🧠:<think>Okay, so I have this problem about two reservoirs, A and B, connected by pipes. The goal is to figure out the total volume of water transferred between them over 24 hours and then determine the net change in each reservoir's volume after that time. Let me try to break this down step by step.First, part 1 is about calculating the total volume transferred from reservoir A to B. The flow rate is given by the function ( f(t) = 500 sinleft(frac{pi t}{12}right) ) cubic meters per hour. Since this is a rate, to find the total volume over 24 hours, I need to integrate this function from t = 0 to t = 24.So, the total volume ( V ) transferred is the integral of ( f(t) ) from 0 to 24:[V = int_{0}^{24} 500 sinleft(frac{pi t}{12}right) dt]I remember that the integral of ( sin(ax) ) is ( -frac{1}{a} cos(ax) ). Let me apply that here.Let me set ( a = frac{pi}{12} ), so the integral becomes:[V = 500 left[ -frac{12}{pi} cosleft(frac{pi t}{12}right) right]_0^{24}]Calculating the bounds:At t = 24:[-frac{12}{pi} cosleft(frac{pi times 24}{12}right) = -frac{12}{pi} cos(2pi) = -frac{12}{pi} times 1 = -frac{12}{pi}]At t = 0:[-frac{12}{pi} cosleft(0right) = -frac{12}{pi} times 1 = -frac{12}{pi}]Subtracting the lower bound from the upper bound:[-frac{12}{pi} - (-frac{12}{pi}) = 0]Wait, that can't be right. If I integrate a sine function over one full period, the integral should be zero because the positive and negative areas cancel out. But in this case, the sine function is modeling the flow rate from A to B. So, does that mean that over 24 hours, the net transfer is zero? That seems counterintuitive because the flow rate is positive when the sine is positive and negative when it's negative. But in reality, the flow rate can't be negative; it's just the direction of flow. Hmm, maybe I need to consider the absolute value of the flow rate?Wait, no, the function ( f(t) ) is defined as the flow rate from A to B. So, when ( f(t) ) is positive, water is flowing from A to B, and when it's negative, water is flowing from B to A. So, the integral gives the net flow from A to B. If the integral is zero, that means over 24 hours, the amount flowing from A to B is equal to the amount flowing back from B to A, resulting in no net change. But the question is asking for the total volume transferred, not the net volume. Hmm, so maybe I need to calculate the total amount of water that moved, regardless of direction.So, if I take the absolute value of the flow rate and integrate that over 24 hours, I would get the total volume transferred. But the problem says "the total volume of water transferred from A to B." So, does that mean only the times when water is flowing from A to B, or the net transfer?Looking back at the problem statement: "calculate the total volume of water transferred from A to B." So, I think it's the net transfer, which would be zero because the integral is zero. But that seems odd because over 24 hours, the sine function completes two full cycles (since the period is 24 hours? Wait, let's check the period.The function is ( sinleft(frac{pi t}{12}right) ). The period ( T ) is given by ( 2pi / (pi/12) ) = 24 hours. So, over 24 hours, it's