 costumes in the film, 45 feature a certain type of lace. Historically, this lace was present in 20% of attire. She wants to know if the proportion in the film is significantly different from the historical proportion at a 5% significance level.Hmm, this sounds like a hypothesis test for a single proportion. I remember that when dealing with proportions, we can use a z-test if certain conditions are met, like the sample size being large enough. Let me recall the steps.First, I need to set up the null and alternative hypotheses. The null hypothesis (H₀) is that the proportion in the film is equal to the historical proportion, and the alternative hypothesis (H₁) is that it's different. So:H₀: p = 0.20 H₁: p ≠ 0.20Next, I need to calculate the sample proportion. She observed 45 out of 300 costumes with lace. So, the sample proportion (p̂) is 45/300. Let me compute that:45 divided by 300 is 0.15. So, p̂ = 0.15.Now, I need to check if the conditions for the z-test are satisfied. The conditions are that np₀ and n(1 - p₀) should both be at least 5, where n is the sample size and p₀ is the hypothesized proportion.Calculating np₀: 300 * 0.20 = 60 Calculating n(1 - p₀): 300 * 0.80 = 240Both 60 and 240 are way larger than 5, so the conditions are satisfied. Great, we can proceed with the z-test.Now, the formula for the z-test statistic is:z = (p̂ - p₀) / sqrt[(p₀ * (1 - p₀)) / n]Plugging in the numbers:p̂ = 0.15 p₀ = 0.20 n = 300So, numerator: 0.15 - 0.20 = -0.05 Denominator: sqrt[(0.20 * 0.80) / 300] First, compute 0.20 * 0.80 = 0.16 Then, 0.16 / 300 ≈ 0.0005333 Square root of that is sqrt(0.0005333) ≈ 0.023094So, z ≈ -0.05 / 0.023094 ≈ -2.165Okay, so the z-score is approximately -2.165. Now, since this is a two-tailed test (because the alternative hypothesis is ≠), I need to find the p-value associated with this z-score.Looking at the standard normal distribution table, a z-score of 2.165 corresponds to a cumulative probability of about 0.9846. Since it's negative, the left tail probability is 1 - 0.9846 = 0.0154. But since it's two-tailed, we double this to get 0.0308.So, the p-value is approximately 0.0308, which is less than the significance level of 0.05. Therefore, we reject the null hypothesis.Wait, but let me double-check my z-score calculation. Maybe I made a mistake in the denominator.Wait, sqrt[(0.20 * 0.80)/300] is sqrt(0.16/300). 0.16 divided by 300 is approximately 0.0005333, and the square root of that is approximately 0.023094. So, 0.05 divided by 0.023094 is approximately 2.165. But since p̂ is less than p₀, it's negative. So, z ≈ -2.165.Yes, that seems correct. So, the p-value is about 0.0308, which is less than 0.05, so we reject H₀. Therefore, the proportion of lace in the film is significantly different from the historical proportion.But wait, let me think again. The p-value is the probability of observing a sample proportion as extreme as the one observed, assuming H₀ is true. Since 0.0308 is less than 0.05, we have sufficient evidence to reject H₀. So, the film's proportion is significantly different.Alternatively, we could have computed the critical z-values for a two-tailed test at 5% significance. The critical z-values are ±1.96. Our calculated z is -2.165, which is less than -1.96, so it falls in the rejection region. Therefore, we reject H₀.So, conclusion for Sub-problem 1: The proportion of lace in the film is significantly different from the historical proportion at the 5% significance level.Sub-problem 2: Confidence Interval for Embroidery StitchesNow, moving on to Sub-problem 2. Dr. Evelyn is comparing the complexity of embroidery on costumes. She has data:- Film costumes: 50 samples, average stitches 1200, standard deviation 150.- Historical: average stitches 1100, standard deviation 140.She wants to construct a 95% confidence interval for the mean number of stitches on film costumes and determine if it overlaps with the historical mean.Alright, so this is about constructing a confidence interval for a single mean, assuming normality. Since the sample size is 50, which is greater than 30, the Central Limit Theorem applies, so the sampling distribution is approximately normal.The formula for a confidence interval is:CI = x̄ ± z*(σ / sqrt(n))But wait, here, we have the sample standard deviation, not the population standard deviation. However, since the population standard deviation is unknown, we should technically use a t-distribution. But the problem says to assume both distributions are approximately normal, so maybe we can use the z-score? Or perhaps it's okay to use the sample standard deviation with z.Wait, let me think. The formula for the confidence interval when σ is unknown is:CI = x̄ ± t*(s / sqrt(n))But since the sample size is 50, which is large, the t-distribution and z-distribution are very similar. The critical value for a 95% confidence interval with large n is approximately 1.96.But to be precise, with n = 50, degrees of freedom = 49. Let me check the t-value for 49 degrees of freedom and 95% confidence. Using a t-table or calculator, the critical t-value is approximately 2.0096. That's slightly higher than 1.96.But the problem says to assume both distributions are approximately normal, so maybe they expect us to use the z-score. Hmm.Alternatively, since the standard deviation is given for the film costumes as 150, which is the sample standard deviation, we might need to use the t-distribution. But given that n is 50, which is moderately large, the difference between t and z is small.But let's see what the question says. It says, "assuming both distributions are approximately normal," so maybe they just want us to use the z-score. Alternatively, perhaps they consider the standard deviation as known? Wait, no, it's given as a sample standard deviation.Wait, hold on. The problem states:- The average number of embroidery stitches on 50 costumes from the film is 1200 with a standard deviation of 150.So, that's the sample mean and sample standard deviation. So, σ is unknown, so we should use the t-distribution.But the problem also mentions that historical records indicate the average was 1100 with a standard deviation of 140. So, the historical standard deviation is given as 140, but that's not directly relevant for constructing the confidence interval for the film.So, focusing on the film data:n = 50 x̄ = 1200 s = 150We need a 95% confidence interval. Since σ is unknown, we use the t-distribution. The critical t-value for 49 degrees of freedom and 95% confidence is approximately 2.0096.So, the formula is:CI = x̄ ± t*(s / sqrt(n))Plugging in the numbers:t ≈ 2.0096 s = 150 n = 50 sqrt(n) = sqrt(50) ≈ 7.0711So, standard error (SE) = 150 / 7.0711 ≈ 21.2132Then, the margin of error (ME) = 2.0096 * 21.2132 ≈ 42.64Therefore, the confidence interval is:1200 ± 42.64 Which is approximately (1200 - 42.64, 1200 + 42.64) So, (1157.36, 1242.64)Alternatively, if we use the z-score of 1.96, let's compute that as well for comparison.ME_z = 1.96 * (150 / sqrt(50)) ≈ 1.96 * 21.2132 ≈ 41.57So, CI_z = 1200 ± 41.57 ≈ (1158.43, 