🧠:<think>Okay, so I have these two questions about modeling fungal infections in crops. Let me take them one at a time.Starting with the first question: I need to find the time ( t ) when half of the plants are infected. The model given is the logistic growth function:[ P(t) = frac{K}{1 + frac{K - P_0}{P_0} e^{-rt}} ]They've given me ( K = 0.9 ), ( P_0 = 0.1 ), and ( r = 0.05 ). I need to find ( t ) when ( P(t) = 0.5 ).Alright, so let me plug in the values into the equation. First, let me write down the equation with the given constants:[ 0.5 = frac{0.9}{1 + frac{0.9 - 0.1}{0.1} e^{-0.05t}} ]Simplify the denominator first. The term ( frac{0.9 - 0.1}{0.1} ) is ( frac{0.8}{0.1} ), which is 8. So the equation becomes:[ 0.5 = frac{0.9}{1 + 8 e^{-0.05t}} ]Now, I need to solve for ( t ). Let me rearrange this equation step by step.First, multiply both sides by the denominator to get rid of the fraction:[ 0.5 times (1 + 8 e^{-0.05t}) = 0.9 ]Calculate the left side:[ 0.5 + 4 e^{-0.05t} = 0.9 ]Subtract 0.5 from both sides:[ 4 e^{-0.05t} = 0.4 ]Divide both sides by 4:[ e^{-0.05t} = 0.1 ]Now, take the natural logarithm of both sides to solve for ( t ):[ ln(e^{-0.05t}) = ln(0.1) ]Simplify the left side:[ -0.05t = ln(0.1) ]I know that ( ln(0.1) ) is approximately ( -2.302585 ). So,[ -0.05t = -2.302585 ]Divide both sides by -0.05:[ t = frac{-2.302585}{-0.05} ]Calculate that:[ t = 46.0517 ]So, approximately 46.05 days. Since the question doesn't specify rounding, I can keep it at two decimal places or maybe round to the nearest whole number. Let me see, 46.05 is closer to 46 than 47, so maybe 46 days.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Plugged in ( P(t) = 0.5 ), ( K = 0.9 ), ( P_0 = 0.1 ), correct.2. Calculated ( (0.9 - 0.1)/0.1 = 8 ), that's right.3. Set up the equation ( 0.5 = 0.9 / (1 + 8 e^{-0.05t}) ), correct.4. Multiplied both sides by denominator: 0.5*(1 + 8 e^{-0.05t}) = 0.9, correct.5. 0.5 + 4 e^{-0.05t} = 0.9, yes.6. Subtract 0.5: 4 e^{-0.05t} = 0.4, correct.7. Divide by 4: e^{-0.05t} = 0.1, correct.8. Take ln: -0.05t = ln(0.1) ≈ -2.302585, correct.9. Divide: t ≈ 46.05, yes.Looks solid. So, the time when half the plants are infected is approximately 46.05 days.Moving on to the second question: Calculating the yield when ( t = 50 ) days. The yield is given by:[ Y(t) = Y_0 (1 - P(t)) ]Where ( Y_0 = 1000 ) kg. So, I need to find ( Y(50) ).First, I need to compute ( P(50) ) using the logistic growth function. Let me write down the function again with the given constants:[ P(t) = frac{0.9}{1 + 8 e^{-0.05t}} ]So, plug in ( t = 50 ):[ P(50) = frac{0.9}{1 + 8 e^{-0.05 times 50}} ]Calculate the exponent first: ( -0.05 times 50 = -2.5 ). So,[ P(50) = frac{0.9}{1 + 8 e^{-2.5}} ]Now, I need to compute ( e^{-2.5} ). I remember that ( e^{-2} ) is about 0.1353, and ( e^{-3} ) is about 0.0498. Since 2.5 is halfway, maybe around 0.0821? Let me check with a calculator.Actually, ( e^{-2.5} ) is approximately 0.082085.So, plug that in:[ P(50) = frac{0.9}{1 + 8 times 0.082085} ]Calculate the denominator:8 * 0.082085 = 0.65668So, denominator is 1 + 0.65668 = 1.65668Thus,[ P(50) = frac{0.9}{1.65668} ]Compute that division:0.9 divided by 1.65668. Let me do this step by step.1.65668 goes into 0.9 how many times? Since 1.65668 * 0.5 = 0.82834, which is less than 0.9.Subtract 0.82834 from 0.9: 0.07166Now, bring down a zero: 0.71661.65668 goes into 0.7166 approximately 0.43 times (since 1.65668 * 0.4 = 0.662672, which is close to 0.7166).So, total is approximately 0.5 + 0.43 = 0.93.Wait, that seems high. Wait, actually, maybe I should use a calculator approach.Alternatively, 0.9 / 1.65668 ≈ 