🧠:<think>Okay, so I have this problem where Alex is trying to show Jamie the benefits of choosing tennis over golf by using some math models. There are two parts to this problem. Let me try to tackle them one by one.Starting with the first part: They have a logistic function modeling the probability of winning a tennis match based on training hours, T. The formula is given as:[ P(T) = frac{1}{1 + e^{-k(T - T_0)}} ]They want Jamie to have at least a 70% chance of winning, which is 0.7 in probability terms. The constants are given as k = 0.1 and T0 = 20. So, I need to find the minimum T such that P(T) is at least 0.7.Hmm, okay. So, let me write down the equation:[ 0.7 = frac{1}{1 + e^{-0.1(T - 20)}} ]I need to solve for T here. Let me rearrange this equation step by step. First, I can take reciprocals on both sides to get rid of the fraction:[ frac{1}{0.7} = 1 + e^{-0.1(T - 20)} ]Calculating 1/0.7, which is approximately 1.4286. So,[ 1.4286 = 1 + e^{-0.1(T - 20)} ]Subtract 1 from both sides:[ 0.4286 = e^{-0.1(T - 20)} ]Now, to solve for the exponent, I can take the natural logarithm (ln) of both sides. Remember that ln(e^x) = x.So,[ ln(0.4286) = -0.1(T - 20) ]Calculating ln(0.4286). Let me recall that ln(1) is 0, ln(e^{-1}) is about -1, and 0.4286 is roughly 1/e, since e is about 2.718, so 1/e is about 0.3679. Hmm, 0.4286 is a bit higher. Maybe I should just compute it.Using a calculator, ln(0.4286) is approximately -0.8473. So,[ -0.8473 = -0.1(T - 20) ]Multiply both sides by -1 to make it positive:[ 0.8473 = 0.1(T - 20) ]Now, divide both sides by 0.1:[ 8.473 = T - 20 ]So, adding 20 to both sides:[ T = 20 + 8.473 ][ T ≈ 28.473 ]Since training hours can't be a fraction, Jamie would need to train at least 29 hours to have a 70% chance of winning. But wait, let me check if 28.473 is enough. If T is 28.473, then plugging back into the original equation:[ P(28.473) = frac{1}{1 + e^{-0.1(28.473 - 20)}} ][ = frac{1}{1 + e^{-0.1(8.473)}} ][ = frac{1}{1 + e^{-0.8473}} ][ e^{-0.8473} ≈ 0.4286 ]So,[ P ≈ frac{1}{1 + 0.4286} ≈ frac{1}{1.4286} ≈ 0.7 ]So, 28.473 hours gives exactly 70%. Since you can't train a fraction of an hour, Jamie needs to train at least 29 hours to exceed 70%. So, the minimum number of training hours required is 29.Alright, that was part one. Now, moving on to part two.They are comparing earnings from tennis and golf. The earnings from tennis per match are a linear function: E_t(T) = aT + b. The constants given are a = 50, b = 1000. So, E_t(T) = 50T + 1000.The golf earnings are constant at C = 2000 per match. They want the tennis earnings to exceed the golf earnings by at least 500. So, the condition is:E_t(T) ≥ C + 500Which translates to:50T + 1000 ≥ 2000 + 500Simplify the right side:50T + 1000 ≥ 2500Subtract 1000 from both sides:50T ≥ 1500Divide both sides by 50:T ≥ 30So, Jamie needs to train at least 30 hours for the tennis earnings per match to exceed golf earnings by 500.Wait, let me double-check that. If T = 30,E_t(30) = 50*30 + 1000 = 1500 + 1000 = 2500Golf earnings are 2000, so 2500 - 2000 = 500. So, yes, exactly 500 more. If T is 30, it's equal. But the problem says "exceed by at least 500," so 30 is the minimum. If T is more than 30, the difference increases. So, 