1241.57)So, the t-interval is slightly wider than the z-interval, but both are close.But since the problem says to assume the distribution is approximately normal, maybe they expect us to use the z-score. However, technically, since σ is unknown, we should use t. But given that n is 50, the difference is minimal.But let's proceed with the t-interval as more accurate.So, the 95% confidence interval is approximately (1157.36, 1242.64).Now, we need to determine if this interval overlaps with the historical mean of 1100.Looking at the interval, the lower bound is approximately 1157.36, which is higher than 1100. So, the interval does not overlap with 1100.Therefore, the historical mean of 1100 is not within the confidence interval for the film's mean. This suggests that the film's mean number of embroidery stitches is significantly different from the historical mean at the 95% confidence level.Wait, but let me think again. The confidence interval is for the film's mean, and we're checking if the historical mean falls within this interval. Since it doesn't, that implies that the film's mean is significantly different from the historical mean.Alternatively, if we were to perform a hypothesis test, we could compare the two means, but since the question specifically asks to construct a confidence interval and check for overlap, this approach suffices.So, conclusion for Sub-problem 2: The 95% confidence interval for the film's mean embroidery stitches is approximately (1157.36, 1242.64), which does not overlap with the historical mean of 1100. Therefore, the film's mean is significantly different from the historical mean.Final AnswerSub-problem 1: The proportion is significantly different. Sub-problem 2: The confidence interval does not overlap with the historical mean.So, in boxed form:Sub-problem 1: boxed{text{Reject } H_0} Sub-problem 2: boxed{text{Confidence interval does not overlap with historical mean}}</think>

answer:<think>Okay, so I have this problem where I need to evaluate the performance of a secret evaluator. The evaluator assesses reports, and their performance is based on two scores: accuracy (A) and thoroughness (T). Both are on a scale from 0 to 100. The first part of the problem says that the correlation between accuracy and thoroughness is a quadratic function given by T(A) = -0.1A² + 10A. I need to find the maximum possible thoroughness score and the corresponding accuracy score. Hmm, quadratic functions. I remember that a quadratic function in the form of f(x) = ax² + bx + c has its vertex at x = -b/(2a). Since the coefficient of A² is negative (-0.1), the parabola opens downward, which means the vertex is the maximum point. So, the maximum thoroughness occurs at A = -b/(2a). In this case, a = -0.1 and b = 10. So plugging into the formula, A = -10/(2*(-0.1)) = -10/(-0.2) = 50. So the accuracy score that gives the maximum thoroughness is 50. Now, to find the maximum thoroughness score, I plug A = 50 back into the equation T(A) = -0.1*(50)^2 + 10*(50). Let's compute that. First, 50 squared is 2500. Multiply that by -0.1: -0.1*2500 = -250. Then, 10*50 is 500. So, T = -250 + 500 = 250. Wait, that can't be right because the thoroughness score is supposed to be on a scale from 0 to 100. 250 is way beyond that. Did I make a mistake?Let me check the function again: T(A) = -0.1A² + 10A. If A is 50, then T = -0.1*(2500) + 500 = -250 + 500 = 250. Hmm, but the scale is 0-100. Maybe the function is defined such that T(A) can exceed 100? Or perhaps it's a theoretical maximum regardless of the scale? Wait, the problem says the scores are measured on a scale from 0 to 100, but the function might not necessarily cap at 100. So, maybe the maximum thoroughness is indeed 250, but in reality, it's capped at 100? Or perhaps the function is just a model, and the scores can go beyond 100? Wait, the problem doesn't specify that T(A) is bounded by 100, it just says the scores are on a scale from 0 to 100. So, perhaps the function is just a model, and the maximum thoroughness is 250, but in practice, it can't exceed 100. Hmm, but the question is asking for the maximum possible thoroughness score according to the function, not considering the scale. So, maybe it's 250. But that seems odd because 250 is way beyond the scale. Maybe I misinterpreted the function. Let me check the function again: T(A) = -0.1A² + 10A. So, when A is 0, T is 0. When A is 100, T would be -0.1*(10000) + 1000 = -1000 + 1000 = 0. So, at A=100, T=0. So, the function peaks at A=50, T=250. So, according to the function, the maximum thoroughness is 250, but in reality, the evaluator can't have a thoroughness score higher than 100. Wait, maybe I need to cap the thoroughness score at 100. So, if the function gives 250, but the maximum possible is 100, then the maximum thoroughness is 100. But then, when does T(A) = 100? Let's solve for A when T(A) = 100. So, 100 = -0.1A² + 10A. Rearranging: -0.1A² + 10A - 100 = 0. Multiply both sides by -10 to eliminate decimals: A² - 100A + 1000 = 0. Using quadratic formula: A = [100 ± sqrt(10000 - 4000)] / 2 = [100 ± sqrt(6000)] / 2. sqrt(6000) is approximately 77.46. So, A ≈ (100 ± 77.46)/2. So, two solutions: (100 + 77.46)/2 ≈ 177.46/2 ≈ 88.73, and (100 - 77.46)/2 ≈ 22.54/2 ≈ 11.27. So, T(A) = 100 when A ≈ 11.27 or A ≈ 88.73. But since the function peaks at A=50 with T=250, which is beyond the scale, the actual maximum thoroughness score would be 100, achieved at A≈11.27 and A≈88.73. But the question is asking for the maximum possible thoroughness score according to the function, not considering the scale. So, maybe it's 250. But that seems contradictory because the scale is 0-100. Wait, perhaps the function is defined such that T(A) is capped at 100. So, the maximum thoroughness is 100, and it occurs at two points: A≈11.27 and A≈88.73. But the function itself, without considering the scale, would have a maximum at 250. I think the problem is expecting us to find the maximum thoroughness according to the function, regardless of the scale. So, the maximum thoroughness is 250 at A=50. But that seems odd because the scale is 0-100. Maybe the function is scaled differently? Or perhaps it's a typo? Wait, let me check the function again: T(A) = -0.1A² + 10A. If A is 50, T is 250. But if A is 100, T is 0. So, the function is a downward opening parabola with vertex at (50, 250). So, the maximum thoroughness is indeed 250, but in reality, the evaluator can't have a thoroughness score higher than 100. So, maybe the function is just a model, and the maximum thoroughness is 250, but in practice, it's limited to 100. But the question is asking for the maximum possible thoroughness score, so I think it's 250. But that doesn't make sense because the scale is 0-100. Maybe I need to adjust the function? Or perhaps the function is correct, and the scores can go beyond 100? Wait, the problem says the scores are measured on a scale from 0 to 100, but it doesn't say that the function is bounded by that scale. So, maybe the function is just a mathematical model, and the maximum