🧠:<think>Okay, so I have this problem about Sporting Cristal's performance over 10 seasons. It's divided into two parts, and I need to solve both. Let me take it step by step.Starting with the first part: I need to find the average number of goals conceded per match and then the average goal differential. They gave me the total number of matches, total goals scored, and a function for goals conceded. Alright, so Sporting Cristal played 380 matches over 10 seasons. They scored 720 goals in total. The function for goals conceded is ( C(x) = 0.02x^2 + 0.5x + 280 ), where ( x ) is the number of matches played. Since they played 380 matches, I need to plug that into the function to find the total goals conceded.Let me calculate ( C(380) ). So, substituting x with 380:( C(380) = 0.02*(380)^2 + 0.5*(380) + 280 ).First, compute ( 380^2 ). 380 times 380. Let me do that step by step. 38 squared is 1444, so 380 squared is 144400. So, 0.02 times 144400 is... 0.02 is 2%, so 2% of 144400 is 2888.Next term: 0.5 times 380 is 190.Then, the constant term is 280.Adding them all together: 2888 + 190 + 280. Let's compute that. 2888 + 190 is 3078, and 3078 + 280 is 3358. So, total goals conceded is 3358.Wait, that seems high. Let me double-check my calculations. 380 squared is indeed 144400. 0.02 times 144400 is 2888. 0.5 times 380 is 190. 2888 + 190 is 3078, plus 280 is 3358. Yeah, that's correct.So, total goals conceded is 3358 over 380 matches. Therefore, the average goals conceded per match is total conceded divided by number of matches. So, 3358 / 380.Let me compute that. 3358 divided by 380. Let's see, 380 times 8 is 3040. 3358 minus 3040 is 318. 380 goes into 318 zero times. So, 8. So, 8.0? Wait, no, 318 is less than 380, so it's 8.0 plus 318/380.Wait, maybe a better way is to divide both numerator and denominator by 2. 3358 / 380 is the same as 1679 / 190. Let me compute 1679 divided by 190. 190 times 8 is 1520. 1679 minus 1520 is 159. So, 8 and 159/190. 159 divided by 190 is approximately 0.8368. So, total is approximately 8.8368. So, about 8.84 goals conceded per match on average.Wait, that seems really high. Goals conceded per match average of almost 9? That would mean they're conceding almost a goal per match, but 8.84 is actually like 8.84 goals per match? Wait, that can't be right because 3358 goals over 380 matches is indeed 8.84 per match. But in reality, football teams don't concede 8-9 goals per match on average. That seems way too high. Maybe I made a mistake in interpreting the function.Wait, let me check the function again. It's ( C(x) = 0.02x^2 + 0.5x + 280 ). So, when x is 380, it's 0.02*(380)^2 + 0.5*380 + 280. So, 0.02*(144400) is 2888, 0.5*380 is 190, plus 280. So, 2888 + 190 is 3078, plus 280 is 3358. So, that's correct. So, 3358 goals conceded over 380 matches. So, 3358 / 380 is indeed 8.8368. So, approximately 8.84 goals conceded per match.Wait, that seems extremely high. Maybe the function is not in goals per match but total goals? But the function is given as C(x) = ..., where x is the number of matches. So, it's total goals conceded. So, over 380 matches, they conceded 3358 goals. So, 3358 / 380 is 8.84 per match. That seems way too high for a football team. Maybe the function is misinterpreted? Or perhaps it's a typo in the problem.Wait, let me think again. Maybe the function is in terms of something else? Or perhaps the function is goals conceded per match? But the wording says "the total number of goals conceded over these matches is represented by the function C(x) = 0.02x^2 + 0.5x + 280". So, yeah, total goals conceded is C(x). So, x is number of matches, so C(x) is total goals. So, 3358 is