🧠:<think>Okay, so I have this problem where a guitarist wants to incorporate live loops into their DJ set. The guitarist plays riffs that are exactly 12 beats long at a tempo of 120 beats per minute. But the DJ's tracks vary in tempo from 100 to 140 beats per minute. The goal is to make sure the 12-beat loop aligns perfectly with the tracks so that the transitions are seamless.Let me break down the first question: Determine the set of all possible tempos (in beats per minute) for the DJ tracks such that the guitarist's 12-beat loop aligns perfectly with a whole number of measures, assuming each measure has 4 beats.Hmm, okay. So, the guitarist's loop is 12 beats long. Each measure is 4 beats, so 12 beats would be 3 measures. But the DJ's track has a different tempo. So, we need to find the tempos where the 12-beat loop of the guitarist aligns with a whole number of measures on the DJ track.Wait, so the 12-beat loop should fit into the DJ track's measure structure. Since each measure is 4 beats, the loop should be a multiple of 4 beats in terms of the DJ's tempo.But the loop is 12 beats at 120 BPM. So, first, let's find out how long the loop is in seconds. At 120 BPM, each beat is 0.5 seconds. So, 12 beats would be 6 seconds.Now, the DJ track has a tempo T BPM. So, the number of beats in the DJ track during 6 seconds is (T BPM) * (6 seconds) / 60 seconds per minute. That simplifies to (T * 6) / 60 = T / 10 beats.So, the number of beats in the DJ track during the 6 seconds is T / 10. Since each measure is 4 beats, the number of measures is (T / 10) / 4 = T / 40.We need this number of measures to be a whole number because the loop should align perfectly. So, T / 40 must be an integer. Therefore, T must be a multiple of 40.But the DJ's tempo varies from 100 to 140 BPM. So, we need to find all multiples of 40 within this range.Let's list the multiples of 40: 40, 80, 120, 160, etc. But our range is 100 to 140. So, 120 is the only multiple of 40 in that range.Wait, but 120 is the guitarist's tempo. But the DJ's tempo can be 120 as well, right? So, 120 is included.But let me double-check. If T is 120, then T / 40 is 3, which is an integer. So, that works. What about other multiples? 80 is below 100, so it's not in the range. 160 is above 140, so it's also not in the range. So, 120 is the only tempo in the range 100-140 where the loop aligns perfectly with a whole number of measures.Wait, but maybe I made a mistake here. Let me think again. The loop is 12 beats at 120 BPM, which is 6 seconds. On the DJ track, the number of beats in 6 seconds is T * 6 / 60 = T / 10. So, the number of measures is T / 40. For this to be an integer, T must be a multiple of 40.But 120 is a multiple of 40, so that's good. But are there other tempos where T / 40 is an integer? For example, if T is 100, then T / 40 is 2.5, which is not an integer. If T is 120, it's 3, which is good. If T is 140, it's 3.5, which is not an integer. So, only 120 works.But wait, maybe I'm approaching this incorrectly. Perhaps the loop should align with the DJ's measure structure, meaning that the loop duration in seconds should correspond to an integer number of measures on the DJ track.Let me try another approach. The loop is 12 beats at 120 BPM, which is 6 seconds. On the DJ track, each measure is 4 beats. So, the duration of each measure on the DJ track is (60 / T) * 4 seconds. So, the number of measures the loop spans is 6 / [(60 / T) * 4] = (6 * T) / (60 * 4) = (6T) / 240 = T / 40.So, again, T / 40 must be an integer. So, T must be a multiple of 40. Within 100-140, only 120 is a multiple of 40. So, the set is {120}.But wait, is that the only possibility? Let me think differently. Maybe the loop's duration should be a multiple of the DJ track's measure duration.The loop is 6 seconds. The DJ track's measure duration is (60 / T) * 4 = 240 / T seconds per measure. So, 6 seconds divided by (240 / T) seconds per measure should be an integer.So, 6 / (240 / T) = (6T) / 240 = T / 40. So, again, T must