one full period. So, integrating over one full period, the net flow is zero. So, the net transfer is zero, meaning equal amounts flow both ways.But the question says "the total volume of water transferred from A to B." Hmm, maybe they mean the total amount that moved from A to B, regardless of what came back. So, in that case, I need to integrate the absolute value of ( f(t) ) over 24 hours.But the problem didn't specify absolute value, so perhaps it's just the net transfer, which is zero. But that seems contradictory because the question is about transferring from A to B, so maybe it's just the integral, which is zero. But that would mean no net transfer, but the total amount transferred back and forth is non-zero.Wait, let me think again. The question says "the total volume of water transferred from A to B." So, if water flows from A to B for some time and then back, the total volume transferred from A to B is the integral of the positive parts, and the total transferred from B to A is the integral of the negative parts. But the question specifically asks for the transfer from A to B, so maybe it's just the integral of the positive parts.But how do I know when the flow is positive? Let's see, ( f(t) = 500 sin(pi t / 12) ). The sine function is positive in the first half of its period and negative in the second half. So, over 24 hours, it's positive from t = 0 to t = 12, and negative from t = 12 to t = 24.Therefore, the total volume transferred from A to B is the integral from 0 to 12 of ( 500 sin(pi t / 12) ) dt.Similarly, the volume transferred from B to A is the integral from 12 to 24 of ( -500 sin(pi t / 12) ) dt, but since we're only asked for A to B, we can just compute the first integral.So, let's compute that.[V = int_{0}^{12} 500 sinleft(frac{pi t}{12}right) dt]Using the same integral formula:[V = 500 left[ -frac{12}{pi} cosleft(frac{pi t}{12}right) right]_0^{12}]At t = 12:[-frac{12}{pi} cosleft(frac{pi times 12}{12}right) = -frac{12}{pi} cos(pi) = -frac{12}{pi} (-1) = frac{12}{pi}]At t = 0:[-frac{12}{pi} cos(0) = -frac{12}{pi} (1) = -frac{12}{pi}]Subtracting:[frac{12}{pi} - (-frac{12}{pi}) = frac{24}{pi}]Multiply by 500:[V = 500 times frac{24}{pi} = frac{12000}{pi} approx 3819.72 text{ cubic meters}]So, approximately 3819.72 cubic meters transferred from A to B over 24 hours.But wait, the flow rate is 500 sin(πt/12). So, over 24 hours, it's symmetric. So, the total volume transferred from A to B is equal to the volume transferred from B to A, each being 12000/π. So, the net transfer is zero, but the total transfer is 24000/π, but the question specifically asks for the transfer from A to B, so it's 12000/π.But let me double-check. If I integrate from 0 to 24, I get zero, which is the net transfer. But the question is about the total volume transferred from A to B, which would be the integral over the period when f(t) is positive, which is 0 to 12, giving 12000/π.Alternatively, if I consider the total transfer regardless of direction, it's twice that, but the question is specifically from A to B, so I think it's 12000/π.Let me compute 12000 divided by π:12000 / 3.1416 ≈ 3819.72 cubic meters.So, approximately 