total goals conceded over 380 matches. So, average is 8.84 per match. That seems way too high because in reality, even bad teams don't concede 8-9 goals per match on average. Maybe the function is per season?Wait, the problem says over the last 10 seasons, Sporting Cristal has played a total of 380 matches. So, 38 matches per season? That seems low because usually, football leagues have more matches. For example, in Peru, the league might have 18 teams, each playing 34 matches, but over 10 seasons, that would be 340 matches. But here it's 380, so maybe 38 matches per season? Hmm, maybe.But regardless, according to the problem, the function C(x) is total goals conceded over x matches. So, 380 matches, 3358 goals conceded. So, 3358 / 380 is approximately 8.84 per match. That seems high, but perhaps it's correct for the sake of the problem.So, moving on. The average goals scored per match is total goals scored divided by number of matches. They scored 720 goals over 380 matches. So, 720 / 380.Let me compute that. 720 divided by 380. 380 goes into 720 once, with 340 remaining. So, 1. So, 1.0. Then, 340/380 is approximately 0.8947. So, total is approximately 1.8947 goals per match. So, about 1.895 goals scored per match on average.So, average goals scored per match is approximately 1.895, and average goals conceded per match is approximately 8.84. Therefore, the average goal differential per match is goals scored minus goals conceded, so 1.895 - 8.84.Let me compute that. 1.895 - 8.84 is negative, so it's -6.945. So, approximately -6.945 goals per match. So, Sporting Cristal is averaging a negative goal differential, meaning they're scoring less than they're conceding on average.That seems quite bad, but again, maybe it's correct for the problem's context.So, to recap, average goals scored per match is 720 / 380 ≈ 1.895, average goals conceded per match is 3358 / 380 ≈ 8.84, so average goal differential is approximately -6.945.So, that's part 1 done.Moving on to part 2: Assuming the distribution of goals scored per match follows a Poisson distribution, calculate the probability that Sporting Cristal scores exactly 3 goals in a randomly selected match. We need to use the average number of goals scored per match from part 1 as the parameter λ for the Poisson distribution.So, the Poisson probability mass function is P(k) = (λ^k * e^(-λ)) / k!, where k is the number of occurrences (in this case, goals). So, we need to compute P(3) with λ ≈ 1.895.First, let me note that λ is approximately 1.895, which is roughly 1.9. So, let me use 1.895 for more precision.So, compute P(3) = (1.895^3 * e^(-1.895)) / 3!.Let me compute each part step by step.First, compute 1.895^3.1.895 * 1.895: Let's compute that. 1.895 squared.1.895 * 1.895:Compute 1.8 * 1.8 = 3.241.8 * 0.095 = 0.1710.095 * 1.8 = 0.1710.095 * 0.095 = 0.009025So, adding them up:3.24 + 0.171 + 0.171 + 0.009025 = 3.24 + 0.342 + 0.009025 = 3.591025Wait, that seems incorrect because 1.895 is approximately 1.9, and 1.9 squared is 3.61, so 3.591025 is close. So, 1.895 squared is approximately 3.591025.Then, multiply that by 1.895 again to get the cube.So, 3.591025 * 1.895.Let me compute that.First, 3 * 1.895 = 5.6850.591025 * 1.895.Compute 0.5 * 1.895 = 0.94750.091025 * 1.895.Compute 0.09 * 1.895 = 0.170550.001025 * 1.895 ≈ 0.001941So, adding those:0.9475 + 0.17055 = 1.118051.11805 + 0.001941 ≈ 1.119991So, total is 5.685 + 1.119991 ≈ 6.804991So, approximately 6.805.So, 1.895^3 ≈ 6.805.Next, compute e^(-1.895). e is approximately 2.71828. So, e^(-1.895) is 1 / e^(1.895).Compute e^(1.895). Let me recall that e^1 = 2.71828, e^2 ≈ 7.38906.1.895 is close to 2, so e^1.895 is a bit