0.543.Wait, let me compute it more accurately.Let me write 0.9 divided by 1.65668.Multiply numerator and denominator by 100000 to eliminate decimals:90000 / 165668 ≈ ?Compute 165668 * 0.543 ≈ 165668 * 0.5 = 82834, 165668 * 0.04 = 6626.72, 165668 * 0.003 = 497.004. Adding up: 82834 + 6626.72 = 89460.72 + 497.004 ≈ 89957.724. That's very close to 90000, so 0.543 is a good approximation.So, ( P(50) ≈ 0.543 ).Therefore, the proportion of infected plants is approximately 54.3%.Now, compute the yield:[ Y(50) = 1000 times (1 - 0.543) ]Calculate ( 1 - 0.543 = 0.457 ).Multiply by 1000:[ Y(50) = 1000 times 0.457 = 457 text{ kg} ]So, the yield after 50 days is approximately 457 kg.Wait, let me double-check the calculation for ( P(50) ):[ P(50) = 0.9 / (1 + 8 e^{-2.5}) ]We had ( e^{-2.5} ≈ 0.082085 ), so 8 * 0.082085 ≈ 0.65668.1 + 0.65668 = 1.65668.0.9 / 1.65668 ≈ 0.543, yes.So, 1 - 0.543 = 0.457, times 1000 is 457 kg. That seems correct.Alternatively, maybe I should use more precise values for ( e^{-2.5} ). Let me compute ( e^{-2.5} ) more accurately.The exact value of ( e^{-2.5} ) is approximately 0.082085.So, 8 * 0.082085 = 0.65668.1 + 0.65668 = 1.65668.0.9 / 1.65668: Let me compute this division more precisely.Compute 0.9 / 1.65668.Let me write this as 900 / 1656.68.Divide 900 by 1656.68.1656.68 goes into 900 zero times. So, 0.Add a decimal: 9000 divided by 1656.68.1656.68 * 5 = 8283.4Subtract from 9000: 9000 - 8283.4 = 716.6Bring down a zero: 7166.1656.68 goes into 7166 about 4 times (1656.68 * 4 = 6626.72)Subtract: 7166 - 6626.72 = 539.28Bring down a zero: 5392.81656.68 goes into 5392.8 about 3 times (1656.68 * 3 = 4970.04)Subtract: 5392.8 - 4970.04 = 422.76Bring down a zero: 4227.61656.68 goes into 4227.6 about 2 times (1656.68 * 2 = 3313.36)Subtract: 4227.6 - 3313.36 = 914.24Bring down a zero: 9142.41656.68 goes into 9142.4 about 5 times (1656.68 * 5 = 8283.4)Subtract: 9142.4 - 8283.4 = 859Bring down a zero: 85901656.68 goes into 8590 about 5 times (1656.68 * 5 = 8283.4)Subtract: 8590 - 8283.4 = 306.6So, putting it all together, we have 0.5432... So, approximately 0.5432.Therefore, ( P(50) ≈ 0.5432 ), so ( 1 - P(50) ≈ 0.4568 ).Thus, ( Y(50) = 1000 * 0.4568 ≈ 456.8 ) kg, which is approximately 457 kg.So, that seems consistent.Alternatively, if I use a calculator for ( e^{-2.5} ), it's exactly 0.082085, so 8 * 0.082085 = 0.65668. Then, 1 + 0.65668 = 1.65668. 0.9 / 1.65668 = approximately 0.5432, as above.So, yes, 457 kg is accurate.Therefore, the answers are approximately 46.05 days for the first question and 457 kg for the second.Final Answer1. The time when half of the plants are infected is boxed{46.05} days.2. The yield after 50 days is boxed{457} kg.</think>

🧠:<think>Alright, so I have this problem about a marine wildlife rescue center, and I need to solve two parts. Let me take it step by step.First, the center has success rates for rehabilitating different species: dolphins at 85%, sea turtles at 70%, and seals at 75%. They received a grant and expect to increase these rates. The first question is about calculating the expected increase in the number of successfully rehabilitated animals next year if the success rates improve by certain percentages.They mention that this year, they successfully rehabilitated 40 dolphins, 50 sea turtles, and 60 seals. So, I think I need to find out how many more animals they expect to successfully rehabilitate next year with the improved success rates.Let me break it down:1. Dolphins: Current success rate is 85%, and it's expected to increase by 10%. So, the new success rate will be 85% + 10% of 85%. Wait, hold on. Is the 10% an absolute