30 is the required number of training hours.Wait, hold on, but in part one, the required training hours were 29. So, if Jamie is training 30 hours, he would have more than 70% chance of winning, right? Because 30 is more than 28.473. So, that seems consistent.Is there any chance I made a mistake in the calculations? Let me verify.For part one:0.7 = 1 / (1 + e^{-0.1(T -20)})Multiply both sides by denominator:0.7(1 + e^{-0.1(T -20)}) = 10.7 + 0.7e^{-0.1(T -20)} = 1Subtract 0.7:0.7e^{-0.1(T -20)} = 0.3Divide by 0.7:e^{-0.1(T -20)} = 0.3 / 0.7 ≈ 0.4286Take ln:-0.1(T -20) = ln(0.4286) ≈ -0.8473Divide both sides by -0.1:T -20 = 8.473T ≈ 28.473, so 29 hours. That seems correct.For part two:E_t(T) = 50T + 1000We need 50T + 1000 ≥ 2000 + 500So, 50T + 1000 ≥ 250050T ≥ 1500T ≥ 30. Correct.So, both answers seem solid.Final Answer1. The minimum number of training hours required is boxed{29}.2. The number of training hours necessary is boxed{30}.</think>

🧠:<think>Okay, so I need to figure out the total number of chapters in the 5th novel and also find a general expression for the total number of chapters in the nth novel. Plus, I have to prove that the total number of chapters is always greater than the sum of the introduction and development chapters for any n ≥ 1. Hmm, let me break this down step by step.First, let's understand the problem. Each novel has three types of chapters: introduction (I(n)), development (D(n)), and resolution (R(n)). The total number of chapters would be the sum of these three. So, for part a), I need to compute I(5), D(5), and R(5), then add them up. For part b), I need a general formula for the total chapters, which would be I(n) + D(n) + R(n). Then, I have to prove that this total is always greater than I(n) + D(n), which means R(n) must be greater than zero. But since R(n) is the product of I(n) and D(n), and both I(n) and D(n) are positive for all n, R(n) will always be positive. So, that might be straightforward, but maybe I need to think more about it.Starting with part a). Let's compute each component for n=5.1. I(n) = 2^n + 1. So, I(5) = 2^5 + 1. 2^5 is 32, so I(5) = 32 + 1 = 33.2. D(n) is the Fibonacci sequence. The Fibonacci sequence is defined as F(n) = F(n-1) + F(n-2) with F(1)=1 and F(2)=1. So, let's list out the Fibonacci numbers up to n=5.- F(1) = 1- F(2) = 1- F(3) = F(2) + F(1) = 1 + 1 = 2- F(4) = F(3) + F(2) = 2 + 1 = 3- F(5) = F(4) + F(3) = 3 + 2 = 5So, D(5) = F(5) = 5.3. R(n) = I(n) * D(n). So, R(5) = I(5) * D(5) = 33 * 5. Let me compute that: 33*5 is 165.Therefore, the total number of chapters in the 5th novel is I(5) + D(5) + R(5) = 33 + 5 + 165.Let me add those up: 33 + 5 is 38, and 38 + 165 is 203. So, the total is 203 chapters.Wait, let me double-check my calculations to make sure I didn't make a mistake.I(5) = 2^5 + 1 = 32 + 1 = 33. Correct.Fibonacci numbers: F(1)=1, F(2)=1, F(3)=2, F(4)=3, F(5)=5. Correct.R(5) = 33 * 5 = 165. Correct.Total chapters: 33 + 5 + 165 = 203. Yep, that seems right.Okay, so part a) is 203 chapters.Moving on to part b). I need a general expression for the total number of chapters in the nth novel. Let's denote the total as T(n). So,T(n) = I(n) + D(n) + R(n)Given that I(n) = 2^n + 1, D(n) = F(n), and R(n) = I(n) * D(n). So