less than e^2.Compute e^1.895. Let me use the Taylor series or maybe approximate it.Alternatively, use the fact that ln(6.6) is approximately 1.887, so e^1.887 ≈ 6.6. So, 1.895 is a bit higher, so e^1.895 ≈ 6.6 * e^(0.008). Since e^0.008 ≈ 1 + 0.008 + (0.008)^2/2 ≈ 1.008032.So, e^1.895 ≈ 6.6 * 1.008032 ≈ 6.6 * 1.008 ≈ 6.6528.Therefore, e^(-1.895) ≈ 1 / 6.6528 ≈ 0.1503.Alternatively, using a calculator, e^1.895 is approximately 6.652, so e^(-1.895) ≈ 0.1503.So, e^(-1.895) ≈ 0.1503.Now, 3! is 6.So, putting it all together:P(3) = (6.805 * 0.1503) / 6.First, compute 6.805 * 0.1503.6 * 0.1503 = 0.90180.805 * 0.1503 ≈ 0.1209So, total is approximately 0.9018 + 0.1209 ≈ 1.0227.Then, divide by 6: 1.0227 / 6 ≈ 0.17045.So, approximately 0.1705.Therefore, the probability is approximately 17.05%.Wait, let me verify that calculation again because 6.805 * 0.1503.Compute 6 * 0.1503 = 0.90180.805 * 0.1503: Let's compute 0.8 * 0.1503 = 0.12024, and 0.005 * 0.1503 = 0.0007515. So, total is 0.12024 + 0.0007515 ≈ 0.1209915.So, total is 0.9018 + 0.1209915 ≈ 1.0227915.Divide by 6: 1.0227915 / 6 ≈ 0.170465.So, approximately 0.1705, or 17.05%.So, the probability is approximately 17.05%.Alternatively, if I use more precise calculations:Compute 1.895^3:1.895 * 1.895 = 3.5910253.591025 * 1.895:Let me compute 3.591025 * 1.895.3 * 1.895 = 5.6850.591025 * 1.895:Compute 0.5 * 1.895 = 0.94750.091025 * 1.895:0.09 * 1.895 = 0.170550.001025 * 1.895 ≈ 0.001941So, 0.9475 + 0.17055 = 1.11805 + 0.001941 ≈ 1.119991So, total is 5.685 + 1.119991 ≈ 6.804991So, 1.895^3 ≈ 6.804991e^(-1.895) ≈ 0.1503So, 6.804991 * 0.1503 ≈ 1.02279Divide by 6: 1.02279 / 6 ≈ 0.170465So, approximately 0.1705, or 17.05%.Therefore, the probability is approximately 17.05%.Alternatively, using a calculator for more precision:Compute λ = 1.895Compute P(3) = (1.895^3 * e^(-1.895)) / 6Compute 1.895^3:1.895 * 1.895 = 3.5910253.591025 * 1.895:Let me compute 3.591025 * 1.895:3.591025 * 1 = 3.5910253.591025 * 0.8 = 2.872823.591025 * 0.09 = 0.323192253.591025 * 0.005 = 0.017955125Adding them up: 3.591025 + 2.87282 = 6.463845 + 0.32319225 = 6.78703725 + 0.017955125 ≈ 6.804992375So, 1.895^3 ≈ 6.804992375e^(-1.895): Let me use a calculator for more precision.e^1.895 ≈ 6.65227So, e^(-1.895) ≈ 1 / 6.65227 ≈ 0.150301So, 6.804992375 * 0.150301 ≈ 6.804992375 * 0.150301Compute 6 * 0.150301 = 0.9018060.804992375 * 0.150301 ≈ 0.120999So, total ≈ 0.901806 + 0.120999 ≈ 1.022805Divide by 6: 1.022805 / 6 ≈ 0.1704675So, approximately 0.1704675, which is about 17.05%.So, the probability is approximately 17.05%.Therefore, the final answer for part 2 is approximately 17.05%.But let me check if I can express it more precisely. Maybe using more decimal places.Alternatively, use the exact value:λ = 1.895P(3) = (1.895^3 * e^(-1.895)) / 6Compute 1.895^3:1.895 * 1.895 = 3.5910253.591025 * 1.895:Let me compute 3.591025 * 1.895:= 3.591025 * (1 + 0.8 + 0.09 + 0.005)= 3.591025 + 2.87282 + 0.32319225 + 0.017955125= 3.591025 + 2.87282 = 6.4638456.463845 + 0.32319225 = 6.787037256.78703725 + 0.017955125 ≈ 6.804992375So, 1.895^3 ≈ 6.804992375e^(-1.895) ≈ 0.150301So, 6.804992375 * 0.150301 ≈ 1.022805Divide by 6: 1.022805 / 6 ≈ 0.1704675So, approximately 0.1704675, which is 17.04675%.So, rounding to four decimal places, 0.1705, or 17.05%.Alternatively, if we want to express it as a fraction or more precise decimal, but I think 17.05% is sufficient.So, to summarize:1. Average goals scored per match: 720 / 380 ≈ 1.895Average goals conceded per match: 3358 / 380 ≈ 8.84Average goal differential: 1.895 - 8.84 ≈ -6.9452. Probability of scoring exactly 3 goals: Approximately 17.05%So, I think that's it.</think>