be a multiple of 40. So, same conclusion.Therefore, the set of possible tempos is {120}.Wait, but the problem says "the set of all possible tempos... Express your answer as an interval or a set of discrete values." So, it's a discrete set, and the only value is 120.But let me check if there are other tempos where T / 40 is an integer. For example, if T is 100, 100 / 40 = 2.5, not integer. T=110: 110/40=2.75, nope. T=120: 3, yes. T=130: 3.25, nope. T=140: 3.5, nope. So, only 120.So, the answer to part 1 is {120}.Now, part 2: During a performance, the guitarist improvises and decides to play a riff that lasts exactly 24 seconds. Given that the tempo of the DJ track is set at 132 beats per minute, calculate the number of complete measures that the guitarist's riff will span. Assume each measure is still 4 beats long.Okay, so the riff is 24 seconds long. The DJ track is at 132 BPM. Each measure is 4 beats. So, how many measures is 24 seconds?First, find out how many beats are in 24 seconds at 132 BPM.Beats = (132 beats/minute) * (24 seconds) / 60 seconds/minute = (132 * 24) / 60.Calculate that: 132 * 24 = 3168. 3168 / 60 = 52.8 beats.So, 52.8 beats. Since each measure is 4 beats, the number of measures is 52.8 / 4 = 13.2 measures.But the question asks for the number of complete measures. So, we take the integer part, which is 13 measures.Wait, but let me double-check. 132 BPM means each beat is 60/132 = 5/11 seconds per beat. So, 24 seconds is 24 / (5/11) = 24 * (11/5) = (24*11)/5 = 264/5 = 52.8 beats. So, 52.8 beats / 4 beats per measure = 13.2 measures. So, complete measures are 13.Alternatively, since 24 seconds is the duration, and each measure is 4 beats, which is (4 beats) * (60/132) seconds per beat = 240/132 = 20/11 ≈ 1.818 seconds per measure.So, number of measures is 24 / (20/11) = 24 * (11/20) = (24*11)/20 = 264/20 = 13.2. So, again, 13 complete measures.So, the answer is 13.But wait, let me think if there's another way. Maybe using the loop duration and the DJ's tempo to find the number of measures.The riff is 24 seconds. The DJ's tempo is 132 BPM. So, beats in 24 seconds: 132 * 24 / 60 = 52.8 beats. Measures: 52.8 / 4 = 13.2. So, 13 complete measures.Yes, that's consistent.So, the answers are:1. {120}2. 13But let me make sure I didn't make any calculation errors.For part 1:Loop duration: 12 beats at 120 BPM. 12 beats / 120 BPM = 0.1 minutes = 6 seconds.DJ track tempo T. In 6 seconds, the number of beats is T * 6 / 60 = T / 10. Number of measures: T / 40. Must be integer. So, T must be multiple of 40. In 100-140, only 120.Yes.For part 2:24 seconds at 132 BPM: 132 * 24 / 60 = 52.8 beats. 52.8 / 4 = 13.2 measures. So, 13 complete measures.Yes.I think that's correct.</think>

🧠:<think>Okay, so I have these two sub-problems to solve related to Poisson distributions in quality control. Let me take them one at a time and think through each step carefully.Starting with Sub-problem 1:We have a manufacturer with a new component where the number of defects follows a Poisson distribution. The average number of defects per unit, λ, is 0.1. They inspect a batch of 100 units, and we need to find the probability that there are no more than 5 defective units in this batch.Hmm, okay. So, first, I remember that the Poisson distribution is used to model the number of events happening in a fixed interval of time or space. In this case, the number of defective units in a batch. The formula for the Poisson probability mass function is:P(X = k) = (λ^k * e^(-λ)) / k!Where:- P(X = k) is the probability of k occurrences,- λ is the average rate (mean number of occurrences),- e is the base of the natural logarithm,- k! is the factorial of k.But wait, in this problem, we're dealing with a batch of 100 units, each with an average of 0.1 defects. So, the total average number of defects in the batch would be λ_total = 100 * 0.1 = 10. Is that right? Because if each unit has an average of 0.1 defects, then 100 units would have 10 on average.Yes, that makes sense. So, instead