thoroughness is indeed 250. So, to answer part 1: the maximum thoroughness is 250, achieved when accuracy is 50. Moving on to part 2: The overall performance score P is a weighted sum, P = w1*A + w2*T. We need to determine w1 and w2, given that the optimal performance occurs when w1 = 2w2, and the evaluator achieves the maximum thoroughness score found in part 1. Also, the optimal performance score should be 180. Wait, so when the evaluator achieves the maximum thoroughness, which is 250, and the corresponding accuracy is 50. So, P = w1*50 + w2*250. And we know that w1 = 2w2. So, substituting w1 = 2w2 into the equation: P = 2w2*50 + w2*250 = 100w2 + 250w2 = 350w2. We are told that the optimal performance score is 180, so 350w2 = 180. Solving for w2: w2 = 180 / 350 = 18/35 ≈ 0.5143. Then, w1 = 2w2 = 2*(18/35) = 36/35 ≈ 1.0286. But wait, weights usually sum to 1 or are positive numbers. Here, w1 ≈1.0286 and w2≈0.5143, so their sum is about 1.5429. Is that acceptable? The problem doesn't specify that the weights need to sum to 1, just that they are weights. So, maybe it's okay. Alternatively, maybe I made a mistake in interpreting the optimal performance. The optimal performance occurs when w1 = 2w2, and the evaluator is at maximum thoroughness, which is 250, with A=50. So, P = 180 = w1*50 + w2*250, and w1 = 2w2. So, substituting: 180 = 2w2*50 + w2*250 = 100w2 + 250w2 = 350w2. So, w2 = 180/350 = 18/35, which simplifies to 18/35. So, w1 = 36/35. So, the weights are w1 = 36/35 and w2 = 18/35. But let me double-check: 36/35 is approximately 1.0286, and 18/35 is approximately 0.5143. So, P = 1.0286*50 + 0.5143*250 ≈ 51.43 + 128.57 ≈ 180. Yes, that adds up. So, the weights are w1 = 36/35 and w2 = 18/35. But wait, is there another way to interpret the problem? Maybe the optimal performance is when the weights are set such that the maximum of P occurs at the maximum thoroughness. But the problem says the optimal performance score occurs when w1 = 2w2 and the evaluator achieves the maximum thoroughness score. So, it's given that at maximum thoroughness, the performance is 180, and w1 = 2w2. So, my previous calculation seems correct. So, to recap: 1. Maximum thoroughness is 250 at A=50. 2. Weights are w1 = 36/35 and w2 = 18/35. But wait, 36/35 is more than 1, which might be unusual for weights, but unless specified otherwise, it's acceptable. Alternatively, maybe the weights should be normalized such that w1 + w2 = 1. But the problem doesn't say that. It just says they are weights. So, I think the answer is w1 = 36/35 and w2 = 18/35. But let me check if the problem expects the weights to be in a certain form. It says "determine the values of w1 and w2", so fractions are fine. So, final answers: 1. Maximum thoroughness is 250 at A=50. 2. w1 = 36/35 and w2 = 18/35. Wait, but 36/35 is 1.02857 and 18/35 is 0.5142857. So, if I write them as fractions, they are 36/35 and 18/35. Alternatively, maybe I should express them in simplest form. 36/35 and 18/35 are already in simplest form. So, I think that's it.</think>

answer:<think>First, I need to determine the amount of the bank loan, which is 40% of the total project cost of 10 million. This gives a loan amount of 4 million.Next, I'll calculate the total amount to be repaid on the bank loan using the compound interest formula. The formula is A = P(1 + r)^n, where P is the principal amount, r is the annual interest rate, and n is the number of years. Plugging in the values, A = 4,000,000(1 + 0.05)^5.After calculating the total repayment amount, I'll subtract the principal to find the total interest paid over the 5-year period.For the second part, the remaining funding needed is 6 million, which is covered by a private investor and a crowdfunding platform. The private investor provides twice the amount of the crowdfunding platform. Let's denote the crowdfunding amount as C. Therefore, the private investor contributes 2C, and the total funding from both sources is 3C, which equals 6 million. Solving for C gives the amount needed from the crowdfunding platform.Finally, I'll calculate the total amount to be repaid to the crowdfunding platform using the compound interest formula again, with an 8% annual interest rate over 3 years.</think>

answer:<think>First, I need to determine how much money Alex has left after purchasing the concert ticket. The total budget is 500, and the concert ticket costs 120, so the remaining budget is 500 - 120 = 380.Next, I'll set up an inequality to find out the maximum number of vinyl records Alex can buy with the remaining 380. Each vinyl record costs 40, so the inequality is 40x ≤ 380. Solving for x gives x ≤ 9.5. Since Alex can't buy half a vinyl record, the maximum number of records he can purchase is 9.Now, to find out how much Alex donates, I'll calculate 20% of the remaining budget after buying the concert ticket and the vinyl records. The remaining budget after these purchases is 380 - (9 × 40) = 380 - 360 = 20. Therefore, the donation amount is 20% of 20, which is 4.</think>

answer:<think>Alright, so I'm trying to solve this problem about designing a difficulty algorithm for a retro-style game. It's split into two sub-problems, and I need to figure out both. Let me start with Sub-problem 1.Sub-problem 1: Modeling Difficulty with a Sine FunctionThe problem states that the difficulty function is given by ( D(t) = A sin(Bt) + C ). The requirements are that the difficulty oscillates between 1 and 5, the average difficulty is 3, and one full oscillation happens every 10 minutes. I need to find the constants A, B, and C.First, I remember that the sine function oscillates between -1 and 1. So, if we have ( sin(Bt) ), it will go from -1 to 1. To make the difficulty oscillate between 1 and 5, we need to scale and shift this sine wave.Let me think about the amplitude. The sine function has a range of 2 (from -1 to 1). If we want the difficulty to have a range of 4 (from 1 to 5), the amplitude A should be half of that range because the sine function's amplitude is the distance from the centerline to the peak. So, the total range is 2A. Therefore, 2A = 4, which means A = 2.Next, the average difficulty is 3. Since the sine function oscillates around its midline, which is determined by the constant C. The midline should be the average difficulty. So, C = 3.Now, for the period. The function should complete one full oscillation every 10 minutes. The period of a sine function ( sin(Bt) ) is ( frac{2pi}{B} ). So, setting this equal to 10 minutes:( frac{2pi}{B} = 10 )Solving for B:( B = frac{2pi}{10} = frac{pi}{5} )So, putting it all together, the function is:( D(t) = 2 sinleft(frac{pi}{5} tright) + 3 )Let me double-check. The amplitude is 2, so the sine wave goes from -2 to +2, and adding 3 shifts it to 1 to 5. The average is 3, which is correct. The period is 10 minutes, so that's also correct. I think that's all for Sub-problem 1.Sub-problem 2: Adjusting Difficulty with Gaussian NoiseNow, the adjusted difficulty is ( D_{adj}(t) = D(t) + N(0, 0.25) ). We need to find the probability that ( D_{adj}(t) > 5 ).First, let's understand what ( D_{adj}(t) ) is. It's the original difficulty plus a Gaussian random variable with mean 0 and variance 0.25 (so standard deviation is 0.5). Given that ( D(t) ) is between 1 and 5, the adjusted difficulty can go above 5 or below 1. But we're concerned with the probability that it exceeds 5.So, ( D_{adj}(t) = D(t) + X ), where ( X sim N(0, 0.25) ). We need ( P(D_{adj}(t) > 5) = P(D(t) + X > 5) ).Since ( D(t) ) can vary, but we want the maximum probability that ( D(t) + X > 5 ). Wait, but actually, ( D(t) ) is a function of time, but the Gaussian noise is independent of time. So, for any given t, ( D(t) ) is a specific value, and X is a random variable. So, the probability that ( D(t) + X > 5 ) depends on the value of ( D(t) ) at that time.But the problem says "at any given time t", so we might need to consider the maximum possible probability, or perhaps the probability when ( D(t) ) is at its maximum.Wait, let me think. Since ( D(t) ) oscillates between 1 and 5, the maximum value of ( D(t) ) is 5. So, when ( D(t) = 5 ), what is the probability that ( D_{adj}(t) = 5 + X > 5 )? That's equivalent to ( X > 0 ). Since X is a Gaussian with mean 0, the probability that X > 0 is 0.5.But wait, is that the maximum probability? Because when ( D(t) ) is less than 5, say 4, then ( D_{adj}(t) = 4 + X ). The probability that 4 + X > 5 is the probability that X > 1. Since X has a standard deviation of 0.5, 1 is 2 standard deviations above the mean.So, the probability that X > 1 is the same as the probability that a standard normal variable Z > (1 - 0)/0.5 = 2. From standard normal tables, P(Z > 2) ≈ 0.0228.But wait, the question is asking for the probability that ( D_{adj}(t) ) exceeds 5 at any given time t. So, depending on the value of ( D(t) ), the probability changes. But since the question is general, not specifying a particular t, perhaps we need to find the maximum probability over all t, or maybe the expected probability.Alternatively, maybe it's asking for the probability that, for some t, ( D_{adj}(t) > 5 ). But that might be more complicated, involving the maximum of a stochastic process.Wait, the problem says "the probability that the adjusted difficulty ( D_{adj}(t) ) exceeds 5 at any given time t". So, for a specific t, what is the probability that ( D_{adj}(t) > 5 ). But since t is arbitrary, and ( D(t) ) varies with t, the probability will vary. So, perhaps the question is asking for the maximum probability over all t, or maybe the average probability.But I think more likely, it's asking for the probability that, for a given t, ( D_{adj}(t) > 5 ). Since ( D(t) ) is a function of t, and X is independent, the probability depends on ( D(t) ). So, perhaps we need to express it in terms of ( D(t) ), but the problem might be expecting a numerical answer, so maybe it's considering the worst case, when ( D(t) ) is at its maximum.Wait, let me read the problem again: "Calculate the probability that the adjusted difficulty ( D_{adj}(t) ) exceeds 5 at any given time ( t )."Hmm, "at any given time t" might mean for a specific t, but since t is arbitrary, perhaps we need to find the maximum probability over all t. Because if t is such that ( D(t) ) is 5, then the probability is 0.5, but if ( D(t) ) is less, the probability is lower.Alternatively, maybe the question is considering the probability over all t, but that would be a different approach, involving integrating over t or something.Wait, perhaps the question is simply asking, for a given t, what is the probability that ( D_{adj}(t) > 5 ). But since ( D(t) ) is a function of t, and X is independent, the probability is ( P(X > 5 - D(t)) ). But since D(t) varies, the probability varies. So, unless we have a specific t, we can't give a numerical answer. But the problem says "at any given time t", so maybe it's considering the maximum probability, which occurs when ( D(t) ) is as large as possible, i.e., 5. So, when ( D(t) = 5 ), ( P(D_{adj}(t) > 5) = P(X > 0) = 0.5 ).But wait, that seems high. Alternatively, maybe the question is considering the probability that, for some t, ( D_{adj}(t) > 5 ). But that would be a different calculation, involving the maximum of the process over t, which is more complex.Wait, perhaps I'm overcomplicating. Let me think again. The adjusted difficulty is ( D(t) + X ), where X is N(0, 0.25). So, for any given t, ( D_{adj}(t) ) is a random variable with mean ( D(t) ) and variance 0.25. So, the probability that ( D_{adj}(t) > 5 ) is the same as ( P(D(t) + X > 5) = P(X > 5 - D(t)) ).Since X is N(0, 0.25), we can standardize it:( P(X > 5 - D(t)) = Pleft( Z > frac{5 - D(t)}{0.5} right) ), where Z is the standard normal variable.So, ( P = 1 - Phileft( frac{5 - D(t)}{0.5} right) ), where ( Phi ) is the standard normal CDF.But since ( D(t) ) varies between 1 and 5, the term ( frac{5 - D(t)}{0.5} ) varies between 0 and 8.So, when ( D(t) = 5 ), ( P = 1 - Phi(0) = 0.5 ).When ( D(t) = 4 ), ( P = 1 - Phi(2) ≈ 1 - 0.9772 = 0.0228 ).When ( D(t) = 3 ), ( P = 1 - Phi(4) ≈ 0 ) (since Φ(4) is almost 1).Similarly, for ( D(t) ) less than 5, the probability decreases.But the problem is asking for "the probability that the adjusted difficulty ( D_{adj}(t) ) exceeds 5 at any given time t". So, perhaps it's asking for the maximum possible probability, which is 0.5 when ( D(t) = 5 ).Alternatively, maybe it's asking for the expected probability over all t. Since ( D(t) ) oscillates between 1 and 5, the expected value of ( P(D_{adj}(t) > 5) ) would require integrating over the period.But that seems more complicated, and the problem might be expecting the maximum probability, which is 0.5.Wait, but 0.5 seems high because when ( D(t) = 5 ), adding a Gaussian with mean 0 and SD 0.5, the probability of exceeding 5 is 0.5. But in reality, when ( D(t) ) is 5, the difficulty can't go below 5, but it can go above. Wait, no, because ( D(t) ) is 5, and X can be positive or negative. So, ( D_{adj}(t) ) can be 5 + X, which can be greater than 5 or less than 5. So, the probability that it's greater than 5 is indeed 0.5.But wait, the original difficulty is 5, and the adjusted difficulty is 5 + X. So, the probability that it's above 5 is 0.5, same as below.But the problem is about exceeding 5, so it's 0.5.But wait, let me think again. If ( D(t) = 5 ), then ( D_{adj}(t) = 5 + X ). So, the probability that ( D_{adj}(t) > 5 ) is the same as ( X > 0 ), which is 0.5.But if ( D(t) ) is less than 5, say 4, then ( D_{adj}(t) = 4 + X ). The probability that 4 + X > 5 is ( P(X > 1) ). Since X ~ N(0, 0.25), 1 is 2 standard deviations above the mean. So, ( P(X > 1) = P(Z > 2) ≈ 0.0228 ).So, the probability varies depending on ( D(t) ). But the question is asking for "the probability that the adjusted difficulty ( D_{adj}(t) ) exceeds 5 at any given time t". So, perhaps it's asking for the maximum probability over all t, which is 0.5.Alternatively, maybe it's asking for the probability that, for some t, ( D_{adj}(t) > 5 ). But that would be a different calculation, involving the probability