3819.72 cubic meters.But let me see if the problem expects an exact value or a numerical approximation. The problem says "calculate the total volume," so maybe leave it in terms of π.So, 12000/π cubic meters.Alternatively, since 12000/π is approximately 3819.72, but perhaps the problem expects an exact value.So, moving on to part 2.Reservoir A loses water at a rate ( g(t) = 30 + 20cosleft(frac{pi t}{6}right) ) cubic meters per hour. Reservoir B gains water from rain at a constant rate of 80 cubic meters per hour.We need to find the net change in both reservoirs after 24 hours, starting full.So, for reservoir A, the net change is the loss due to g(t) plus the net transfer from A to B. Wait, but in part 1, we found that the net transfer from A to B is zero, but the total transfer is 3819.72 out and the same back in. But for the net change, it's the net transfer plus the loss from g(t).Wait, no. Let me think carefully.Reservoir A has two factors affecting its volume: the loss due to g(t) and the net transfer from A to B.Similarly, reservoir B has two factors: the gain from rain and the net transfer from A to B.But wait, in part 1, we found that the net transfer from A to B over 24 hours is zero. So, the net change in A is just the loss due to g(t), and the net change in B is just the gain from rain.But let me verify.Reservoir A:- Losing water at rate g(t) = 30 + 20 cos(πt/6)- Also, water is flowing out to B and flowing back from B. The net flow is zero, as per part 1.So, the net change in A is the integral of g(t) over 24 hours, subtracted from the initial volume.But since they start full, the net change is just the total loss.Similarly, reservoir B:- Gaining 80 cubic meters per hour from rain.- Net flow from A is zero, so no net gain or loss from that.Therefore, the net change in B is just the total rainwater collected, which is 80 * 24.But wait, let me think again. The flow from A to B is f(t), which has a net of zero, but the actual flow is oscillating. So, does that mean that the net change is only due to g(t) for A and only due to rain for B?Wait, no. Because the flow from A to B is part of the system. So, the net change in A is the loss due to g(t) plus the net outflow from A to B. But since the net outflow is zero, the net change is just the loss from g(t).Similarly, for B, the net change is the inflow from rain plus the net inflow from A to B, which is zero. So, net change is just rain.But wait, actually, the flow from A to B is part of the system, so the net change in A is the loss from g(t) plus the outflow from A to B minus the inflow from B to A. But since the net outflow is zero, it's just the loss from g(t).Similarly, for B, net change is the inflow from rain plus the net inflow from A to B, which is zero, so just the rain.But let me make sure.The total change in A is:ΔA = -∫₀²⁴ g(t) dt - ∫₀²⁴ f(t) dtBut since ∫₀²⁴ f(t) dt = 0, ΔA = -∫₀²⁴ g(t) dtSimilarly, for B:ΔB = ∫₀²⁴ 80 dt + ∫₀²⁴ f(t) dtBut again, ∫₀²⁴ f(t) dt = 0, so ΔB = ∫₀²⁴ 80 dtTherefore, the net change in A is the negative of the integral of g(t) over 24 hours, and the net change in B is the integral of 80 over 24 hours.So, let's compute