substituting these in,T(n) = (2^n + 1) + F(n) + (2^n + 1) * F(n)Let me factor this expression. Notice that (2^n + 1) is a common factor in the last two terms. So,T(n) = (2^n + 1) + F(n) + (2^n + 1) * F(n) = (2^n + 1) + F(n) * [1 + (2^n + 1)] = (2^n + 1) + F(n) * (2^n + 2) = (2^n + 1) + (2^n + 2) * F(n)Alternatively, we can factor 2^n + 1 as a common term:T(n) = (2^n + 1) * (1 + F(n)) + F(n)Wait, let me check that:(2^n + 1) * (1 + F(n)) = (2^n + 1) + (2^n + 1)*F(n)But T(n) is (2^n + 1) + F(n) + (2^n + 1)*F(n) = (2^n + 1) + F(n)*(1 + 2^n + 1) = (2^n + 1) + F(n)*(2^n + 2). Hmm, so maybe it's better to leave it as:T(n) = (2^n + 1) + F(n) + (2^n + 1) * F(n)Alternatively, factor out (2^n + 1):T(n) = (2^n + 1)(1 + F(n)) + F(n)Wait, let me compute:(2^n + 1)(1 + F(n)) = (2^n + 1) + (2^n + 1)F(n)So, T(n) = (2^n + 1)(1 + F(n)) + F(n) - (2^n + 1)F(n) + F(n). Hmm, no, that complicates it more. Maybe it's better to just leave it as:T(n) = (2^n + 1) + F(n) + (2^n + 1)F(n)Alternatively, factor F(n):T(n) = (2^n + 1) + F(n)(1 + 2^n + 1) = (2^n + 1) + F(n)(2^n + 2)Yes, that seems better. So,T(n) = (2^n + 1) + (2^n + 2)F(n)Alternatively, factor 2^n:T(n) = 2^n(1 + F(n)) + 1 + 2F(n)But I don't know if that's any simpler. Maybe the first expression is fine.Alternatively, let's see if we can write it as:T(n) = (2^n + 1)(1 + F(n)) + F(n) - (2^n + 1)F(n) + F(n). Wait, that might not be helpful.Alternatively, perhaps it's better to just express T(n) as:T(n) = (2^n + 1) + F(n) + (2^n + 1)F(n) = (2^n + 1)(1 + F(n)) + F(n) - (2^n + 1)F(n) + F(n). Hmm, no, that seems circular.Wait, perhaps I can factor (2^n + 1) as follows:T(n) = (2^n + 1) + F(n) + (2^n + 1)F(n) = (2^n + 1)(1 + F(n)) + F(n) - (2^n + 1)F(n) + F(n). Hmm, no, that's not helpful.Wait, maybe I can factor F(n):T(n) = (2^n + 1) + F(n)(1 + 2^n + 1) = (2^n + 1) + F(n)(2^n + 2)Yes, that seems correct. So,T(n) = (2^n + 1) + (2^n + 2)F(n)Alternatively, factor 2 from the second term:T(n) = (2^n + 1) + 2(2^{n-1} + 1)F(n)But I don't know if that's helpful. Maybe it's better to leave it as:T(n) = (2^n + 1) + (2^n + 2)F(n)Alternatively, we can write it as:T(n) = (2^n + 1)(1 + F(n)) + F(n) - (2^n + 1)F(n) + F(n). Wait, that's not helpful.Alternatively, perhaps it's better to just leave it as:T(n) = I(n) + D(n) + R(n) = (2^n + 1) + F(n) + (2^n + 1)F(n)Which can be written as:T(n) = (2^n + 1)(1 + F(n)) + F(n)Wait, let me check:(2^n + 1)(1 + F(n)) = (2^n + 1) + (2^n + 1)F(n)So, T(n) = (2^n + 1)(1 + F(n)) + F(n) - (2^n + 1)F(n) + F(n). Hmm, no, that's not helpful.Wait, perhaps it's better to just accept that T(n) = (2^n + 1) + F(n) + (2^n + 1)F(n). So, that's the expression.Alternatively, factor F(n):T(n) = (2^n + 1) + F(n)(1 + 2^n + 1) = (2^n + 1) + F(n)(2^n + 2)Yes, that seems correct.So, T(n) = (2^n + 1) + (2^n + 2)F(n)Alternatively, factor 2 from the second term:T(n) = (2^n + 1) + 2(2^{n-1} + 1)F(n)But I don't know if that's any better.Alternatively, factor (2^n + 1):T(n) = (2^n + 1)(1 + F(n)) + F(n) - (2^n + 1)F(n) + F(n). Hmm, no, that's not helpful.I think the expression T(n) = (2^n + 1) + (2^n + 2)F(n) is as simplified as it can get. So, that's the general expression.Now, the second part of b) is to prove that T(n) > I(n) + D(n) for all n ≥ 1.Given