that the maximum of ( D_{adj}(t) ) exceeds 5 over the entire game duration. But since the game is infinite, that probability would approach 1, because the Gaussian noise will eventually cause ( D_{adj}(t) ) to exceed 5.But the problem doesn't specify a time frame, so maybe it's just asking for the probability at a given t, which depends on ( D(t) ). But since the question is phrased as "at any given time t", perhaps it's considering the worst-case scenario, which is when ( D(t) ) is at its maximum, giving a probability of 0.5.Alternatively, maybe the question is considering the probability that, for a randomly chosen t, ( D_{adj}(t) > 5 ). In that case, we would need to average over all t. Since ( D(t) ) oscillates between 1 and 5, we can model this as a periodic function and compute the expected probability.Let me try that approach.The expected probability would be the average of ( P(D_{adj}(t) > 5) ) over one period, which is 10 minutes.So, ( E[P] = frac{1}{10} int_{0}^{10} P(D(t) + X > 5) dt ).But ( P(D(t) + X > 5) = P(X > 5 - D(t)) ).Since X is N(0, 0.25), this is ( 1 - Phileft( frac{5 - D(t)}{0.5} right) ).So, ( E[P] = frac{1}{10} int_{0}^{10} left[ 1 - Phileft( frac{5 - D(t)}{0.5} right) right] dt ).But ( D(t) = 2 sinleft( frac{pi}{5} t right) + 3 ).So, ( 5 - D(t) = 5 - 3 - 2 sinleft( frac{pi}{5} t right) = 2 - 2 sinleft( frac{pi}{5} t right) = 2(1 - sinleft( frac{pi}{5} t right)) ).Thus, ( frac{5 - D(t)}{0.5} = frac{2(1 - sin(frac{pi}{5} t))}{0.5} = 4(1 - sin(frac{pi}{5} t)) ).So, ( E[P] = frac{1}{10} int_{0}^{10} left[ 1 - Phi(4(1 - sin(frac{pi}{5} t))) right] dt ).This integral looks complicated. Maybe we can make a substitution. Let me set ( u = frac{pi}{5} t ), so when t=0, u=0; t=10, u=2π. Then, dt = ( frac{5}{pi} du ).So, ( E[P] = frac{1}{10} cdot frac{5}{pi} int_{0}^{2pi} left[ 1 - Phi(4(1 - sin u)) right] du ).Simplifying, ( E[P] = frac{1}{2pi} int_{0}^{2pi} left[ 1 - Phi(4(1 - sin u)) right] du ).This integral is still quite complex because it involves the standard normal CDF composed with a function of u. I don't think there's an analytical solution, so we might need to approximate it numerically.But since this is a thought process, I can consider that the integral would require numerical methods. However, perhaps the problem is expecting a simpler answer, considering the maximum probability when ( D(t) = 5 ), which is 0.5.Alternatively, maybe the problem is considering that the adjusted difficulty can exceed 5 only when ( D(t) ) is close to 5, and the probability is low otherwise. But without more context, it's hard to say.Wait, another approach: since ( D(t) ) is a sine wave with amplitude 2, centered at 3, and the Gaussian noise has a standard deviation of 0.5, the probability that ( D_{adj}(t) > 5 ) is equivalent to ( X > 5 - D(t) ). The maximum value of ( 5 - D(t) ) is 2 (when ( D(t) = 3 )), but wait, no. Wait, when ( D(t) ) is at its minimum, 1, ( 5 - D(t) = 4 ). When ( D(t) ) is at its maximum, 5, ( 5 - D(t) = 0 ).Wait, no, that's not right. ( 5 - D(t) ) is the threshold for X. So, when ( D(t) ) is 5, ( 5 - D(t) = 0 ), so ( X > 0 ), which is 0.5. When ( D(t) ) is 4, ( 5 - D(t) = 1 ), so ( X > 1 ), which is about 0.0228. When ( D(t) ) is 3, ( 5 - D(t) = 2 ), so ( X > 2 ), which is about 0.0228 (wait, no, 2 standard deviations is about 0.0228, but since the standard deviation is 0.5, 2 standard deviations is 1, so ( X > 2 ) would be even smaller, like 0.0032 or something.Wait, let me correct that. The standard deviation is 0.5, so 1 standard deviation is 0.5. So, when ( D(t) = 4 ), ( 5 - D(t) = 1 ), which is 2 standard deviations above the mean (since 1 / 0.5 = 2). So, ( P(X > 1) = P(Z > 2) ≈ 0.0228 ).When ( D(t) = 3 ), ( 5 - D(t) = 2 ), which is 4 standard deviations above the mean (2 / 0.5 = 4). So, ( P(X > 2) ≈ 0.0032 ).When ( D(t) = 1 ), ( 5 - D(t) = 4 ), which is 8 standard deviations above the mean. The probability of that is practically zero.So, the probability that ( D_{adj}(t) > 5 ) varies depending on ( D(t) ). The maximum probability is 0.5 when ( D(t) = 5 ), and it decreases as ( D(t) ) decreases.But the problem is asking for "the probability that the adjusted difficulty ( D_{adj}(t) ) exceeds 5 at any given time t". So, perhaps it's considering the maximum possible probability, which is 0.5.Alternatively, if we consider the expected probability over all t, we'd have to integrate over the period, but that's complicated. Since the problem is likely expecting a simple answer, I think the maximum probability is 0.5.But wait, let me think again. When ( D(t) = 5 ), the adjusted difficulty is ( 5 + X ), so the probability of exceeding 5 is 0.5. But when ( D(t) ) is less than 5, the probability is less. So, the maximum probability is 0.5, but the average probability is much lower.But the problem doesn't specify whether it's asking for the maximum probability or the average. Since it's phrased as "at any given time t", I think it's asking for the maximum probability, which is 0.5.But wait, another perspective: since ( D(t) ) is a function of t, and X is independent, the probability that ( D_{adj}(t) > 5 ) for some t is 1, because over an infinite time, the Gaussian noise will eventually cause ( D_{adj}(t) ) to exceed 5. But the problem doesn't specify a time frame, so maybe it's asking for the probability at a specific t, which is 0.5 when ( D(t) = 5 ).But I'm not sure. Maybe the problem is expecting the probability when ( D(t) ) is at its maximum, so the answer is 0.5.Alternatively, perhaps the problem is considering that the adjusted difficulty can't exceed 5, but that's not the case because the Gaussian noise can add to it.Wait, let me think about the distribution of ( D_{adj}(t) ). For a given t, ( D_{adj}(t) ) is normally distributed with mean ( D(t) ) and variance 0.25. So, the probability that it exceeds 5 is ( P(N(D(t), 0.25) > 5) ).This is equivalent to ( P(Z > (5 - D(t))/0.5) ), where Z is standard normal.So, the probability is ( 1 - Phi( (5 - D(t))/0.5 ) ).But since ( D(t) ) varies, the probability varies. The maximum probability occurs when ( D(t) ) is as small as possible, but wait, no. Wait, when ( D(t) ) is as large as possible, ( 5 - D(t) ) is as small as possible, so ( (5 - D(t))/0.5 ) is as small as possible, making ( Phi ) as large as possible, so ( 1 - Phi ) as small as possible. Wait, that's the opposite.Wait, no. Let me clarify:When ( D(t) ) is large (close to 5), ( 5 - D(t) ) is small (close to 0), so ( (5 - D(t))/0.5 ) is small, so ( Phi ) is close to 0.5, so ( 1 - Phi ) is close to 0.5.When ( D(t) ) is small (close to 1), ( 5 - D(t) ) is large (4), so ( (5 - D(t))/0.5 = 8 ), which is way in the tail of the normal distribution, so ( Phi(8) ) is practically 1, so ( 1 - Phi(8) ) is practically 0.Wait, so the probability ( P(D_{adj}(t) > 5) ) is highest when ( D(t) ) is highest, because then the threshold ( 5 - D(t) ) is smallest, making it easier for X to exceed it.So, the maximum probability is when ( D(t) = 5 ), giving ( P = 0.5 ).Therefore, the probability that the adjusted difficulty exceeds 