🧠:<think>Okay, so I have this problem about the population growth of Compton, and I need to figure out the growth constant k and then use that to find the population in 1950 and compare it to the historical record. Let me start by understanding what's given and what I need to find.First, the problem states that the population in 1920 was 5,000, and in 2020 it was 95,000. It also mentions that the growth is exponential, following the model P(t) = P0 * e^(kt). Here, P0 is the initial population, which is 5,000 in 1920. t is the time in years after 1920, so from 1920 to 2020, that's 100 years. Therefore, t = 100 when P(t) = 95,000.So, for part 1, I need to solve for k. Let me write down the equation:95,000 = 5,000 * e^(k * 100)Hmm, okay. I can divide both sides by 5,000 to simplify this equation. Let me do that:95,000 / 5,000 = e^(100k)Calculating the left side: 95,000 divided by 5,000 is 19. So,19 = e^(100k)Now, to solve for k, I need to take the natural logarithm of both sides. Remember, ln(e^x) = x, so:ln(19) = 100kTherefore, k = ln(19) / 100Let me compute ln(19). I know that ln(10) is approximately 2.3026, and ln(20) is about 2.9957. Since 19 is just below 20, ln(19) should be a bit less than 2.9957. Maybe around 2.9444? Let me check with a calculator.Wait, actually, I can use a calculator for more precision. Let me compute ln(19):ln(19) ≈ 2.9444So, k ≈ 2.9444 / 100 ≈ 0.029444So, k is approximately 0.029444 per year.Let me double-check my steps to make sure I didn't make a mistake. I started with the exponential model, plugged in the known values for P(t) and P0, solved for k by dividing both sides by P0, then took the natural log. That seems correct.Now, moving on to part 2. I need to calculate the population in 1950 using this model and then compare it to the historical record of 20,000. First, let's figure out how many years after 1920 the year 1950 is. 1950 minus 1920 is 30 years, so t = 30.Using the model P(t) = 5,000 * e^(kt), with k ≈ 0.029444 and t = 30:P(30) = 5,000 * e^(0.029444 * 30)Let me compute the exponent first: 0.029444 * 30 = 0.88332So, e^0.88332. Let me calculate that. I know that e^0.8 is approximately 2.2255, e^0.9 is about 2.4596. So, 0.88332 is between 0.8 and 0.9, closer to 0.9.Let me use a calculator for more accuracy. e^0.88332 ≈ 2.418So, P(30) ≈ 5,000 * 2.418 ≈ 12,090Wait, that seems low compared to the historical record of 20,000. Hmm, let me check my calculations again.Wait, maybe I made a mistake in calculating e^0.88332. Let me compute it step by step.First, 0.88332. Let me recall that ln(2.418) is approximately 0.88332, so e^0.88332 is indeed approximately 2.418. So, 5,000 * 2.418 is 12,090.But the historical record says 20,000 in 1950. So, the model predicts 12,090, which is significantly lower. That seems like a big discrepancy. Maybe I made a mistake in calculating k?Wait, let me go back to part 1. I had 95,000 = 5,000 * e^(100k). Dividing both sides by 5,000 gives 19 = e^(100k). Taking natural log, ln(19) = 100k, so k = ln(19)/100 ≈ 2.9444/100 ≈ 0.029444. That seems correct.Alternatively, maybe the model isn't perfect, and the growth isn't perfectly exponential? Or perhaps the growth rate changed over time, making the model less accurate for intermediate years. But the problem assumes exponential growth, so I have to go with that.So, according to the model, the population in 1950 would be approximately 12,090, but the actual historical record is 20,000. To find the percentage error, I can use the formula:Percentage Error = |(Predicted - Actual)| / Actual * 100%So, plugging in the numbers:Predicted = 12,090Actual = 20,000Difference = |12,090 - 20,000| = 7,910Percentage Error = (7,910 / 20,000) * 100% ≈ 39.55%So, the model underestimates the population in 1950 by approximately 39.55%.Wait, that seems quite a large error. Maybe I should check my calculations again.Let me recompute e^(0.029444 * 30). 