that T(n) = I(n) + D(n) + R(n), and R(n) = I(n) * D(n). So, T(n) = I(n) + D(n) + I(n)D(n). Therefore, T(n) - (I(n) + D(n)) = I(n)D(n). Since I(n) and D(n) are both positive for all n ≥ 1, their product is positive. Therefore, T(n) - (I(n) + D(n)) > 0, which implies T(n) > I(n) + D(n).Wait, that seems too straightforward. Let me think again.Given that R(n) = I(n) * D(n), which is positive because both I(n) and D(n) are positive. Therefore, adding R(n) to I(n) + D(n) will make T(n) larger than I(n) + D(n). So, yes, T(n) = I(n) + D(n) + R(n) > I(n) + D(n) for all n ≥ 1.Alternatively, to make it more formal, since R(n) > 0, then T(n) = I(n) + D(n) + R(n) > I(n) + D(n).So, that's the proof.Wait, but maybe I should elaborate a bit more.Given that for all n ≥ 1:- I(n) = 2^n + 1, which is always greater than 1, since 2^n ≥ 2 for n ≥ 1.- D(n) = F(n), which is the Fibonacci sequence starting at F(1)=1, F(2)=1, and increasing thereafter. So, D(n) ≥ 1 for all n ≥ 1.Therefore, R(n) = I(n) * D(n) ≥ (2 + 1)*1 = 3 for n=1, and increases from there.Thus, R(n) is always positive, so T(n) = I(n) + D(n) + R(n) is always greater than I(n) + D(n).Therefore, the total number of chapters is always greater than the sum of introduction and development chapters.So, that's the proof.Wait, but maybe I should check for n=1 to make sure.For n=1:I(1) = 2^1 + 1 = 3D(1) = F(1) = 1R(1) = 3 * 1 = 3Total T(1) = 3 + 1 + 3 = 7Sum of I(1) + D(1) = 3 + 1 = 47 > 4, which is true.Similarly, for n=2:I(2) = 2^2 + 1 = 5D(2) = F(2) = 1R(2) = 5 * 1 = 5T(2) = 5 + 1 + 5 = 11Sum I(2) + D(2) = 5 + 1 = 611 > 6, true.n=3:I(3)=2^3+1=9D(3)=F(3)=2R(3)=9*2=18T(3)=9+2+18=29Sum I(3)+D(3)=9+2=1129>11, true.So, it holds for these cases. Therefore, the proof is valid.So, to summarize:a) The total number of chapters in the 5th novel is 203.b) The general expression for the total number of chapters in the nth novel is T(n) = (2^n + 1) + (2^n + 2)F(n), and since R(n) is always positive, T(n) is always greater than I(n) + D(n).Wait, but in the general expression, I think I can write it more neatly. Let me see:T(n) = I(n) + D(n) + R(n) = (2^n + 1) + F(n) + (2^n + 1)F(n)Factor out (2^n + 1):T(n) = (2^n + 1)(1 + F(n)) + F(n) - (2^n + 1)F(n) + F(n). Hmm, no, that's not helpful.Wait, actually, let me compute it again:T(n) = (2^n + 1) + F(n) + (2^n + 1)F(n)= (2^n + 1) + F(n) + (2^n + 1)F(n)= (2^n + 1) + F(n)(1 + 2^n + 1)= (2^n + 1) + F(n)(2^n + 2)Yes, that's correct. So, T(n) = (2^n + 1) + (2^n + 2)F(n)Alternatively, factor 2 from the second term:T(n) = (2^n + 1) + 2(2^{n-1} + 1)F(n)But I don't know if that's any better.Alternatively, factor (2^n + 1):T(n) = (2^n + 1)(1 + F(n)) + F(n) - (2^n + 1)F(n) + F(n). Hmm, no, that's not helpful.I think the expression T(n) = (2^n + 1) + (2^n + 2)F(n) is as simplified as it can get.Alternatively, maybe we can write it as:T(n) = (2^n + 1)(1 + F(n)) + F(n) - (2^n + 1)F(n) + F(n). Hmm, no, that's not helpful.Wait, perhaps it's better to just leave it as T(n) = (2^n + 1) + (2^n + 2)F(n). So, that's the general expression.Therefore, the answers are:a) 203 chapters.b) T(n) = (2^n + 1) + (2^n + 2)F(n), and T(n) > I(n) + D(n) because R(n) is positive.Wait, but maybe I can write T(n) in terms of F(n) and 2^n more