5 at any given time t is 0.5.But wait, that seems counterintuitive because when ( D(t) = 5 ), the adjusted difficulty is 5 + X, which is symmetric around 5, so half the time it's above 5, half below. So, the probability is indeed 0.5.But if the question is asking for the probability that, over time, the adjusted difficulty ever exceeds 5, then it's almost certain, but that's not what the question is asking. It's asking for the probability at any given time t, which is 0.5 when ( D(t) = 5 ).So, I think the answer is 0.5, or 50%.But let me double-check. If ( D(t) = 5 ), then ( D_{adj}(t) = 5 + X ), where X ~ N(0, 0.25). So, the distribution of ( D_{adj}(t) ) is N(5, 0.25). The probability that a normal variable exceeds its mean is 0.5. So yes, that's correct.Therefore, the probability is 0.5.But wait, the problem says "the probability that the adjusted difficulty ( D_{adj}(t) ) exceeds 5 at any given time t". So, for a specific t, it's 0.5 when ( D(t) = 5 ), but for other t's, it's less. But since the question is asking for "at any given time t", perhaps it's considering the maximum possible probability, which is 0.5.Alternatively, if we consider that for any t, the probability is ( P(X > 5 - D(t)) ), and since ( D(t) ) can be as high as 5, the maximum probability is 0.5.So, I think the answer is 0.5, or 50%.</think>

answer:<think>Alright, so I've got this problem about a network of 50 supporters for Barack Obama's 2008 campaign. It's represented as a graph where each node is a supporter and edges are connections between them. There are two parts to the problem. Let me try to tackle them one by one.Starting with part 1: The degree distribution follows a truncated power-law distribution, which is given by P(k) = Ck^{-α} for k_min ≤ k ≤ k_max. They've provided k_min = 3, k_max = 12, and the average degree is 5. I need to find the scaling exponent α.Okay, so first, I remember that in a power-law distribution, the normalization constant C is chosen so that the sum of P(k) over all possible k equals 1. So, for this truncated case, C is the sum from k = k_min to k_max of k^{-α}.So, C = 1 / Σ (from k=3 to 12) k^{-α}Then, the average degree <k> is given by Σ (k * P(k)) from k=3 to 12. That should equal 5.So, <k> = Σ (k * C * k^{-α}) from k=3 to 12 = 5Simplify that: <k> = C * Σ (k^{1 - α}) from k=3 to 12 = 5So, substituting C from earlier, we have:(1 / Σ (k^{-α} from 3 to 12)) * Σ (k^{1 - α} from 3 to 12) = 5Let me denote S1 = Σ (k^{-α} from 3 to 12) and S2 = Σ (k^{1 - α} from 3 to 12). So, S2 / S1 = 5.Therefore, I need to find α such that S2 = 5 * S1.Hmm, this seems like an equation I can't solve analytically because it's a sum over k of different exponents. I might need to use numerical methods or trial and error to find α.Let me write out S1 and S2:S1 = 3^{-α} + 4^{-α} + 5^{-α} + ... + 12^{-α}S2 = 3^{1 - α} + 4^{1 - α} + 5^{1 - α} + ... + 12^{1 - α}Which can also be written as:S2 = 3 * 3^{-α} + 4 * 4^{-α} + ... + 12 * 12^{-α} = Σ (k * k^{-α}) from 3 to 12 = Σ k^{1 - α}So, S2 is just the sum of k^{1 - α} from 3 to 12.Given that, I can compute S1 and S2 for different α and find when S2/S1 = 5.Let me try some values for α.First, let's try α = 2.Compute S1:3^{-2} = 1/9 ≈ 0.11114^{-2} = 1/16 ≈ 0.06255^{-2} = 1/25 = 0.046^{-2} ≈ 0.02787^{-2} ≈ 0.02048^{-2} ≈ 0.01569^{-2} ≈ 0.012310^{-2} = 0.0111^{-2} ≈ 0.008312^{-2} ≈ 0.0069Adding these up:0.1111 + 0.0625 = 0.1736+0.04 = 0.2136+0.0278 ≈ 0.2414+0.0204 ≈ 0.2618+0.0156 ≈ 0.2774+0.0123 ≈ 0.2897+0.01 ≈ 0.2997+0.0083 ≈ 0.308+0.0069 ≈ 0.3149So, S1 ≈ 0.3149 when α=2.Now, S2:3^{1 - 2} = 3^{-1} ≈ 0.33334^{-1} = 0.255^{-1} = 0.26^{-1} ≈ 0.16677^{-1} ≈ 0.14298^{-1} = 0.1259^{-1} ≈ 0.111110^{-1} = 0.111^{-1} ≈ 0.090912^{-1} ≈ 0.0833Adding these:0.3333 + 0.25 = 0.5833+0.2 = 0.7833+0.1667 ≈ 0.95+0.1429 ≈ 1.0929+0.125 ≈ 1.2179+0.1111 ≈ 1.329+0.1 ≈ 1.429+0.0909 ≈ 1.5199+0.0833 ≈ 1.6032So, S2 ≈ 1.6032 when α=2.Then, S2/S1 ≈ 1.6032 / 0.3149 ≈ 5.09.Hmm, that's pretty close to 5. The ratio is approximately 5.09, which is just slightly above 5.So, maybe α is slightly less than 2 because if I decrease α, the terms in S1 and S2 will change.Wait, let's see: If α decreases, say to 1.9, then the exponents become less negative, so the terms in S1 and S2 will be larger.Wait, let me think: For S1, when α decreases, each term k^{-α} increases, so S1 increases. Similarly, S2 is the sum of k^{1 - α}, so as α decreases, 1 - α increases, so each term in S2 increases as well.But how does S2/S1 behave? Let's see:If α decreases, S1 increases and S2 increases. But which one increases more?At α=2, S2/S1 ≈5.09.If I decrease α slightly, say to 1.9, let's compute S1 and S2.Compute S1 for α=1.9:3^{-1.9} ≈ e^{-1.9 ln3} ≈ e^{-1.9*1.0986} ≈ e^{-2.087} ≈ 0.1244^{-1.9} ≈ e^{-1.9 ln4} ≈ e^{-1.9*1.386} ≈ e^{-2.623} ≈ 0.0725^{-1.9} ≈ e^{-1.9 ln5} ≈ e^{-1.9*1.609} ≈ e^{-3.057} ≈ 0.0466^{-1.9} ≈ e^{-1.9 ln6} ≈ e^{-1.9*1.792} ≈ e^{-3.405} ≈ 0.0337^{-1.9} ≈ e^{-1.9 ln7} ≈ e^{-1.9*1.946} ≈ e^{-3.697} ≈ 0.0248^{-1.9} ≈ e^{-1.9 ln8} ≈ e^{-1.9*2.079} ≈ e^{-3.95} ≈ 0.0199^{-1.9} ≈ e^{-1.9 ln9} ≈ e^{-1.9*2.197} ≈ e^{-4.174} ≈ 0.01510^{-1.9} ≈ e^{-1.9 ln10} ≈ e^{-1.9*2.302} ≈ e^{-4.374} ≈ 0.01211^{-1.9} ≈ e^{-1.9 ln11} ≈ e^{-1.9*2.398} ≈ e^{-4.556} ≈ 0.01012^{-1.9} ≈ e^{-1.9 ln12} ≈ e^{-1.9*2.485} ≈ e^{-4.721} ≈ 0.008Adding these up:0.124 + 0.072 = 0.196+0.046 = 0.242+0.033 = 0.275+0.024 = 0.299+0.019 = 0.318+0.015 = 0.333+0.012 = 0.345+0.010 = 0.355+0.008 = 0.363So, S1 ≈ 0.363 when α=1.9.Now, S2 for α=1.9 is Σ k^{1 - 1.9} = Σ k^{-0.9}.Compute each term:3^{-0.9} ≈ e^{-0.9 ln3} ≈ e^{-0.9*1.0986} ≈ e^{-0.989} ≈ 0.3724^{-0.9} ≈ e^{-0.9 ln4} ≈ e^{-0.9*1.386} ≈ e^{-1.247} ≈ 0.2875^{-0.9} ≈ e^{-0.9 ln5} ≈ e^{-0.9*1.609} ≈ e^{-1.448} ≈ 0.2356^{-0.9} ≈ e^{-0.9 ln6} ≈ e^{-0.9*1.792} ≈ e^{-1.613} ≈ 0.2017^{-0.9} ≈ e^{-0.9 ln7} ≈ e^{-0.9*1.946} ≈ e^{-1.751} ≈ 0.1748^{-0.9} ≈ e^{-0.9 ln8} ≈ e^{-0.9*2.079} ≈ e^{-1.871} ≈ 0.1549^{-0.9} ≈ e^{-0.9 ln9} ≈ e^{-0.9*2.197} ≈ e^{-1.977} ≈ 0.13910^{-0.9} ≈ e^{-0.9 ln10} ≈ e^{-0.9*2.302} ≈ e^{-2.072} ≈ 0.12711^{-0.9} ≈ e^{-0.9 ln11} ≈ e^{-0.9*2.398} ≈ e^{-2.158} ≈ 0.11512^{-0.9} ≈ e^{-0.9 ln12} ≈ e^{-0.9*2.485} ≈ e^{-2.236} ≈ 0.105Adding these up:0.372 + 0.287 = 0.659+0.235 = 0.894+0.201 = 1.095+0.174 = 1.269+0.154 = 1.423+0.139 = 1.562+0.127 = 1.689+0.115 = 1.804+0.105 = 1.909So, S2 ≈ 1.909 when α=1.9.Then, S2/S1 ≈ 1.909 / 0.363 ≈ 5.26.Hmm, that's higher than 5.09 when α=2. Wait, but we wanted S2/S1 = 5. So, when α=2, it's 5.09, which is just above 5. When α=1.9, it's 5.26, which is even higher. That suggests that as α decreases, S2/S1 increases, which is the opposite of what I thought earlier.Wait, maybe I made a mistake in reasoning. Let me think again.When α increases, the exponents become more negative, so the terms in S1 and S2 decrease. So, S1 and S2 both decrease as α increases. But how does their ratio behave?At α=2, S2/S1≈5.09At α=3, let's compute S1 and S2.S1 for α=3:3^{-3}=1/27≈0.0374^{-3}=1/64≈0.01565^{-3}=0.0086^{-3}≈0.00467^{-3}≈0.00298^{-3}=0.001959^{-3}≈0.0013710^{-3}=0.00111^{-3}≈0.0007512^{-3}≈0.00057Adding up:0.037 + 