these.First, for reservoir A:Compute ∫₀²⁴ [30 + 20 cos(πt/6)] dtLet me split the integral:∫₀²⁴ 30 dt + ∫₀²⁴ 20 cos(πt/6) dtFirst integral:∫₀²⁴ 30 dt = 30t |₀²⁴ = 30*24 - 30*0 = 720Second integral:∫₀²⁴ 20 cos(πt/6) dtLet me compute this:Let u = πt/6, so du = π/6 dt, dt = 6/π duLimits: when t=0, u=0; t=24, u= π*24/6 = 4πSo,20 ∫₀^{4π} cos(u) * (6/π) du = (120/π) ∫₀^{4π} cos(u) duIntegral of cos(u) is sin(u):(120/π) [sin(u)]₀^{4π} = (120/π)(sin(4π) - sin(0)) = (120/π)(0 - 0) = 0So, the second integral is zero.Therefore, the total integral for g(t) is 720 + 0 = 720 cubic meters.So, the net change in A is -720 cubic meters.For reservoir B:ΔB = ∫₀²⁴ 80 dt = 80*24 = 1920 cubic meters.Therefore, after 24 hours, reservoir A has lost 720 cubic meters, and reservoir B has gained 1920 cubic meters.But wait, let me think again. The flow from A to B is f(t), which has a net of zero, so it doesn't contribute to the net change. The loss from A is due to g(t), and the gain in B is due to rain. So, yes, that seems correct.But let me double-check the integral of g(t):g(t) = 30 + 20 cos(πt/6)The integral over 24 hours:30*24 = 720The integral of cos(πt/6) over 24 hours:The period of cos(πt/6) is 12 hours, so over 24 hours, it's two full periods. The integral over each period is zero, so over two periods, it's still zero. Therefore, the integral is 720.So, net change in A is -720, and in B is +1920.But wait, the problem says "the net change in the volume of water in both reservoirs after 24 hours, assuming they both start full."So, starting full, reservoir A has 10,000 cubic meters, and B has 7,000.After 24 hours, A has 10,000 - 720 = 9,280B has 7,000 + 1920 = 8,920But the question asks for the net change, not the final volume. So, net change is -720 for A and +1920 for B.But let me see if I missed anything. The flow from A to B is f(t), which has a net of zero, so it doesn't affect the net change. The loss from A is g(t), which is 720, and the gain in B is 1920.Yes, that seems correct.So, summarizing:1. Total volume transferred from A to B: 12000/π ≈ 3819.72 cubic meters.2. Net change in A: -720 cubic meters.Net change in B: +1920 cubic meters.But wait, the problem says "consider only the effects of the functions f(t), g(t), and the rainwater collection system." So, I think that's all accounted for.Therefore, the answers are:1. 12000/π cubic meters.2. Net change in A: -720, net change in B: +1920.But let me write them in the required format.</think>

🧠:<think>Alright, so I have this problem about modeling the publication trends of two countries, A and B, using exponential functions. The functions are given for each country, and I need to find two things: first, the year when the number of publications in Country A equals that in Country B, and second, the total number of publications combined over the entire century from 1900 to 2000.Starting with the first part: I need to find the year ( t ) when ( P_A(t) = P_B(t) ). The functions are:[ P_A(t) = 150 e^{0.02(t-1900)} + 300 e^{-0.03(t-1950)} ][ P_B(t) = 200 e^{0.015(t-1920)} + 250 e^{-0.025(t-1970)} ]So, I need to solve the equation:[ 150 e^{0.02(t-1900)} + 300 e^{-0.03(t-1950)} = 200 