increase or a percentage increase? Hmm, the problem says "increase these rates by a certain percentage," so I think it's a percentage increase. So, for dolphins, the new rate is 85% * 1.10.Similarly, for sea turtles, it's 70% increased by 15%, so 70% * 1.15. For seals, 75% increased by 5%, so 75% * 1.05.Wait, but hold on. The number of animals they are rehabilitating is given as 40 dolphins, 50 sea turtles, and 60 seals this year. So, does that mean they are trying to rehabilitate 40 dolphins this year, and next year, with the improved success rates, they will have more successes? Or is it that they are expecting to have more animals to rehabilitate next year?Wait, the problem says, "they expect to increase these rates by a certain percentage." So, the number of animals they are trying to rehabilitate might remain the same, but the success rates go up, leading to more successful rehabilitations.So, for each species, the number of successful rehabilitations this year is given, but next year, with the improved success rates, the number will be higher. So, I need to calculate the expected number of successful rehabilitations next year and then find the increase.Wait, but the problem says, "calculate the expected increase in the number of successfully rehabilitated animals next year." So, I think I need to compute the difference between next year's successful numbers and this year's.So, let's compute next year's success numbers:For dolphins: Current success rate is 85%, so 40 dolphins * 85% = 34 successfully rehabilitated this year. Next year, the success rate increases by 10%, so new rate is 85% * 1.10 = 93.5%. So, next year's successful rehabilitations would be 40 * 93.5% = 37.4. So, the increase is 37.4 - 34 = 3.4, which is approximately 3 more dolphins.Wait, but 40 * 93.5% is 40 * 0.935 = 37.4, yes. So, the increase is 3.4, which is about 3 or 4. But since we can't have a fraction of an animal, maybe we round it? Or perhaps we can keep it as a decimal for now.Similarly, for sea turtles: This year, they successfully rehabilitated 50 * 70% = 35. Next year, the success rate increases by 15%, so new rate is 70% * 1.15 = 80.5%. So, next year's successful rehabilitations would be 50 * 80.5% = 40.25. So, the increase is 40.25 - 35 = 5.25, which is about 5 or 5.25.For seals: This year, 60 * 75% = 45 successfully rehabilitated. Next year, success rate increases by 5%, so new rate is 75% * 1.05 = 78.75%. So, next year's successful rehabilitations would be 60 * 78.75% = 47.25. The increase is 47.25 - 45 = 2.25.So, total increase is 3.4 + 5.25 + 2.25 = 10.9. So, approximately 11 more animals successfully rehabilitated next year.But wait, let me double-check my calculations because I might have made a mistake.For dolphins: 85% success rate, 40 dolphins. So, 40 * 0.85 = 34. Next year, 85% * 1.10 = 93.5%, so 40 * 0.935 = 37.4. So, increase is 3.4.Sea turtles: 50 * 0.70 = 35. Next year, 70% * 1.15 = 80.5%, so 50 * 0.805 = 40.25. Increase is 5.25.Seals: 60 * 0.75 = 45. Next year, 75% * 1.05 = 78.75%, so 60 * 0.7875 = 47.25. Increase is 2.25.Total increase: 3.4 + 5.25 + 2.25 = 10.9. So, approximately 11 animals.But the problem says "calculate the expected increase in the number of successfully rehabilitated animals next year." So, I think 10.9 is the exact number, but since we can't have a fraction of an animal, we might need to round it to 11.Alternatively, maybe the question expects us to use the current number of animals being rehabilitated (40, 50, 60) and apply the increased success rates to those numbers, then subtract the current successful numbers.Wait, that's what I did. So, 40 dolphins, 