0.0156 = 0.0526+0.008 = 0.0606+0.0046 ≈0.0652+0.0029≈0.0681+0.00195≈0.07+0.00137≈0.0714+0.001≈0.0724+0.00075≈0.07315+0.00057≈0.0737So, S1≈0.0737 when α=3.S2 for α=3 is Σ k^{-2} from 3 to12.Which is the same as S1 when α=2, which was ≈0.3149.Wait, no, S2 when α=3 is Σ k^{1 - 3}=Σ k^{-2}= same as S1 when α=2, which was ≈0.3149.Wait, no, actually, S2 when α=3 is Σ k^{-2} from 3 to12, which is the same as S1 when α=2, which was ≈0.3149.So, S2≈0.3149 when α=3.Thus, S2/S1≈0.3149 / 0.0737≈4.27.So, at α=3, the ratio is ≈4.27, which is less than 5.So, as α increases from 2 to 3, S2/S1 decreases from ~5.09 to ~4.27.At α=2, ratio≈5.09At α=1.9, ratio≈5.26Wait, so as α decreases, the ratio increases.But we need the ratio to be exactly 5.At α=2, it's 5.09, which is just above 5.So, maybe α is slightly above 2? Wait, but when α increases, the ratio decreases.Wait, no, when α increases, S2/S1 decreases, as seen from α=2 to 3.So, to get S2/S1=5, which is just slightly above 5.09 at α=2, we need to decrease α slightly below 2 to make the ratio higher.Wait, but at α=2, ratio≈5.09, which is higher than 5.Wait, but we need S2/S1=5, which is less than 5.09.So, actually, if we increase α slightly above 2, the ratio would decrease from 5.09 towards 4.27 as α increases to 3.Wait, so if at α=2, ratio=5.09, and we need ratio=5, which is less than 5.09, we need to increase α slightly above 2.Wait, that seems contradictory to my earlier thought.Wait, let's think about the function f(α)=S2/S1.We have:At α=2, f(α)=5.09At α=3, f(α)=4.27So, f(α) is a decreasing function of α.Therefore, to get f(α)=5, which is less than 5.09, we need to increase α beyond 2.Wait, but 5 is less than 5.09, so since f(α) is decreasing, we need a higher α to get a lower ratio.Wait, but 5 is less than 5.09, so to get a lower ratio, we need a higher α.Wait, but 5 is less than 5.09, so if we increase α, f(α) decreases further below 5.09, which would make it less than 5.Wait, but we need f(α)=5, which is less than 5.09, so we need to increase α beyond 2.Wait, but when α increases, f(α) decreases, so to reach 5, which is less than 5.09, we need to increase α.Wait, that seems correct.So, let's try α=2.1.Compute S1 and S2 for α=2.1.First, S1=Σ k^{-2.1} from 3 to12.Compute each term:3^{-2.1}=e^{-2.1 ln3}≈e^{-2.1*1.0986}=e^{-2.307}≈0.1004^{-2.1}=e^{-2.1 ln4}=e^{-2.1*1.386}=e^{-2.901}≈0.0545^{-2.1}=e^{-2.1 ln5}=e^{-2.1*1.609}=e^{-3.379}≈0.0346^{-2.1}=e^{-2.1 ln6}=e^{-2.1*1.792}=e^{-3.763}≈0.0247^{-2.1}=e^{-2.1 ln7}=e^{-2.1*1.946}=e^{-4.087}≈0.0178^{-2.1}=e^{-2.1 ln8}=e^{-2.1*2.079}=e^{-4.366}≈0.0139^{-2.1}=e^{-2.1 ln9}=e^{-2.1*2.197}=e^{-4.614}≈0.00910^{-2.1}=e^{-2.1 ln10}=e^{-2.1*2.302}=e^{-4.834}≈0.00811^{-2.1}=e^{-2.1 ln11}=e^{-2.1*2.398}=e^{-5.036}≈0.00612^{-2.1}=e^{-2.1 ln12}=e^{-2.1*2.485}=e^{-5.218}≈0.005Adding these up:0.100 + 0.054 = 0.154+0.034 = 0.188+0.024 = 0.212+0.017 = 0.229+0.013 = 0.242+0.009 = 0.251+0.008 = 0.259+0.006 = 0.265+0.005 = 0.270So, S1≈0.270 when α=2.1.Now, S2=Σ k^{1 - 2.1}=Σ k^{-1.1} from 3 to12.Compute each term:3^{-1.1}=e^{-1.1 ln3}=e^{-1.1*1.0986}=e^{-1.208}≈0.2994^{-1.1}=e^{-1.1 ln4}=e^{-1.1*1.386}=e^{-1.525}≈0.2185^{-1.1}=e^{-1.1 ln5}=e^{-1.1*1.609}=e^{-1.770}≈0.1706^{-1.1}=e^{-1.1 ln6}=e^{-1.1*1.792}=e^{-2.0}≈0.1357^{-1.1}=e^{-1.1 ln7}=e^{-1.1*1.946}=e^{-2.140}≈0.1178^{-1.1}=e^{-1.1 ln8}=e^{-1.1*2.079}=e^{-2.287}≈0.1019^{-1.1}=e^{-1.1 ln9}=e^{-1.1*2.197}=e^{-2.417}≈0.08910^{-1.1}=e^{-1.1 ln10}=e^{-1.1*2.302}=e^{-2.532}≈0.07911^{-1.1}=e^{-1.1 ln11}=e^{-1.1*2.398}=e^{-2.638}≈0.07012^{-1.1}=e^{-1.1 ln12}=e^{-1.1*2.485}=e^{-2.733}≈0.064Adding these up:0.299 + 0.218 = 0.517+0.170 = 0.687+0.135 = 0.822+0.117 = 0.939+0.101 = 1.040+0.089 = 1.129+0.079 = 1.208+0.070 = 1.278+0.064 = 1.342So, S2≈1.342 when α=2.1.Then, S2/S1≈1.342 / 0.270≈4.97.That's very close to 5. So, at α=2.1, the ratio is approximately 4.97, which is just slightly below 5.Previously, at α=2, the ratio was≈5.09.So, we can interpolate between α=2 and α=2.1.At α=2: ratio=5.09At α=2.1: ratio=4.97We need ratio=5.So, let's set up a linear approximation.Let’s denote f(α)=S2/S1.We have f(2)=5.09f(2.1)=4.97We need to find α such that f(α)=5.Assuming f(α) is approximately linear between α=2 and α=2.1.The change in f is 4.97 - 5.09 = -0.12 over Δα=0.1.We need Δf=5 - 5.09= -0.09.So, the fraction is (-0.09)/(-0.12)=0.75.Thus, α=2 + 0.75*0.1=2 +0.075=2.075.So, approximately α≈2.075.To check, let's compute f(2.075).But this might be tedious, but given the linearity assumption, it's approximately 2.075.Alternatively, we can use more precise methods, but for the sake of this problem, maybe α≈2.08.But let me see if I can get a better estimate.Alternatively, let's compute f(2.05).Compute S1 and S2 for α=2.05.First, S1=Σ k^{-2.05} from 3 to12.Compute each term:3^{-2.05}=e^{-2.05 ln3}=e^{-2.05*1.0986}=e^{-2.252}=≈0.1044^{-2.05}=e^{-2.05 ln4}=e^{-2.05*1.386}=e^{-2.835}=≈0.0585^{-2.05}=e^{-2.05 ln5}=e^{-2.05*1.609}=e^{-3.299}=≈0.0366^{-2.05}=e^{-2.05 ln6}=e^{-2.05*1.792}=e^{-3.675}=≈0.0257^{-2.05}=e^{-2.05 ln7}=e^{-2.05*1.946}=e^{-4.000}=≈0.0188^{-2.05}=e^{-2.05 ln8}=e^{-2.05*2.079}=e^{-4.254}=≈0.0149^{-2.05}=e^{-2.05 ln9}=e^{-2.05*2.197}=e^{-4.506}=≈0.01010^{-2.05}=e^{-2.05 ln10}=e^{-2.05*2.302}=e^{-4.719}=≈0.00811^{-2.05}=e^{-2.05 ln11}=e^{-2.05*2.398}=e^{-4.916}=≈0.00612^{-2.05}=e^{-2.05 ln12}=e^{-2.05*2.485}=e^{-5.104}=≈0.005Adding these up:0.104 + 0.058 = 0.162+0.036 = 0.198+0.025 = 0.223+0.018 = 0.241+0.014 = 0.255+0.010 = 0.265+0.008 = 0.273+0.006 = 0.279+0.005 = 0.284So, S1≈0.284 when α=2.05.Now, S2=Σ k^{-1.05} from 3 to12.Compute each term:3^{-1.05}=e^{-1.05 ln3}=e^{-1.05*1.0986}=e^{-1.153}=≈0.3164^{-1.05}=e^{-1.05 ln4}=e^{-1.05*1.386}=e^{-1.455}=≈0.2335^{-1.05}=e^{-1.05 ln5}=e^{-1.05*1.609}=e^{-1.689}=≈0.1846^{-1.05}=e^{-1.05 ln6}=e^{-1.05*1.792}=e^{-1.887}=≈0.1527^{-1.05}=e^{-1.05 ln7}=e^{-1.05*1.946}=e^{-2.043}=≈0.1298^{-1.05}=e^{-1.05 ln8}=e^{-1.05*2.079}=e^{-2.173}=≈0.1139^{-1.05}=e^{-1.05 ln9}=e^{-1.05*2.197}=e^{-2.297}=≈0.10010^{-1.05}=e^{-1.05 ln10}=e^{-1.05*2.302}=e^{-2.417}=≈0.08911^{-1.05}=e^{-1.05 ln11}=e^{-1.05*2.398}=e^{-2.523}=≈0.07912^{-1.05}=e^{-1.05 ln12}=e^{-1.05*2.485}=e^{-2.614}=≈0.070Adding these up:0.316 + 0.233 = 0.549+0.184 = 0.733+0.152 = 0.885+0.129 = 1.014+0.113 = 1.127+0.100 = 1.227+0.089 = 1.316+0.079 = 1.395+0.070 = 1.465So, S2≈1.465 when α=2.05.Then, S2/S1≈1.465 / 0.284≈5.16.Hmm, that's higher than 5.09 at α=2.Wait, that can't be right because when α increases, f(α) should decrease.Wait, but at α=2.05, f(α)=5.16, which is higher than at α=2 (5.09). That contradicts the earlier assumption that f(α) is decreasing with α.Wait, maybe my calculations are off.Wait, at α=2.05, S1=0.284, S2=1.465, so S2/S1≈5.16.But at α=2, S2/S1≈5.09.Wait, so as α increases from 2 to 2.05, f(α) increases from 5.09 to 5.16, which suggests that f(α) is increasing with α, which contradicts earlier.Wait, that can't be. There must be a mistake in calculations.Wait, let me double-check S1 and S2 for α=2.05.Wait, for S1, when α=2.05, each term is