elegantly. Let me think.Alternatively, since R(n) = I(n)D(n) = (2^n + 1)F(n), then T(n) = I(n) + D(n) + R(n) = (2^n + 1) + F(n) + (2^n + 1)F(n) = (2^n + 1)(1 + F(n)) + F(n). Hmm, that's another way to write it.But I think the expression T(n) = (2^n + 1) + (2^n + 2)F(n) is acceptable.Alternatively, factor 2^n:T(n) = 2^n(1 + F(n)) + 1 + 2F(n)But again, not sure if that's better.I think either form is acceptable, but perhaps the first form is better.So, to conclude:a) The total number of chapters in the 5th novel is 203.b) The general expression is T(n) = (2^n + 1) + (2^n + 2)F(n), and since R(n) is positive, T(n) > I(n) + D(n) for all n ≥ 1.</think>

🧠:<think>Okay, so I have these two sub-problems about Alex and his caffeine intake. Let me try to figure them out step by step.Starting with Sub-problem 1: Alex drinks a cup of coffee at 5:30 AM with 150 mg of caffeine. The decay constant k is 0.2 per hour. We need to find out how much caffeine is left by 8:00 AM.First, I need to figure out how much time passes between 5:30 AM and 8:00 AM. Let's see, from 5:30 to 6:30 is one hour, 6:30 to 7:30 is another, and 7:30 to 8:00 is another half hour. So that's a total of 2.5 hours. So t = 2.5 hours.The formula given is C(t) = C0 * e^(-kt). Plugging in the numbers, C0 is 150 mg, k is 0.2, and t is 2.5.So, C(2.5) = 150 * e^(-0.2 * 2.5). Let me compute the exponent first: 0.2 * 2.5 = 0.5. So it's e^(-0.5). I remember that e^(-0.5) is approximately 0.6065. So, 150 * 0.6065. Let me calculate that.150 * 0.6 is 90, and 150 * 0.0065 is about 0.975. So adding them together, 90 + 0.975 = 90.975 mg. So approximately 90.98 mg of caffeine remains at 8:00 AM.Wait, let me double-check that calculation. Maybe I should use a calculator for e^(-0.5). Actually, e^(-0.5) is approximately 0.60653066. So, 150 * 0.60653066. Let me compute 150 * 0.6 = 90, 150 * 0.00653066 ≈ 0.9796. So total is approximately 90 + 0.9796 ≈ 90.9796 mg. So, about 90.98 mg. Yeah, that seems right.So, for Sub-problem 1, the caffeine left is approximately 90.98 mg.Moving on to Sub-problem 2: Alex needs at least 75 mg of caffeine to stay focused. He drinks another cup at 7:00 AM, same 150 mg and k=0.2. We need to calculate the total caffeine at 8:00 AM and see if it's at least 75 mg.So, first, let's figure out the time between each coffee and 8:00 AM.The first coffee is at 5:30 AM, so from 5:30 to 8:00 is 2.5 hours as before. The second coffee is at 7:00 AM, so from 7:00 to 8:00 is 1 hour.So, the first coffee contributes C1 = 150 * e^(-0.2 * 2.5) which we already calculated as approximately 90.98 mg.The second coffee is consumed at 7:00 AM, so it has 1 hour to decay by 8:00 AM. So, C2 = 150 * e^(-0.2 * 1). Let's compute that.First, 0.2 * 1 = 0.2. So, e^(-0.2) is approximately 0.8187. So, 150 * 0.8187 ≈ 150 * 0.8 = 120, and 150 * 0.0187 ≈ 2.805. So total is approximately 120 + 2.805 ≈ 122.805 mg.Wait, that doesn't seem right. Wait, 150 * 0.8187. Let me compute it more accurately. 