0.029444 * 30 is indeed 0.88332. e^0.88332 is approximately 2.418. 5,000 * 2.418 is 12,090. Yes, that's correct.Alternatively, maybe I should use more precise values for ln(19). Let me check ln(19) more accurately. Using a calculator, ln(19) is approximately 2.944438979. So, k = 2.944438979 / 100 ≈ 0.02944438979.So, 0.02944438979 * 30 = 0.8833316937e^0.8833316937 ≈ e^0.8833316937. Let me compute this more accurately.I know that e^0.8833316937 can be calculated using a Taylor series or a calculator. Let me use a calculator:e^0.8833316937 ≈ 2.418145So, 5,000 * 2.418145 ≈ 12,090.725So, approximately 12,090.73, which is about 12,091.So, the model's prediction is 12,091, while the actual population was 20,000. So, the difference is 20,000 - 12,091 = 7,909.Percentage error is (7,909 / 20,000) * 100 ≈ 39.545%, which is approximately 39.55%.So, that seems correct. Therefore, the model underestimates the population in 1950 by about 39.55%.Alternatively, maybe the model is correct, but the growth rate changed over time, or perhaps the initial assumption of exponential growth isn't accurate for Compton's population. But according to the problem, we have to assume exponential growth, so we proceed with that.So, summarizing:1. The growth constant k is approximately 0.029444 per year.2. The model predicts a population of approximately 12,090 in 1950, which is a 39.55% error compared to the historical record of 20,000.I think that's it. I don't see any mistakes in my calculations, so I'll go with these results.</think>

🧠:<think>Alright, so I have this problem about the Orix Buffaloes baseball team, and I need to analyze their performance using some math. Let me try to break it down step by step.First, part 1 says that the Buffaloes have played a total of (n) games in a season. They have (W) wins and (L) losses, so (W + L = n). The probability (P) of them winning a game is given by (P(W) = frac{kW}{n}), where (k) is a constant. I need to derive a differential equation that models the change in the number of wins ((W)) as a function of the number of games played ((n)).Hmm, okay. So, let's think about this. Probability (P) is the chance of winning a game, which is a function of the number of wins (W). So, as they play more games, (W) can increase or stay the same, but not decrease because losses are just the remaining games.Wait, but (n) is the total number of games played, so as time goes on, (n) increases. So, (W) is a function of (n), right? So, (W = W(n)). The probability of winning the next game is (P(W) = frac{kW}{n}). So, each game, the probability they win is proportional to the number of wins they've had so far, scaled by (k) and divided by the total games played.So, if I think about the change in (W) with respect to (n), that's (dW/dn). Since each game they play, they either win or lose. The expected change in (W) when they play one more game is the probability they win that game, which is (P(W) = frac{kW}{n}). So, the rate of change of (W) with respect to (n) should be equal to the probability of winning the next game.Therefore, the differential equation is:[frac{dW}{dn} = frac{kW}{n}]Is that right? Let me double-check. So, for each game played, the expected increase in wins is the probability of winning that game. So yes, (dW/dn = P(W)), which is (frac{kW}{n}). That makes sense.Okay, so part 1 is done. The differential equation is (frac{dW}{dn} = frac{kW}{n}).Now, moving on to part 2. They give me