of considering each unit individually, we can model the total number of defective units in the batch as a Poisson distribution with λ = 10.So, now, the problem is asking for the probability that there are no more than 5 defective units. That is, P(X ≤ 5), where X follows a Poisson distribution with λ = 10.Calculating this directly might be a bit tedious because we'd have to sum the probabilities from X=0 to X=5. But maybe I can use the cumulative distribution function or some approximation.Wait, Poisson distributions can be approximated by normal distributions when λ is large. Since λ = 10 is moderately large, perhaps a normal approximation would work here. Let me recall that for Poisson(λ), the mean and variance are both λ, so the standard deviation is sqrt(λ) = sqrt(10) ≈ 3.1623.So, if I use the normal approximation, I can model X as approximately N(10, 3.1623^2). Then, to find P(X ≤ 5), I can standardize it:Z = (X - μ) / σ = (5 - 10) / 3.1623 ≈ (-5) / 3.1623 ≈ -1.5811Looking up this Z-score in the standard normal distribution table, the probability that Z is less than or equal to -1.5811 is approximately 0.0571. So, about 5.71% chance.But wait, I should consider the continuity correction since we're approximating a discrete distribution with a continuous one. So, instead of P(X ≤ 5), we should calculate P(X ≤ 5.5). Let me adjust that.Z = (5.5 - 10) / 3.1623 ≈ (-4.5) / 3.1623 ≈ -1.423Looking up Z = -1.423, the cumulative probability is approximately 0.0778, or 7.78%.Hmm, so with continuity correction, the probability is about 7.78%. But is this accurate enough? Maybe I should compute the exact Poisson probabilities and sum them up.Calculating P(X ≤ 5) exactly:P(X = 0) = (10^0 * e^-10) / 0! = e^-10 ≈ 0.0000454P(X = 1) = (10^1 * e^-10) / 1! = 10 * e^-10 ≈ 0.000454P(X = 2) = (10^2 * e^-10) / 2! = 100 * e^-10 / 2 ≈ 0.00227P(X = 3) = (10^3 * e^-10) / 6 ≈ 1000 * e^-10 / 6 ≈ 0.00758P(X = 4) = (10^4 * e^-10) / 24 ≈ 10000 * e^-10 / 24 ≈ 0.0189P(X = 5) = (10^5 * e^-10) / 120 ≈ 100000 * e^-10 / 120 ≈ 0.0378Now, summing these up:0.0000454 + 0.000454 ≈ 0.0005+ 0.00227 ≈ 0.00277+ 0.00758 ≈ 0.01035+ 0.0189 ≈ 0.02925+ 0.0378 ≈ 0.06705So, approximately 6.705% chance.Comparing this to the normal approximations: without continuity correction, it was about 5.71%, with continuity correction, 7.78%. The exact value is around 6.7%, so the continuity correction gives a closer estimate but still not exact.Alternatively, maybe I can use the Poisson cumulative distribution function directly. But since I don't have a calculator here, I have to compute it manually as I did above.So, the exact probability is approximately 6.7%. Therefore, the probability that there are no more than 5 defective units in the batch is roughly 6.7%.Wait, let me double-check my calculations for each term.Starting with P(X=0):e^-10 is approximately 0.0000454, correct.P(X=1): 10 * e^-10 ≈ 0.000454, correct.P(X=2): (100 / 2) * e^-10 = 50 * 0.0000454 ≈ 0.00227, correct.P(X=3): (1000 / 6) * e^-10 ≈ 166.6667 * 0.0000454 ≈ 0.00758, correct.P(X=4): (10000 / 24) * e^-10 ≈ 416.6667 * 0.0000454 ≈ 0.0189, correct.P(X=5): (100000 / 120) * e^-10 ≈ 833.3333 * 0.0000454 ≈ 0.0378, correct.Adding them: 0.0000454 + 0.000454 = 0.0005+ 0.00227 = 0.00277+ 0.00758 = 0.01035+ 0.0189 = 0.02925+ 0.0378 = 0.06705Yes, that's about 6.705%, so approximately 6.7%.Alternatively, using a calculator or statistical software, the exact value can be found, but since I'm doing it manually, 6.7% is a reasonable approximation.So, moving on to Sub-problem 2:The quality control process is improved, reducing the average defects per unit to λ = 0.05. We need to calculate the expected number of defective units in a new batch of 200 units and the probability that this batch will contain exactly 7 defective units.First, the expected number. Since each unit has an average of 0.05 defects, for 200 units, the total expected defects would be λ_total = 200 * 0.05 = 10. So, the expected number is 10 defective units.Wait, that's