150 * 0.8 = 120, 150 * 0.0187 = approx 2.805. So, 120 + 2.805 = 122.805 mg. Hmm, that seems high because it's less than 150 but more than half.Wait, but actually, 0.8187 is about 81.87% of the original caffeine. So, 150 * 0.8187 is indeed approximately 122.8 mg. So, that seems correct.So, the total caffeine in Alex's body at 8:00 AM is the sum of the remaining caffeine from both coffees: C1 + C2 ≈ 90.98 + 122.805 ≈ 213.785 mg.Wait, that seems way more than 75 mg. So, 213.785 mg is way above 75 mg. So, Alex's strategy works because he has more than 75 mg.But wait, hold on. Is that correct? Because when you take another dose, does it just add up? Or does the body's caffeine level reset?Wait, no, the model is exponential decay for each dose. So, each dose decays independently. So, the total caffeine is the sum of the remaining from each dose. So, yes, 90.98 + 122.805 ≈ 213.785 mg.But wait, 213 mg is way more than 75 mg, so it definitely meets the requirement. So, Alex's strategy works.But wait, let me think again. Is there any consideration about the half-life or something else? Or is it just additive?I think in pharmacokinetics, when you take multiple doses, each dose contributes its own decay curve, so the total is the sum. So, yeah, adding them up is correct.So, total caffeine is approximately 213.79 mg, which is more than 75 mg. So, Alex can stay focused.Wait, but let me check the calculations again to be sure.First coffee: 5:30 AM to 8:00 AM is 2.5 hours. C1 = 150 * e^(-0.2*2.5) = 150 * e^(-0.5) ≈ 150 * 0.6065 ≈ 90.975 mg.Second coffee: 7:00 AM to 8:00 AM is 1 hour. C2 = 150 * e^(-0.2*1) = 150 * e^(-0.2) ≈ 150 * 0.8187 ≈ 122.805 mg.Total: 90.975 + 122.805 ≈ 213.78 mg.Yes, that's correct. So, Alex has over 200 mg, which is more than enough for his 75 mg requirement.So, summarizing:Sub-problem 1: Approximately 90.98 mg.Sub-problem 2: Total caffeine is approximately 213.78 mg, which is above 75 mg, so his strategy works.I think that's it. I don't see any mistakes in my reasoning.Final AnswerSub-problem 1: boxed{90.98} mgSub-problem 2: boxed{213.78} mg, which meets the requirement.</think>

🧠:<think>Alright, so I'm trying to solve these two sub-problems about trade interactions between European settlers and Native American tribes. The problem involves two goods, A and B, and then a third good, C. The rates of exchange are given by sinusoidal and linear functions, respectively. Let me break this down step by step.First, for sub-problem 1: I need to determine the total amount of good B obtained by exchanging 100 units of good A continuously over the first 12 months. The rate of exchange is given by the function ( R(t) = 3sinleft(frac{pi}{6}tright) + 5 ). Hmm, okay. So, R(t) represents how many units of B you get for each unit of A at time t. Since it's a rate, to find the total amount of B over 12 months, I think I need to integrate R(t) over the time period from 0 to 12. But wait, it's 100 units of A being exchanged continuously. So, does that mean I need to multiply the rate by 100?Let me think. If I have 100 units of A, and each unit of A is being exchanged at a rate R(t), then the total amount of B obtained at any time t would be 100 * R(t). But since this is happening continuously over time, I need to integrate this product over the 12 months.So, the integral setup should be the integral from t = 0 to t = 12 of 100 * R(t) dt. That makes sense because integrating over time will give the total accumulation of B.Therefore, substituting R(t) into