that the season has 144 games, so (n = 144). Initially, after 72 games, they have a 50% win rate. So, when (n = 72), (W = 0.5 times 72 = 36). So, the initial condition is (W(72) = 36). Also, (k = 1.5). I need to find the expected number of wins at the end of the season, which is when (n = 144).So, I have the differential equation:[frac{dW}{dn} = frac{1.5 W}{n}]This is a separable differential equation. Let me rewrite it:[frac{dW}{W} = 1.5 frac{dn}{n}]Integrating both sides:[int frac{1}{W} dW = int 1.5 frac{1}{n} dn]Which gives:[ln|W| = 1.5 ln|n| + C]Exponentiating both sides:[W = C n^{1.5}]Where (C) is the constant of integration. Now, I need to find (C) using the initial condition. When (n = 72), (W = 36):[36 = C times 72^{1.5}]Let me compute (72^{1.5}). (72^{1.5} = 72 times sqrt{72}). Let's calculate (sqrt{72}):[sqrt{72} = sqrt{36 times 2} = 6 sqrt{2} approx 6 times 1.4142 = 8.4852]So, (72^{1.5} = 72 times 8.4852 approx 72 times 8.4852). Let me compute that:72 * 8 = 57672 * 0.4852 ≈ 72 * 0.4 = 28.8; 72 * 0.0852 ≈ 6.1464So, total ≈ 28.8 + 6.1464 ≈ 34.9464Therefore, (72^{1.5} ≈ 576 + 34.9464 ≈ 610.9464)So, (36 = C times 610.9464)Therefore, (C = 36 / 610.9464 ≈ 0.059)Wait, let me compute that more accurately.36 divided by 610.9464. Let me do 36 / 610.9464.First, 610.9464 / 36 ≈ 16.9707, so 36 / 610.9464 ≈ 1 / 16.9707 ≈ 0.0589So, approximately 0.0589.Therefore, the solution is:[W(n) = 0.0589 times n^{1.5}]Now, we need to find (W(144)).Compute (144^{1.5}). (144^{1.5} = 144 times sqrt{144} = 144 times 12 = 1728)So, (W(144) = 0.0589 times 1728)Compute 0.0589 * 1728:First, 0.05 * 1728 = 86.40.0089 * 1728 ≈ Let's compute 0.01 * 1728 = 17.28, so 0.0089 is about 0.9 * 17.28 ≈ 15.552So, total ≈ 86.4 + 15.552 ≈ 101.952So, approximately 101.952 wins.But let me compute it more precisely.0.0589 * 1728:Breakdown:0.05 * 1728 = 86.40.008 * 1728 = 13.8240.0009 * 1728 = 1.5552So, adding up: 86.4 + 13.824 = 100.224; 100.224 + 1.5552 ≈ 101.7792So, approximately 101.78 wins.But wait, let me compute 0.0589 * 1728 step by step.0.0589 * 1728:Multiply 1728 by 0.05: 1728 * 0.05 = 86.4Multiply 1728 by 0.008: 1728 * 0.008 = 13.824Multiply 1728 by 0.0009: 1728 * 0.0009 = 1.5552Now, add them together:86.4 + 13.824 = 100.224100.224 + 1.5552 = 101.7792So, approximately 101.78 wins.But since the number of wins must be an integer, but the model gives a continuous approximation, so we can say approximately 101.78, which is roughly 102 wins.But let me see if I can compute it more accurately.Alternatively, maybe I made a mistake in calculating (C). Let me go back.We had:(36 = C times 72^{1.5})So, (C = 36 / 72^{1.5})Compute 72^{1.5}:72^{1.5} = 72 * sqrt(72) = 72 * 6 * sqrt(2) = 432 * 1.4142 ≈ 432 * 1.4142Compute 432 * 1.4142:First, 400 * 1.4142 = 565.6832 * 1.4142 ≈ 45.2544Total ≈ 565.68 + 45.2544 ≈ 610.9344So, 72^{1.5} ≈ 610.9344Therefore, (C = 36 / 610.9344 ≈ 0.0589), which is what I had before.So, (W(n) = 0.0589 * n^{1.5})Thus, (W(144) = 0.0589 * 144^{1.5})144^{1.5} = 144 * sqrt(144) = 144 * 12 = 1728So, 0.0589 * 1728 ≈ 101.7792So, approximately 101.78 wins.But let me check if the model is correct. The differential equation is (dW/dn = (1.5 W)/n), which is a separable equation leading to (W = C n^{1.5}). That seems correct.But wait, let me think about the initial condition again. At (n = 72), (W = 36). So, plugging into (W = C n^{1.5}):36 = C * 72^{1.5} => C = 36 / 72^{1.5} = 36 / (72 * sqrt(72)) = 36 / (72 * 6 * sqrt(2)) = 36 / (432 * sqrt(2)) = (36 / 432) / sqrt(2) = (1/12) / sqrt(2) = 1 / (12 sqrt(2)) ≈ 1 / 16.9706 ≈ 0.0589, which matches.So, the solution is correct.Therefore, at (n = 144), (W ≈ 101.78). Since the number of wins must be an