interesting. Even though the average per unit decreased, the batch size increased, so the total expected defects remain the same as in Sub-problem 1. Hmm, that's a good observation.Now, the second part is the probability of exactly 7 defective units. Again, since we have 200 units each with λ = 0.05, the total λ is 10. So, we model the number of defective units as Poisson(10).Thus, P(X = 7) = (10^7 * e^-10) / 7!Let me compute that.First, compute 10^7: 10,000,000e^-10 ≈ 0.00004547! = 5040So, P(X=7) = (10,000,000 * 0.0000454) / 5040Compute numerator: 10,000,000 * 0.0000454 = 454So, 454 / 5040 ≈ 0.09008So, approximately 9.008%.Let me verify this calculation step by step.10^7 is 10,000,000, correct.e^-10 ≈ 0.0000454, correct.7! = 7*6*5*4*3*2*1 = 5040, correct.So, 10^7 * e^-10 = 10,000,000 * 0.0000454 = 454, correct.Then, 454 / 5040 ≈ 0.09008, which is approximately 9.01%.Alternatively, using a calculator, the exact value might be slightly different due to more precise e^-10 value.But for our purposes, 9.01% is a good approximation.Wait, let me check with more precise e^-10.e^-10 is approximately 0.00004539993.So, 10^7 * 0.00004539993 = 10,000,000 * 0.00004539993 = 453.9993Then, 453.9993 / 5040 ≈ 0.090079So, approximately 0.090079, which is about 9.0079%, so 9.01%.Therefore, the probability is approximately 9.01%.Alternatively, using the Poisson formula in a calculator or software, it would give the exact value, but manually, 9.01% is accurate enough.So, summarizing Sub-problem 2:Expected number of defective units: 10.Probability of exactly 7 defective units: approximately 9.01%.Wait, but hold on. In Sub-problem 1, the total λ was 10 for 100 units, and in Sub-problem 2, it's also 10 for 200 units. So, the expected number is the same, but the probabilities might differ because the underlying distribution is the same. Wait, no, the underlying distribution is Poisson(10) in both cases, so the probabilities for specific numbers would be the same if the total λ is the same.Wait, but in Sub-problem 1, we were dealing with 100 units, each with λ=0.1, so total λ=10. In Sub-problem 2, 200 units with λ=0.05, total λ=10. So, both are Poisson(10). Therefore, the probability of exactly 7 defective units is the same in both cases, which is approximately 9.01%.But in Sub-problem 1, we were calculating P(X ≤ 5), which is a different probability.So, in both cases, the total λ is 10, so the Poisson distribution parameters are the same. Therefore, any probability calculations based on the total number of defective units would be the same regardless of the batch size, as long as the total λ remains the same.That's an important point. So, whether it's 100 units with λ=0.1 or 200 units with λ=0.05, as long as the total λ is 10, the distribution of the number of defective units is Poisson(10).Therefore, the probability calculations for the number of defective units are solely dependent on the total λ, not on the individual λ per unit or the batch size.That makes sense because the Poisson distribution is additive. If you have multiple independent Poisson processes, their sums are also Poisson with the sum of their λs.So, in this case, each unit is a Poisson process with its own λ, and the total is the sum of all those, which is Poisson(10).Therefore, the distribution is the same regardless of how the λ is distributed across the units.So, in Sub-problem 2, the expected number is 10, same as in Sub-problem 1, and the probability of exactly 7 defective units is the same as it would be in any Poisson(10) distribution.Therefore, my calculations seem consistent.So, to recap:Sub-problem 1: Probability of no more than 5 defective units in a batch of 100 with λ=0.1 per unit is approximately 6.7%.Sub-problem 2: Expected defective units in a batch of 200 with λ=0.05 per unit is 10, and the probability of exactly 7 defective units is approximately 9.01%.I think that's all. I don't see any mistakes in my reasoning, but let me just cross-verify.For Sub-problem 1, using Poisson(10), P(X ≤ 5) ≈ 6.7%. Using the normal approximation with continuity correction gave 7.78%, which is a bit higher, but the exact calculation is more accurate.For Sub-problem 2, since the total λ is the same, the expected value is the same, and the probability for exactly 7 is the same as it would be in any Poisson(10) scenario.Yes, that all makes sense.Final AnswerSub-problem 1: The probability is boxed{0.067}.Sub-problem 2: The expected number is boxed{10} and the probability is boxed{0.090}.</think>