e^{0.015(t-1920)} + 250 e^{-0.025(t-1970)} ]This looks a bit complicated because it's a transcendental equation involving exponentials with different bases and coefficients. I don't think I can solve this algebraically, so I might need to use numerical methods or graphing to approximate the solution.First, let me simplify the exponents a bit to make it easier to handle. Let's define a new variable ( x = t - 1900 ). Then, ( t = x + 1900 ), and the exponents can be rewritten in terms of ( x ).For ( P_A(t) ):- The first term exponent is ( 0.02(t - 1900) = 0.02x ).- The second term exponent is ( -0.03(t - 1950) = -0.03(x + 1900 - 1950) = -0.03(x - 50) = -0.03x + 1.5 ).So, ( P_A(t) = 150 e^{0.02x} + 300 e^{-0.03x + 1.5} ).Similarly, for ( P_B(t) ):- The first term exponent is ( 0.015(t - 1920) = 0.015(x + 1900 - 1920) = 0.015(x - 20) = 0.015x - 0.3 ).- The second term exponent is ( -0.025(t - 1970) = -0.025(x + 1900 - 1970) = -0.025(x - 70) = -0.025x + 1.75 ).So, ( P_B(t) = 200 e^{0.015x - 0.3} + 250 e^{-0.025x + 1.75} ).Now, substituting back into the equation:[ 150 e^{0.02x} + 300 e^{-0.03x + 1.5} = 200 e^{0.015x - 0.3} + 250 e^{-0.025x + 1.75} ]Hmm, this still looks complicated, but maybe I can factor out some constants to make it simpler.Let me rewrite each term:For ( P_A(t) ):- ( 150 e^{0.02x} )- ( 300 e^{-0.03x} e^{1.5} approx 300 times 4.4817 e^{-0.03x} approx 1344.51 e^{-0.03x} )For ( P_B(t) ):- ( 200 e^{0.015x} e^{-0.3} approx 200 times 0.7408 e^{0.015x} approx 148.16 e^{0.015x} )- ( 250 e^{-0.025x} e^{1.75} approx 250 times 5.7546 e^{-0.025x} approx 1438.65 e^{-0.025x} )So, substituting these approximations:[ 150 e^{0.02x} + 1344.51 e^{-0.03x} = 148.16 e^{0.015x} + 1438.65 e^{-0.025x} ]This is still a bit messy, but maybe I can rearrange terms:[ 150 e^{0.02x} - 148.16 e^{0.015x} = 1438.65 e^{-0.025x} - 1344.51 e^{-0.03x} ]Let me compute the left-hand side (LHS) and right-hand side (RHS) separately as functions of ( x ) and see where they intersect.Alternatively, perhaps I can define a function ( f(x) = P_A(t) - P_B(t) ) and find the root of ( f(x) = 0 ).So, ( f(x) = 150 e^{0.02x} + 300 e^{-0.03x + 1.5} - 200 e^{0.015x - 0.3} - 250 e^{-0.025x + 1.75} )I can compute ( f(x) ) for different values of ( x ) (i.e., different years) and see when it crosses zero.Given that ( x = t - 1900 ), and ( t ) ranges from 1900 to 2000, so ( x ) ranges from 0 to 100.I can try plugging in some values of ( x ) to approximate where ( f(x) = 0 ).Let me start by testing some midpoints.First, let's try ( x = 50 ) (i.e., year 1950):Compute each term:- ( 150 e^{0.02*50} = 150 e^{1} approx 150 * 2.7183 ≈ 407.745 )- ( 300 e^{-0.03*50 + 1.5} = 300 e^{-1.5 + 1.5} = 300 e^{0} = 300 * 1 = 300 )- ( 200 e^{0.015*50 - 0.3} = 200 e^{0.75 - 0.3} = 200 e^{0.45} ≈ 200 * 1.5683 ≈ 313.66 )- ( 250 e^{-0.025*50 + 1.75} = 250 e^{-1.25 + 1.75} = 250 e^{0.5} ≈ 250 * 1.6487 ≈ 412.18 )So, ( f(50) = 407.745 + 300 - 313.66 - 412.18 ≈ 707.745 - 725.84 ≈ -18.095 ). So, negative.Now, let's try ( x = 60 ) (year 1960):- ( 150 e^{0.02*60} = 150 e^{1.2} ≈ 150 * 3.3201 ≈ 498.015 )- ( 300 e^{-0.03*60 + 1.5} = 300 e^{-1.8 + 1.5} = 300 e^{-0.3} ≈ 300 * 0.7408 ≈ 222.24 )- ( 200 