50 sea turtles, 60 seals. So, next year, same number of animals being attempted, but higher success rates.So, yes, the increase is 10.9, approximately 11.But let me think again. Is the 10% increase on the success rate or on the number of animals? The problem says "increase these rates by a certain percentage," so it's on the success rate, not on the number of animals. So, the number of animals being rehabilitated is the same, 40, 50, 60.Therefore, the increase in successful rehabilitations is 3.4 + 5.25 + 2.25 = 10.9, which is approximately 11.So, for part 1, the expected increase is 11 animals.Now, moving on to part 2.The parent plans to share stories of successful rehabilitation on a new digital platform, estimating that with each additional story, the number of followers increases by a function ( f(x) = 1000 + 200x - 3x^2 ), where ( x ) is the number of stories shared. Determine the number of stories ( x ) that maximizes the number of followers, and calculate that maximum number of followers.Okay, so this is a quadratic function in terms of ( x ). The function is ( f(x) = -3x^2 + 200x + 1000 ). Since the coefficient of ( x^2 ) is negative (-3), the parabola opens downward, so the vertex is the maximum point.To find the maximum, we can use the vertex formula. For a quadratic function ( ax^2 + bx + c ), the x-coordinate of the vertex is at ( x = -b/(2a) ).Here, ( a = -3 ), ( b = 200 ). So,( x = -200 / (2 * -3) = -200 / (-6) = 200 / 6 ≈ 33.333 ).So, the maximum occurs at ( x ≈ 33.333 ). Since the number of stories must be an integer, we need to check ( x = 33 ) and ( x = 34 ) to see which gives a higher number of followers.Let's compute ( f(33) ):( f(33) = 1000 + 200*33 - 3*(33)^2 ).Compute each term:200*33 = 660033^2 = 1089, so 3*1089 = 3267So, ( f(33) = 1000 + 6600 - 3267 = 7600 - 3267 = 4333 ).Now, ( f(34) ):200*34 = 680034^2 = 1156, so 3*1156 = 3468Thus, ( f(34) = 1000 + 6800 - 3468 = 7800 - 3468 = 4332 ).So, ( f(33) = 4333 ) and ( f(34) = 4332 ). Therefore, the maximum occurs at ( x = 33 ) with 4333 followers.Alternatively, since the vertex is at 33.333, which is closer to 33, so 33 is the optimal number of stories.Therefore, the number of stories ( x ) that maximizes followers is 33, and the maximum number of followers is 4333.Wait, let me double-check my calculations.For ( x = 33 ):200*33 = 66003*(33)^2 = 3*1089 = 3267So, 1000 + 6600 = 7600; 7600 - 3267 = 4333. Correct.For ( x = 34 ):200*34 = 68003*(34)^2 = 3*1156 = 34681000 + 6800 = 7800; 7800 - 3468 = 4332. Correct.So, yes, 33 stories give the maximum followers of 4333.Alternatively, another way is to take the derivative, but since this is a quadratic, vertex formula is sufficient.So, in summary:1. The expected increase in successfully rehabilitated animals is approximately 11.2. The number of stories that maximizes followers is 33, with a maximum of 4333 followers.Final Answer1. The expected increase is boxed{11} animals.2. The number of stories that maximizes followers is boxed{33}, resulting in a maximum of boxed{4333} followers.</think>

🧠:<think>First, I need to calculate the total score for each hunter by summing up the points from each expedition. Each small game is worth 3 points, each medium game is worth 7 points, and each large game is worth 15 points.For Hunter A:- Expedition 1: (3 * 3) + (2 * 7) + (1 * 15) = 9 + 14 + 15 = 38 points- Expedition 2: (5 * 3) + (1 * 7) + (0 * 15) = 15 + 7 + 0 = 22 points- Expedition 3: (2 * 3) + (3 * 7) + (2 * 15) = 6 + 21 + 30 = 57 points- Expedition 4: (4 * 3) + (0 * 7) + (1 * 15) = 12 + 0 + 15 = 27 points- Expedition 5: (1 * 3) + (4 * 7) + (1 * 15) = 3 + 28 + 15 = 46 