🧠:<think>Okay, so I have this problem where a talent acquisition specialist is evaluating sales agents based on their sales figures and interview scores. The goal is to calculate a weighted score for each agent and then rank them to determine who should be prioritized for hiring. Let me try to break this down step by step.First, I need to understand the data given. There are five sales agents, each with their own sales figures and interview scores. The sales figures are in thousands of dollars, and the interview scores are out of 100. The formula provided for the weighted score is:[ W_i = 0.6 times left( frac{S_i}{S_{max}} right) + 0.4 times left( frac{I_i}{100} right) ]Where ( S_i ) is the sales figure for agent ( i ), ( S_{max} ) is the maximum sales figure among all agents, and ( I_i ) is the interview score for agent ( i ).So, the first thing I need to do is identify ( S_{max} ). Looking at the sales figures:- ( S_1 = 120 )- ( S_2 = 150 )- ( S_3 = 180 )- ( S_4 = 200 )- ( S_5 = 170 )The maximum sales figure here is ( S_4 = 200 ). So, ( S_{max} = 200 ).Now, for each agent, I need to compute two parts: the sales component and the interview component. The sales component is 60% of the weighted score, and the interview component is 40%. Both components are normalized by their respective maximums.Let me start with Agent 1.Agent 1:- Sales: 120- Interview: 85Sales component: ( 0.6 times left( frac{120}{200} right) )First, compute ( frac{120}{200} = 0.6 )Then, multiply by 0.6: ( 0.6 times 0.6 = 0.36 )Interview component: ( 0.4 times left( frac{85}{100} right) )Compute ( frac{85}{100} = 0.85 )Multiply by 0.4: ( 0.4 times 0.85 = 0.34 )Total weighted score ( W_1 = 0.36 + 0.34 = 0.70 )Wait, that seems low. Let me double-check my calculations.Sales component: 120/200 is 0.6, multiplied by 0.6 is 0.36. That's correct.Interview component: 85/100 is 0.85, multiplied by 0.4 is 0.34. Correct.Adding them together: 0.36 + 0.34 is indeed 0.70. So, Agent 1 has a weighted score of 0.70.Agent 2:- Sales: 150- Interview: 90Sales component: ( 0.6 times left( frac{150}{200} right) )150/200 = 0.750.6 * 0.75 = 0.45Interview component: ( 0.4 times left( frac{90}{100} right) )90/100 = 0.90.4 * 0.9 = 0.36Total weighted score ( W_2 = 0.45 + 0.36 = 0.81 )That seems higher than Agent 1, which makes sense because both sales and interview scores are higher.Agent 3:- Sales: 180- Interview: 88Sales component: ( 0.6 times left( frac{180}{200} right) )180/200 = 0.90.6 * 0.9 = 0.54Interview component: ( 0.4 times left( frac{88}{100} right) )88/100 = 0.880.4 * 0.88 = 0.352Total weighted score ( W_3 = 0.54 + 0.352 = 0.892 )Hmm, that's even higher. So Agent 3 is looking good so far.Agent 4:- Sales: 200- Interview: 92Sales component: ( 0.6 times left( frac{200}{200} right) )200/200 = 10.6 * 1 = 0.6Interview component: ( 0.4 times left( frac{92}{100} right) )92/100 = 0.920.4 * 0.92 = 0.368Total weighted score ( W_4 = 0.6 + 0.368 = 0.968 )Wow, that's the highest so far. Agent 4 has the maximum sales and a high interview score, so it makes sense.Agent 5:- Sales: 170- Interview: 87Sales component: ( 0.6 times left( frac{170}{200} right) )170/200 = 0.850.6 * 0.85 = 0.51Interview component: ( 0.4 times left( frac{87}{100} right) )87/100 = 0.870.4 * 0.87 = 0.348Total weighted score ( W_5 = 0.51 + 0.348 = 0.858 )Alright, so compiling all the weighted scores:- Agent 1: 0.70- Agent 2: 0.81- Agent 3: 0.892- Agent 4: 0.968- Agent 5: 0.858Now, to rank them from highest to lowest:1. Agent 4: 0.9682. Agent 3: 0.8923. Agent 