integer, we can round it to 102 wins.But wait, let me think again. The model assumes that the probability of winning each game is (P = (1.5 W)/n). So, as they play more games, their probability of winning each subsequent game depends on their current number of wins.Wait, but when n increases, the denominator increases, but the numerator also increases. So, the probability might stabilize or change in a certain way.But in our solution, (W(n) = C n^{1.5}), so as (n) increases, (W) increases faster than linearly, which makes sense because the probability is increasing as (W) increases.But let me check if the model makes sense. If they start with 36 wins in 72 games, which is 50%, and then as they play more games, their probability of winning each game is 1.5 * W / n. So, initially, when n=72, W=36, so P = 1.5 * 36 / 72 = 0.75. So, their probability of winning the next game is 75%. That seems high, but maybe that's how the model is set up.Wait, but if they have a 50% win rate after 72 games, but the probability of winning the next game is 75%, that suggests they are improving. So, their win rate is increasing as they play more games.But in reality, baseball teams don't necessarily improve as they play more games, but this model suggests that their probability of winning increases with more wins, which could be a positive feedback loop.Anyway, proceeding with the model.So, the solution is (W(n) = C n^{1.5}), with (C ≈ 0.0589). Therefore, at (n=144), (W ≈ 101.78), which is approximately 102 wins.But let me check if I can express this more precisely without approximating.We have (C = 36 / (72^{1.5}) = 36 / (72 * sqrt(72)) = 36 / (72 * 6 * sqrt(2)) = 36 / (432 sqrt(2)) = (36 / 432) / sqrt(2) = (1/12) / sqrt(2) = 1 / (12 sqrt(2)))So, (C = 1 / (12 sqrt(2)))Therefore, (W(n) = (1 / (12 sqrt(2))) * n^{1.5})So, at (n=144):(W(144) = (1 / (12 sqrt(2))) * 144^{1.5})Compute 144^{1.5} = 144 * sqrt(144) = 144 * 12 = 1728So, (W(144) = (1 / (12 sqrt(2))) * 1728 = 1728 / (12 sqrt(2)) = 144 / sqrt(2))Simplify 144 / sqrt(2):Multiply numerator and denominator by sqrt(2):144 sqrt(2) / 2 = 72 sqrt(2)So, (W(144) = 72 sqrt(2))Compute 72 sqrt(2):sqrt(2) ≈ 1.414272 * 1.4142 ≈ 72 * 1.4142Compute 70 * 1.4142 = 98.9942 * 1.4142 = 2.8284Total ≈ 98.994 + 2.8284 ≈ 101.8224So, (W(144) ≈ 101.8224), which is approximately 101.82, which rounds to 102 wins.Therefore, the expected number of wins at the end of the season is approximately 102.But let me see if there's another way to approach this without approximating so much.Wait, the differential equation is (dW/dn = (1.5 W)/n), which is a linear differential equation, and we solved it correctly.Alternatively, we can write it as:[frac{dW}{dn} = frac{3}{2} frac{W}{n}]Which is a standard form, and the solution is (W = C n^{3/2}), which is what we have.So, everything checks out.Therefore, the expected number of wins is approximately 101.82, which is about 102 wins.But since the question says "expected number of wins", it's okay to have a non-integer value, but in reality, you can't have a fraction of a win. However, in the context of expected value, it's acceptable.So, the answer is approximately 101.82, which we can write as (72 sqrt{2}), but numerically, it's about 101.82.But let me compute (72 sqrt{2}) more accurately.sqrt(2) ≈ 1.4142135672 * 1.41421356 ≈70 * 1.41421356 = 98.99494922 * 1.41421356 = 2.82842712Total ≈ 98.9949492 + 2.82842712 ≈ 101.8233763So, approximately 101.8234 wins.So, rounding to the nearest whole number, it's 102 wins.But the question says "expected number of wins", so maybe we can leave it as (72 sqrt{2}), but that's approximately 101.82.Alternatively, maybe the exact value is (72 sqrt{2}), which is about 101.823.But let me see if I can express it in terms of exact values.We had (W(n) = (1 / (12 sqrt(2))) * n^{1.5})At (n=144):(W(144) = (1 / (12 sqrt(2))) * 144^{1.5})But 144^{1.5} = 144 * 12 = 1728So, (W(144) = 1728 / (12 sqrt(2)) = 144 / sqrt(2) = 72 * 2 / sqrt(2) = 72 sqrt(2))Yes, because (2 / sqrt(2) = sqrt(2)). So, (144 / sqrt(2) = 72 * 2 / sqrt(2) = 72 sqrt(2))Therefore, (W(144) = 72 sqrt(2)), which is approximately 101.823.So, the exact value is (72 sqrt{2}), which is about 101.823.But since the question asks for the expected number of wins, and it's a mathematical model, we can present the exact value or the approximate decimal.But in the context of the problem, they might expect the exact form, which is (72 sqrt{2}), but let me check if that's necessary.Alternatively, since the initial condition was given at n=72 with W=36, and k=1.5, and the season is 144 games, the answer is (72 sqrt{2}), which is approximately 101.823.But let me see if I can write it as a fraction times sqrt(2). 72 is 72, so 72 sqrt(2) is the exact value.But perhaps the question expects a numerical value, so approximately 101.82, which is about 102.But let me check if I made any mistakes in the calculations.Wait, when I solved for C, I had:C = 36 / (72^{1.5}) = 36 / (72 * sqrt(72)) = 36 / (72 * 6 * sqrt(2)) = 36 / (432 sqrt(2)) = (36 / 432) / sqrt(2) = (1/12) / sqrt(2) = 1 / (12 sqrt(2))Yes, that's correct.Then, W(n) = (1 / (12 sqrt(2))) * n^{1.5}At n=144, W(144) = (1 / (12 sqrt(2))) * 144^{1.5} = (1 / (12 sqrt(2))) * 1728 = 1728 / (12 sqrt(2)) = 144 / sqrt(2) = 72 sqrt(2)Yes, that's correct.So, the exact value is 72 sqrt(2), which is approximately 101.823.Therefore, the expected number of wins is 72 sqrt(2), or approximately 101.82.But since the question says "find the expected number of wins", and it's a mathematical model, it's acceptable to present the exact value, which is 72 sqrt(2), but if they want a numerical value, it's approximately 101.82.But let me see if the question expects an exact answer or a numerical one.The problem says "find the expected number of wins (W) at the end of the season." It doesn't specify, but in mathematical problems, sometimes exact forms are preferred, especially if they are simple.But 72 sqrt(2) is a simple exact form, so maybe that's acceptable. Alternatively, if they want a numerical value, 101.82 is fine.But let me check if 72 sqrt(2) is indeed the correct exact value.Yes, because:W(n) = C n^{1.5}, with C = 1 / (12 sqrt(2))So, W(144) = (1 / (12 sqrt(2))) * 144^{1.5} = (1 / (12 sqrt(2))) * 1728 = 1728 / (12 sqrt(2)) = 144 / sqrt(2) = 72 * 2 / sqrt(2) = 72 sqrt(2)Yes, that's correct.Therefore, the exact expected number of wins is 72 sqrt(2), which is approximately 101.82.So, I think that's the answer.Final AnswerThe expected number of wins at the end of the season is boxed{72sqrt{2}}.</think>

🧠:<think>First, I need to calculate the total distance Juan drove during the race. The track length is 4.279 kilometers, and he completed 50 laps. Multiplying the number of laps by the track length will give the total distance in kilometers.Next, I'll convert the total distance from kilometers to miles using the conversion factor of 1 kilometer equals approximately 0.621371 miles.For the average speed in kilometers per hour, I'll start by converting the average lap time of 1 minute and 25 seconds into hours. Then, I'll use the formula for speed, which is distance divided by time, to find his average speed in km/h.Finally, to determine his average speed in miles per hour, I'll use the total distance in miles and the same time in hours, applying the speed formula again.</think>