pointsTotal for Hunter A: 38 + 22 + 57 + 27 + 46 = 190 pointsFor Hunter B:- Expedition 1: (4 * 3) + (1 * 7) + (2 * 15) = 12 + 7 + 30 = 49 points- Expedition 2: (2 * 3) + (3 * 7) + (1 * 15) = 6 + 21 + 15 = 42 points- Expedition 3: (5 * 3) + (1 * 7) + (0 * 15) = 15 + 7 + 0 = 22 points- Expedition 4: (1 * 3) + (5 * 7) + (1 * 15) = 3 + 35 + 15 = 53 points- Expedition 5: (3 * 3) + (2 * 7) + (3 * 15) = 9 + 14 + 45 = 68 pointsTotal for Hunter B: 49 + 42 + 22 + 53 + 68 = 234 pointsComparing the totals, Hunter B has a higher score (234 points) than Hunter A (190 points), so Hunter B is the better hunter based on total score.Next, to determine the bonus for the highest average score per expedition, I'll calculate the average for each hunter by dividing their total score by the number of expeditions (5).Hunter A's average: 190 / 5 = 38 points per expeditionHunter B's average: 234 / 5 = 46.8 points per expeditionSince Hunter B has a higher average score, they receive the bonus.</think>

🧠:<think>Alright, so I'm trying to help this aspiring teenage football player analyze Antoine Semenyo's performance data. The goal is to figure out the probability of scoring a goal per shot and then use that to predict the expected number of goals if the teenager takes 100 shots. Let me break this down step by step.First, the problem mentions that Semenyo played 34 matches and scored 17 goals with a total of 85 shots. The number of goals scored, G, follows a binomial distribution with n trials (shots) and probability p of success (scoring a goal). Okay, so for part 1, I need to calculate the maximum likelihood estimate (MLE) of p. Hmm, I remember that in binomial distribution, the MLE for p is the ratio of the number of successes to the number of trials. In this case, the number of successes is the goals scored, which is 17, and the number of trials is the total shots, which is 85. So, I think the MLE for p is just 17 divided by 85. Let me write that down:MLE of p = G / n = 17 / 85.Let me compute that. 17 divided by 85. Hmm, 85 goes into 17 zero times, but wait, 85 is larger than 17, so it's 0.2. Because 85 times 0.2 is 17. So, p is 0.2. That seems straightforward.Wait, is there more to it? I think MLE for binomial is indeed the sample proportion, which is just the number of successes over the number of trials. So, yeah, 17/85 simplifies to 1/5, which is 0.2. So, p is 0.2. Got it.Moving on to part 2. The aspiring player wants to predict his own performance. He's going to take 100 shots next season, and we need to find the expected number of goals he will score using the properties of the binomial distribution.I recall that for a binomial distribution, the expected value or mean is given by n*p, where n is the number of trials and p is the probability of success. So, in this case, n is 100 shots, and p is the MLE we found earlier, which is 0.2.So, the expected number of goals is 100 * 0.2. Let me calculate that. 100 times 0.2 is 20. So, the expected number of goals is 20.Wait, does that make sense? If he's taking 100 shots and each shot has a 20% chance of scoring, then on average, he should score 20 goals. That seems reasonable.Let me just double-check my steps. For part 1, MLE of p is 17/85, which is 0.2. For part 2, expected goals is 100 * 0.2 = 20. Yeah, that seems correct.I don't think I made any mistakes here. It's pretty straightforward once you remember that the MLE for p in a binomial distribution is just the sample proportion, and the expectation is n*p. So, I think I'm confident with these answers.Final Answer1. The MLE of ( p ) is boxed{0.2}.2. The expected number of goals is boxed{20}.</think>