5: 0.8584. Agent 2: 0.815. Agent 1: 0.70So, Agent 4 has the highest weighted score and should be prioritized for hiring. Agent 3 comes next, followed by Agent 5, then Agent 2, and finally Agent 1.Let me just verify my calculations once more to make sure I didn't make any arithmetic errors.For Agent 1: 120/200=0.6, 0.6*0.6=0.36; 85/100=0.85, 0.4*0.85=0.34; 0.36+0.34=0.70. Correct.Agent 2: 150/200=0.75, 0.6*0.75=0.45; 90/100=0.9, 0.4*0.9=0.36; 0.45+0.36=0.81. Correct.Agent 3: 180/200=0.9, 0.6*0.9=0.54; 88/100=0.88, 0.4*0.88=0.352; 0.54+0.352=0.892. Correct.Agent 4: 200/200=1, 0.6*1=0.6; 92/100=0.92, 0.4*0.92=0.368; 0.6+0.368=0.968. Correct.Agent 5: 170/200=0.85, 0.6*0.85=0.51; 87/100=0.87, 0.4*0.87=0.348; 0.51+0.348=0.858. Correct.All calculations seem accurate. So, the ranking is as above, with Agent 4 being the top candidate.Final Answer1. The weighted scores are: - Agent 1: boxed{0.70} - Agent 2: boxed{0.81} - Agent 3: boxed{0.892} - Agent 4: boxed{0.968} - Agent 5: boxed{0.858}2. The ranking from highest to lowest is Agent 4, Agent 3, Agent 5, Agent 2, Agent 1. Therefore, the agent to prioritize for hiring is boxed{Agent 4}.</think>

🧠:<think>Okay, so I'm trying to figure out these two database overload problems. Let me start with the first one about Database A.1. Database A's Overload Probability: - It says Database A's load is modeled by a Poisson process with a rate of λ_A requests per second. The maximum it can handle is M_A requests per second. I need to find the probability that it's overloaded at least once in 10 minutes. Hmm, Poisson processes have independent increments, right? So the number of events in non-overlapping intervals are independent. The number of requests in a time interval of length t follows a Poisson distribution with parameter λ = λ_A * t. But wait, overload occurs when the load exceeds M_A. So, in any given second, if the number of requests is more than M_A, it's overloaded. But since it's a Poisson process, the number of requests in each second is Poisson distributed with parameter λ_A. Wait, actually, the load is the rate, so maybe it's the number of requests per second. So, if the rate λ_A is such that the number of requests in a second is Poisson(λ_A). So, the probability that in a single second, the load exceeds M_A is P(N > M_A), where N ~ Poisson(λ_A). But the question is about at least once in 10 minutes. So, 10 minutes is 600 seconds. So, we have 600 independent seconds, each with a probability p = P(N > M_A) of overloading. So, the probability that it never overloads in 600 seconds is (1 - p)^600. Therefore, the probability that it overloads at least once is 1 - (1 - p)^600. But wait, is that correct? Because in a Poisson process, the number of events in each interval is independent, so yes, each second is independent. So, p = P(N > M_A) = 1 - P(N ≤ M_A). So, the probability of overload in at least one second is 1 - [P(N ≤ M_A)]^600. Alternatively, since the Poisson process is continuous, maybe we can model it as the supremum of the process over the interval exceeding M_A. But I think the approach with discrete seconds is an approximation, but since we're dealing with per-second intervals, it might be acceptable. Alternatively, maybe we should model the maximum load over the interval. But for a Poisson process, the maximum rate isn't straightforward because it's a counting process. The load is the rate, so maybe we need to consider the instantaneous rate, but Poisson processes have jumps, so the load at any point is just the number of events in an infinitesimal interval. Wait, maybe I'm overcomplicating. The question says the load is modeled by a Poisson process with rate λ_A. So, the number of requests in time t is Poisson(λ_A * t). So, the load per second is λ_A, but the actual number of requests in each second is Poisson distributed. So, the maximum load in a second is the maximum number of requests in any second. But actually, the load is the rate, so maybe the question is about the rate