the integral, it becomes:[int_{0}^{12} 100 times left(3sinleft(frac{pi}{6}tright) + 5right) dt]I can factor out the 100 to make it simpler:[100 times int_{0}^{12} left(3sinleft(frac{pi}{6}tright) + 5right) dt]That seems right. So, that's the integral setup for the first part.Moving on to sub-problem 2: Now, I need to calculate the total amount of good C obtained by exchanging the good B obtained from the previous step over the same 12 months. The rate of exchange from B to C is given by ( S(t) = 2t + 1 ).Wait, so S(t) is the rate at which B is exchanged for C. That is, for each unit of B, you get S(t) units of C at time t. But in the first problem, we found the total amount of B obtained over 12 months. So, do I need to now exchange all that B into C over the same period?But hold on, the problem says "continuously over the first 12 months," so it's not just exchanging the total B at the end, but exchanging the B as it's being obtained each month. So, similar to the first problem, I need to integrate the rate of exchange of B to C multiplied by the rate of obtaining B from A.Wait, maybe I need to think in terms of a double integral or something. Let me clarify.In the first problem, we had a rate R(t) which gave us the amount of B per unit A at time t. So, for each infinitesimal time dt, exchanging 100 units of A gives 100 * R(t) dt units of B. Then, for each of those B units, we can exchange them at the rate S(t) to get C.So, the total amount of C would be the integral over time of (100 * R(t) * S(t)) dt, right? Because for each dt, you get 100 * R(t) dt of B, and then each of those is converted to C at S(t), so total C is 100 * R(t) * S(t) dt.Therefore, the integral setup would be:[int_{0}^{12} 100 times R(t) times S(t) dt]Substituting R(t) and S(t):[int_{0}^{12} 100 times left(3sinleft(frac{pi}{6}tright) + 5right) times (2t + 1) dt]Again, I can factor out the 100:[100 times int_{0}^{12} left(3sinleft(frac{pi}{6}tright) + 5right) times (2t + 1) dt]So, that should be the integral setup for the second part.Wait, let me double-check. In the first problem, we integrated 100 * R(t) over 12 months to get total B. Then, for each infinitesimal amount of B obtained at time t, we exchange it at rate S(t) to get C. So, the total C is the integral of (100 * R(t) * S(t)) dt over 12 months. Yes, that makes sense.Alternatively, if we had first calculated the total B, say B_total, and then multiplied by the average S(t) over the year, that would be incorrect because S(t) is changing over time. Instead, we need to consider the exchange happening continuously, so we have to integrate the product of R(t) and S(t) over time.Therefore, my integral setups seem correct.So, to recap:1. For the first sub-problem, the integral is 100 times the integral of R(t) from 0 to 12.2. For the second sub-problem, the integral is 100 times the integral of R(t)*S(t) from 0 to 12.I think that's it. I don't see any mistakes in this reasoning.Final Answer1. The integral setup for the total amount of good B is boxed{100 int_{0}^{12} left(3sinleft(frac{pi}{6}tright) + 5right) dt}.2. The integral setup for the total amount of good C is boxed{100 int_{0}^{12} left(3sinleft(frac{pi}{6}tright) + 5right)(2t + 1) dt}.</think>