k^{-2.05}.3^{-2.05}=e^{-2.05*1.0986}=e^{-2.252}=≈0.1044^{-2.05}=e^{-2.05*1.386}=e^{-2.835}=≈0.0585^{-2.05}=e^{-2.05*1.609}=e^{-3.299}=≈0.0366^{-2.05}=e^{-2.05*1.792}=e^{-3.675}=≈0.0257^{-2.05}=e^{-2.05*1.946}=e^{-4.000}=≈0.0188^{-2.05}=e^{-2.05*2.079}=e^{-4.254}=≈0.0149^{-2.05}=e^{-2.05*2.197}=e^{-4.506}=≈0.01010^{-2.05}=e^{-2.05*2.302}=e^{-4.719}=≈0.00811^{-2.05}=e^{-2.05*2.398}=e^{-4.916}=≈0.00612^{-2.05}=e^{-2.05*2.485}=e^{-5.104}=≈0.005Adding these: 0.104+0.058=0.162; +0.036=0.198; +0.025=0.223; +0.018=0.241; +0.014=0.255; +0.010=0.265; +0.008=0.273; +0.006=0.279; +0.005=0.284. So S1≈0.284.Now, S2=Σ k^{-1.05}.3^{-1.05}=e^{-1.05*1.0986}=e^{-1.153}=≈0.3164^{-1.05}=e^{-1.05*1.386}=e^{-1.455}=≈0.2335^{-1.05}=e^{-1.05*1.609}=e^{-1.689}=≈0.1846^{-1.05}=e^{-1.05*1.792}=e^{-1.887}=≈0.1527^{-1.05}=e^{-1.05*1.946}=e^{-2.043}=≈0.1298^{-1.05}=e^{-1.05*2.079}=e^{-2.173}=≈0.1139^{-1.05}=e^{-1.05*2.197}=e^{-2.297}=≈0.10010^{-1.05}=e^{-1.05*2.302}=e^{-2.417}=≈0.08911^{-1.05}=e^{-1.05*2.398}=e^{-2.523}=≈0.07912^{-1.05}=e^{-1.05*2.485}=e^{-2.614}=≈0.070Adding these: 0.316+0.233=0.549; +0.184=0.733; +0.152=0.885; +0.129=1.014; +0.113=1.127; +0.100=1.227; +0.089=1.316; +0.079=1.395; +0.070=1.465. So S2≈1.465.Thus, S2/S1≈1.465 / 0.284≈5.16.Wait, so at α=2.05, the ratio is higher than at α=2, which suggests that f(α) is increasing with α, which contradicts the earlier assumption.But earlier, at α=2.1, f(α)=4.97, which is lower than at α=2.Wait, that must mean that the function f(α) is not monotonic? Or perhaps my calculations are incorrect.Wait, let me check α=2.1 again.At α=2.1, S1≈0.270, S2≈1.342, so S2/S1≈4.97.Wait, but at α=2.05, S2/S1≈5.16, which is higher than at α=2.1.This suggests that f(α) first increases and then decreases as α increases, which is possible if the function has a maximum somewhere.Wait, but that complicates things. Alternatively, perhaps my calculations are off.Wait, let me try α=2.075.Compute S1 and S2 for α=2.075.First, S1=Σ k^{-2.075} from 3 to12.Compute each term:3^{-2.075}=e^{-2.075*1.0986}=e^{-2.280}=≈0.1014^{-2.075}=e^{-2.075*1.386}=e^{-2.876}=≈0.0565^{-2.075}=e^{-2.075*1.609}=e^{-3.343}=≈0.0366^{-2.075}=e^{-2.075*1.792}=e^{-3.716}=≈0.0247^{-2.075}=e^{-2.075*1.946}=e^{-4.037}=≈0.0188^{-2.075}=e^{-2.075*2.079}=e^{-4.312}=≈0.0139^{-2.075}=e^{-2.075*2.197}=e^{-4.563}=≈0.00910^{-2.075}=e^{-2.075*2.302}=e^{-4.777}=≈0.00811^{-2.075}=e^{-2.075*2.398}=e^{-5.000}=≈0.00612^{-2.075}=e^{-2.075*2.485}=e^{-5.166}=≈0.005Adding these up:0.101 + 0.056 = 0.157+0.036 = 0.193+0.024 = 0.217+0.018 = 0.235+0.013 = 0.248+0.009 = 0.257+0.008 = 0.265+0.006 = 0.271+0.005 = 0.276So, S1≈0.276 when α=2.075.Now, S2=Σ k^{-1.075} from 3 to12.Compute each term:3^{-1.075}=e^{-1.075*1.0986}=e^{-1.181}=≈0.3074^{-1.075}=e^{-1.075*1.386}=e^{-1.490}=≈0.2255^{-1.075}=e^{-1.075*1.609}=e^{-1.730}=≈0.1776^{-1.075}=e^{-1.075*1.792}=e^{-1.927}=≈0.1447^{-1.075}=e^{-1.075*1.946}=e^{-2.090}=≈0.1248^{-1.075}=e^{-1.075*2.079}=e^{-2.232}=≈0.1059^{-1.075}=e^{-1.075*2.197}=e^{-2.363}=≈0.09110^{-1.075}=e^{-1.075*2.302}=e^{-2.476}=≈0.07911^{-1.075}=e^{-1.075*2.398}=e^{-2.583}=≈0.07212^{-1.075}=e^{-1.075*2.485}=e^{-2.675}=≈0.064Adding these up:0.307 + 0.225 = 0.532+0.177 = 0.709+0.144 = 0.853+0.124 = 0.977+0.105 = 1.082+0.091 = 1.173+0.079 = 1.252+0.072 = 1.324+0.064 = 1.388So, S2≈1.388 when α=2.075.Then, S2/S1≈1.388 / 0.276≈5.03.That's very close to 5. So, at α≈2.075, the ratio is≈5.03, which is just slightly above 5.Previously, at α=2.1, the ratio was≈4.97.So, we can interpolate between α=2.075 and α=2.1.At α=2.075: ratio=5.03At α=2.1: ratio=4.97We need ratio=5.The difference between 5.03 and 4.97 is 0.06 over Δα=0.025.We need to find α where ratio=5, which is 0.03 above 4.97.So, the fraction is 0.03 / 0.06=0.5.Thus, α=2.1 - 0.5*0.025=2.1 -0.0125=2.0875.So, approximately α≈2.0875.To check, let's compute f(2.0875).But this is getting too detailed, and for the sake of time, I think α≈2.09 is a good approximation.Alternatively, using linear approximation between α=2.075 (5.03) and α=2.1 (4.97):The desired ratio is 5, which is 0.03 below 5.03.The total change from 5.03 to 4.97 is -0.06 over Δα=0.025.So, to get a change of -0.03, we need Δα=0.025*(0.03/0.06)=0.0125.Thus, α=2.075 +0.0125=2.0875.So, α≈2.0875.Rounding to two decimal places, α≈2.09.Therefore, the scaling exponent α is approximately 2.09.Now, moving to part 2: Calculate the expected number of influencers, defined as supporters with degree ≥8.Given the degree distribution P(k)=Ck^{-α}, with α≈2.09, k_min=3, k_max=12.The expected number is N * Σ P(k) for k=8 to12, where N=50.First, compute C=1 / Σ (k^{-α} from 3 to12).We already computed Σ k^{-α} for α=2.09, but let's compute it accurately.Alternatively, since we know that at α≈2.09, S1≈0.276 (from earlier calculation at α=2.075, which was close to 2.09).Wait, actually, at α=2.075, S1≈0.276.But let's compute S1 for α=2.09.Compute S1=Σ k^{-2.09} from 3 to12.Compute each term:3^{-2.09}=e^{-2.09*1.0986}=e^{-2.297}=≈0.1004^{-2.09}=e^{-2.09*1.386}=e^{-2.903}=≈0.0545^{-2.09}=e^{-2.09*1.609}=e^{-3.365}=≈0.0356^{-2.09}=e^{-2.09*1.792}=e^{-3.743}=≈0.0247^{-2.09}=e^{-2.09*1.946}=e^{-4.050}=≈0.0178^{-2.09}=e^{-2.09*2.079}=e^{-4.333}=≈0.0139^{-2.09}=e^{-2.09*2.197}=e^{-4.603}=≈0.00910^{-2.09}=e^{-2.09*2.302}=e^{-4.813}=≈0.00811^{-2.09}=e^{-2.09*2.398}=e^{-5.000}=≈0.00612^{-2.09}=e^{-2.09*2.485}=e^{-5.196}=≈0.005Adding these up:0.100 + 0.054 = 0.154+0.035 = 0.189+0.024 = 0.213+0.017 = 0.230+0.013 = 0.243+0.009 = 0.252+0.008 = 0.260+0.006 = 0.266+0.005 = 0.271So, S1≈0.271 when α=2.09.Thus, C=1/0.271≈3.69.Now, compute the probability that a node has degree ≥8, which is Σ P(k) from k=8 to12.Compute each term:P(8)=C*8^{-2.09}=3.69*(8^{-2.09})=3.69*e^{-2.09*ln8}=3.69*e^{-2.09*2.079}=3.69*e^{-4.333}=3.69*0.013≈0.048Similarly,P(9)=3.69*9^{-2.09}=3.69*e^{-2.09*2.197}=3.69*e^{-4.603}=3.69*0.009≈0.033P(10)=3.69*10^{-2.09}=3.69*e^{-2.09*2.302}=3.69*e^{-4.813}=3.69*0.008≈0.0295P(11)=3.69*11^{-2.09}=3.69*e^{-2.09*2.398}=3.69*e^{-5.000}=3.69*0.006≈0.0221P(12)=3.69*12^{-2.09}=3.69*e^{-2.09*2.485}=3.69*e^{-5.196}=3.69*0.005≈0.01845Adding these up:0.048 + 0.033 = 0.081+0.0295 ≈0.1105+0.0221≈0.1326+0.01845≈0.15105So, the total probability is≈0.15105.Thus, the expected number of influencers is N * 0.15105=50*0.15105≈7.5525.So, approximately 7.55, which we can round to 8.But let's check the exact calculation.Alternatively, compute each P(k):Compute P(8)=C*8^{-α}=3.69*(8^{-2.09}).Compute 8^{-2.09}=e^{-2.09*ln8}=e^{-2.09*2.079}=e^{-4.333}=≈0.013.So, P(8)=3.69*0.013≈0.048.Similarly,P(9)=3.69*9^{-2.09}=3.69*e^{-2.09*2.197}=e^{-4.603}=≈0.009, so P(9)=3.69*0.009≈0.033.P(10)=3.69*10^{-2.09}=3.69*e^{-2.09*2.302}=e^{-4.813}=≈0.008, so P(10)=3.69*0.008≈0.0295.P(11)=3.69*11^{-2.09}=3.69*e^{-2.09*2.398}=e^{-5.000}=≈0.006, so P(11)=3.69*0.006≈0.0221.P(12)=3.69*12^{-2.09}=3.69*e^{-2.09*2.485}=e^{-5.196}=≈0.005, so P(12)=3.69*0.005≈0.01845.Adding these: 0.048+0.033=0.081; +0.0295=0.1105; +0.0221=0.1326; +0.01845=0.15105.So, total≈0.15105.Thus, expected number=50*0.15105≈7.5525≈7.55.So, approximately 7.55, which is about 8 when rounded.Therefore, the expected number of influencers is approximately 8.</think>