exceeding M_A. But the rate is λ_A, which is constant. So, if λ_A > M_A, then it's always overloaded. But that can't be, because the question is about the probability of being overloaded at least once. So, maybe the load varies over time, but it's a Poisson process, so the number of requests in each second is Poisson(λ_A). So, the load in each second is the number of requests, which is a random variable. Therefore, the load in each second is N ~ Poisson(λ_A). So, the probability that in a given second, the load exceeds M_A is P(N > M_A). So, over 600 seconds, the probability that it never exceeds M_A is [P(N ≤ M_A)]^600. Therefore, the probability of at least one overload is 1 - [P(N ≤ M_A)]^600. So, to compute this, I need to calculate P(N ≤ M_A) where N ~ Poisson(λ_A). Then raise it to the power of 600 and subtract from 1. But wait, if λ_A is the average rate per second, then the load is the number of requests per second, which is Poisson distributed. So, yes, that makes sense. So, the formula is: P(overload at least once) = 1 - [P(N ≤ M_A)]^600 Where N ~ Poisson(λ_A) Alternatively, if λ_A is the average per second, and M_A is the maximum per second, then yes. So, that's the approach for the first part.2. Database B's Overload Intervals: - The load is given by L_B(t) = α sin(ω t) + β. We need to find the time intervals within 24 hours where L_B(t) > T_B. Given α = 50, ω = π/12, β = 100, T_B = 130. So, L_B(t) = 50 sin(π t / 12) + 100. We need to solve for t where 50 sin(π t / 12) + 100 > 130. Let's write the inequality: 50 sin(π t / 12) + 100 > 130 Subtract 100: 50 sin(π t / 12) > 30 Divide by 50: sin(π t / 12) > 0.6 So, sin(x) > 0.6, where x = π t / 12. The general solution for sin(x) > 0.6 is: x ∈ (arcsin(0.6) + 2π k, π - arcsin(0.6) + 2π k) for integer k. Compute arcsin(0.6): arcsin(0.6) ≈ 0.6435 radians. So, the solution intervals for x are: (0.6435 + 2π k, π - 0.6435 + 2π k) = (0.6435 + 2π k, 2.4981 + 2π k) Now, since x = π t / 12, we can solve for t: t ∈ ( (0.6435 + 2π k) * (12 / π), (2.4981 + 2π k) * (12 / π) ) Simplify: t ∈ ( (0.6435 * 12 / π) + (2π k * 12 / π), (2.4981 * 12 / π) + (2π k * 12 / π) ) Simplify each term: 0.6435 * 12 / π ≈ (7.722) / 3.1416 ≈ 2.458 hours 2.4981 * 12 / π ≈ (29.9772) / 3.1416 ≈ 9.54 hours 2π k * 12 / π = 24 k So, the intervals are: t ∈ (2.458 + 24 k, 9.54 + 24 k) hours Since we're looking within a 24-hour period, k can be 0 or 1. For k=0: t ∈ (2.458, 9.54) hours For k=1: t ∈ (26.458, 33.54) hours, but since we're only considering up to 24 hours, this interval doesn't apply. Wait, actually, the period of the sine function is 2π / ω = 2π / (π/12) = 24 hours. So, the function repeats every 24 hours. Therefore, within a single 24-hour period, the solution is t ∈ (2.458, 9.54) hours. But let me double-check the calculation: x = π t / 12 So, solving sin(x) > 0.6: x ∈ (arcsin(0.6), π - arcsin(0.6)) + 2π k So, x ∈ (0.6435, 2.4981) + 2π k Then, t = (x * 12) / π So, t ∈ (0.6435 * 12 / π, 2.4981 * 12 / π) + (2π k * 12 / π) Which simplifies to: t ∈ ( (0.6435 * 12)/π, (2.4981 * 12)/π ) + 24 k Calculating: 0.6435 * 12 ≈ 7.722, divided by π ≈ 2.458 hours 2.4981 * 12 ≈ 29.9772, divided by π ≈ 9.54 hours So, within 24 hours, the load exceeds T_B from approximately 2.458 hours to 9.54 hours. Converting 2.458 hours to minutes: 0.458 * 60 ≈ 27.5 minutes, so 2:27.5 AM. Similarly, 9.54 hours is 9 hours and 0.54*60 ≈ 32.4 minutes, so 9:32.4 AM. So, the overload occurs between approximately 2:28 AM and 9:32 AM each day. Wait, but sine functions are periodic, so this interval repeats every 24 hours. So, within a 24-hour period, it's only this one interval where the load exceeds T_B. Let me confirm the math: sin(π t / 12) > 0.6 So, t is in the first and second quadrants where sine is positive. So, the solution is between arcsin(0.6) and π - arcsin(0.6), which is what I had. So, yes, the time intervals are approximately from 2.458 hours to 9.54 hours, which is about 2:28 AM to 9:32 AM. So, that's the solution for the second part.</think>