e^{0.015*60 - 0.3} = 200 e^{0.9 - 0.3} = 200 e^{0.6} ≈ 200 * 1.8221 ≈ 364.42 )- ( 250 e^{-0.025*60 + 1.75} = 250 e^{-1.5 + 1.75} = 250 e^{0.25} ≈ 250 * 1.2840 ≈ 321.00 )So, ( f(60) = 498.015 + 222.24 - 364.42 - 321.00 ≈ 720.255 - 685.42 ≈ 34.835 ). Positive.So, between ( x = 50 ) (f ≈ -18.095) and ( x = 60 ) (f ≈ 34.835), the function crosses zero. So, the root is somewhere between 1950 and 1960.Let me try ( x = 55 ) (year 1955):- ( 150 e^{0.02*55} = 150 e^{1.1} ≈ 150 * 3.0042 ≈ 450.63 )- ( 300 e^{-0.03*55 + 1.5} = 300 e^{-1.65 + 1.5} = 300 e^{-0.15} ≈ 300 * 0.8607 ≈ 258.21 )- ( 200 e^{0.015*55 - 0.3} = 200 e^{0.825 - 0.3} = 200 e^{0.525} ≈ 200 * 1.6918 ≈ 338.36 )- ( 250 e^{-0.025*55 + 1.75} = 250 e^{-1.375 + 1.75} = 250 e^{0.375} ≈ 250 * 1.4545 ≈ 363.63 )So, ( f(55) = 450.63 + 258.21 - 338.36 - 363.63 ≈ 708.84 - 701.99 ≈ 6.85 ). Still positive.So, between ( x = 50 ) (f ≈ -18.095) and ( x = 55 ) (f ≈ 6.85). Let's try ( x = 53 ) (year 1953):- ( 150 e^{0.02*53} = 150 e^{1.06} ≈ 150 * 2.886 ≈ 432.9 )- ( 300 e^{-0.03*53 + 1.5} = 300 e^{-1.59 + 1.5} = 300 e^{-0.09} ≈ 300 * 0.9139 ≈ 274.17 )- ( 200 e^{0.015*53 - 0.3} = 200 e^{0.795 - 0.3} = 200 e^{0.495} ≈ 200 * 1.6406 ≈ 328.12 )- ( 250 e^{-0.025*53 + 1.75} = 250 e^{-1.325 + 1.75} = 250 e^{0.425} ≈ 250 * 1.5296 ≈ 382.4 )So, ( f(53) = 432.9 + 274.17 - 328.12 - 382.4 ≈ 707.07 - 710.52 ≈ -3.45 ). Negative.So, between ( x = 53 ) (f ≈ -3.45) and ( x = 55 ) (f ≈ 6.85). Let's try ( x = 54 ) (year 1954):- ( 150 e^{0.02*54} = 150 e^{1.08} ≈ 150 * 2.944 ≈ 441.6 )- ( 300 e^{-0.03*54 + 1.5} = 300 e^{-1.62 + 1.5} = 300 e^{-0.12} ≈ 300 * 0.8869 ≈ 266.07 )- ( 200 e^{0.015*54 - 0.3} = 200 e^{0.81 - 0.3} = 200 e^{0.51} ≈ 200 * 1.665 ≈ 333 )- ( 250 e^{-0.025*54 + 1.75} = 250 e^{-1.35 + 1.75} = 250 e^{0.4} ≈ 250 * 1.4918 ≈ 372.95 )So, ( f(54) = 441.6 + 266.07 - 333 - 372.95 ≈ 707.67 - 705.95 ≈ 1.72 ). Positive.So, between ( x = 53 ) (f ≈ -3.45) and ( x = 54 ) (f ≈ 1.72). Let's try ( x = 53.5 ) (year 1953.5):But since we can't have half years, maybe approximate. Alternatively, use linear approximation between x=53 and x=54.At x=53: f=-3.45At x=54: f=1.72The change in f is 1.72 - (-3.45) = 5.17 over 1 unit of x.We need to find delta_x where f=0:delta_x = (0 - (-3.45)) / 5.17 ≈ 3.45 / 5.17 ≈ 0.667So, x ≈ 53 + 0.667 ≈ 53.667, which is approximately 1953.667, so around 1953.67.But since we can't have a fraction of a year, we can check if the crossing is between 1953 and 1954.Alternatively, perhaps use a better approximation.Let me compute f(53.5):x=53.5- ( 150 e^{0.02*53.5} = 150 e^{1.07} ≈ 150 * 2.915 ≈ 437.25 )- ( 300 e^{-0.03*53.5 + 1.5} = 300 e^{-1.605 + 1.5} = 300 e^{-0.105} ≈ 300 * 0.8997 ≈ 269.91 )- ( 200 e^{0.015*53.5 - 0.3} = 200 e^{0.8025 - 0.3} = 200 e^{0.5025} ≈ 200 * 1.653 ≈ 330.6 )- ( 250 e^{-0.025*53.5 + 1.75} = 250 e^{-1.3375 + 1.75} = 250 e^{0.4125} ≈ 250 * 1.510 ≈ 377.5 )So, ( f(53.5) = 437.25 + 269.91 - 330.6 - 377.5 ≈ 707.16 - 708.1 ≈ -0.94 ). Still negative.So, f(53.5) ≈ -0.94We need f=0 between x=53.5 and x=54.At x=53.5: f=-0.94At x=54: f=1.72Change in f: 1.72 - (-0.94) = 2.66 over 0.5 units of x.So, to find delta_x from x=53.5 where f=0:delta_x = (0 - (-0.94)) / 2.66 ≈ 0.94 / 2.66 ≈ 0.353So, x ≈ 53.5 + 0.353 ≈ 53.853, which is approximately 1953.85, so around 1953.85.So, approximately 1953.85, which is roughly 1953.85, so about 1953.85. Since we can't have a fraction of a year, we can say either 1953 or 1954, but since the crossing is closer to 1954, maybe 1954.But let's check f(53.85):Wait, maybe instead of going further, let's accept that it's approximately 1953.85, so around 1954.Alternatively, perhaps use a calculator or more precise method, but since I'm doing this manually, I'll go with approximately 1954.So, the answer to part 1 is approximately the year 1954.Moving on to part 2: Calculate the integral ( int_{1900}^{2000} T(t) , dt ) where ( T(t) = P_A(t) + P_B(t) ).So, ( T(t) = 150 e^{0.02(t-1900)} + 300 e^{-0.03(t-1950)} + 200 e^{0.015(t-1920)} + 250 e^{-0.025(t-1970)} )So, the integral is:[ int_{1900}^{2000} [150 e^{0.02(t-1900)} + 300 e^{-0.03(t-1950)} + 200 e^{0.015(t-1920)} + 250 e^{-0.025(t-1970)}] dt ]We can split this integral into four separate integrals:[ 150 int_{1900}^{2000} e^{0.02(t-1900)} dt + 300 int_{1900}^{2000} e^{-0.03(t-1950)} dt + 200 int_{1900}^{2000} e^{0.015(t-1920)} dt + 250 int_{1900}^{2000} e^{-0.025(t-1970)} dt ]Each of these integrals can be solved using the integral of ( e^{kt} ), which is ( frac{1}{k} e^{kt} ).Let me compute each integral one by one.First integral: ( 150 int_{1900}^{2000} e^{0.02(t-1900)} dt )Let me make a substitution: let ( u = t - 1900 ), so when t=1900, u=0; t=2000, u=100. Then, dt = du.So, the integral becomes:( 150 int_{0}^{100} e^{0.02u} du = 150 left[ frac{e^{0.02u}}{0.02} right]_0^{100} = 150 times frac{1}{0.02} [e^{2} - 1] = 150 times 50 [e^{2} - 1] = 7500 [e^{2} - 1] )Compute this numerically:( e^2 ≈ 7.3891 ), so ( e^2 - 1 ≈ 6.3891 )Thus, first integral ≈ 7500 * 6.3891 ≈ 7500 * 6.3891 ≈ let's compute:7500 * 6 = 45,0007500 * 0.3891 ≈ 7500 * 0.3 = 2,250; 7500 * 0.0891 ≈ 668.25So, total ≈ 45,000 + 2,250 + 668.25 ≈ 47,918.25Second integral: ( 300 int_{1900}^{2000} e^{-0.03(t-1950)} dt )Again, substitution: let ( v = t - 1950 ), so when t=1900, v= -50; t=2000, v=50.So, the integral becomes:( 300 int_{-50}^{50} e^{-0.03v} dv = 300 left[ frac{e^{-0.03v}}{-0.03} right]_{-50}^{50} = 300 times left( frac{e^{-1.5} - e^{1.5}}{-0.03} right) )Wait, let me compute step by step:The integral of ( e^{-0.03v} dv ) is ( frac{e^{-0.03v}}{-0.03} ).So, evaluating from v=-50 to v=50:( frac{e^{-0.03*50} - e^{-0.03*(-50)}}{-0.03} = frac{e^{-1.5} - e^{1.5}}{-0.03} )So, the integral becomes:( 300 times frac{e^{-1.5} - e^{1.5}}{-0.03} = 300 times frac{e^{1.5} - e^{-1.5}}{0.03} )Compute this:( e^{1.5} ≈ 4.4817 ), ( e^{-1.5} ≈ 0.2231 )So, ( e^{1.5} - e^{-1.5} ≈ 4.4817 - 0.2231 ≈ 4.2586 )Thus, integral ≈ 300 * (4.2586 / 0.03) ≈ 300 * 141.953 ≈ 42,585.9Third integral: ( 200 int_{1900}^{2000} e^{0.015(t-1920)} dt )Substitution: let ( w = t - 1920 ), so when t=1900, w= -20; t=2000, w=80.So, the integral becomes:( 200 int_{-20}^{80} e^{0.015w} dw = 200 left[ frac{e^{0.015w}}{0.015} right]_{-20}^{80} = 200 times frac{1}{0.015} [e^{1.2} - e^{-0.3}] )Compute this:( e^{1.2} ≈ 3.3201 ), ( e^{-0.3} ≈ 0.7408 )So, ( e^{1.2} - e^{-0.3} ≈ 3.3201 - 0.7408 ≈ 2.5793 )Thus, integral ≈ 200 * (2.5793 / 0.015) ≈ 200 * 171.953 ≈ 34,390.6Fourth integral: ( 250 int_{1900}^{2000} e^{-0.025(t-1970)} dt )Substitution: let ( z = t - 1970 ), so when t=1900, z= -70; t=2000, z=30.So, the integral becomes:( 250 int_{-70}^{30} e^{-0.025z} dz = 250 left[ frac{e^{-0.025z}}{-0.025} right]_{-70}^{30} = 250 times left( frac{e^{-0.75} - e^{1.75}}{-0.025} right) )Wait, let me compute step by step:The integral of ( e^{-0.025z} dz ) is ( frac{e^{-0.025z}}{-0.025} ).Evaluating from z=-70 to z=30:( frac{e^{-0.025*30} - e^{-0.025*(-70)}}{-0.025} = frac{e^{-0.75} - e^{1.75}}{-0.025} )So, the integral becomes:( 250 times frac{e^{-0.75} - e^{1.75}}{-0.025} = 250 times frac{e^{1.75} - e^{-0.75}}{0.025} )Compute this:( e^{1.75} ≈ 5.7546 ), ( e^{-0.75} ≈ 0.4724 )So, ( e^{1.75} - e^{-0.75} ≈ 5.7546 - 0.4724 ≈ 5.2822 )Thus, integral ≈ 250 * (5.2822 / 0.025) ≈ 250 * 211.288 ≈ 52,822Now, summing up all four integrals:First integral ≈ 47,918.25Second integral ≈ 42,585.9Third integral ≈ 34,390.6Fourth integral ≈ 52,822Total ≈ 47,918.25 + 42,585.9 + 34,390.6 + 52,822 ≈Let me add them step by step:47,918.25 + 42,585.9 = 90,504.1590,504.15 + 34,390.6 = 124,894.75124,894.75 + 52,822 = 177,716.75So, the total number of publications over the century is approximately 177,716.75.But since we're dealing with counts of publications, which are discrete, we can round this to the nearest whole number, so approximately 177,717.However, let me double-check the calculations because sometimes when dealing with exponentials, small errors can accumulate.First integral: 7500*(e² -1) ≈ 7500*(7.3891 -1) ≈ 7500*6.3891 ≈ 47,918.25. Correct.Second integral: 300*(e^{1.5} - e^{-1.5})/0.03 ≈ 300*(4.4817 - 0.2231)/0.03 ≈ 300*(4.2586)/0.03 ≈ 300*141.953 ≈ 42,585.9. Correct.Third integral: 200*(e^{1.2} - e^{-0.3})/0.015 ≈ 200*(3.3201 - 0.7408)/0.015 ≈ 200*(2.5793)/0.015 ≈ 200*171.953 ≈ 34,390.6. Correct.Fourth integral: 250*(e^{1.75} - e^{-0.75})/0.025 ≈ 250*(5.7546 - 0.4724)/0.025 ≈ 250*(5.2822)/0.025 ≈ 250*211.288 ≈ 52,822. Correct.So, sum is indeed approximately 177,716.75, which we can round to 177,717.Therefore, the total number of publications over the century is approximately 177,717.Final Answer1. The year when the number of publications in Country A equals that in Country B is boxed{1954}.2. The total number of publications over the century is boxed{177717}.</think>

🧠:<think>Alright, so I have this problem where Alex wants to maximize his training effectiveness. He has three types of exercises: running, strength training, and flexibility. The effectiveness is given by the function ( E(x, y, z) = 3x^2 + 2xy + y^2 + z ). He can't spend more than 10 hours a week on training, so ( x + y + z leq 10 ). First, I need to figure out how to maximize ( E(x, y, z) ) given that constraint. I remember from my calculus class that when you have a function to maximize with a constraint, you can use the method of Lagrange multipliers. So, maybe I should set that up.Let me write the Lagrangian function. It would be the effectiveness function minus a multiplier times the constraint. So,( mathcal{L}(x, y, z, lambda) = 3x^2 + 2xy + y^2 + z - lambda(x + y + z - 10) )Now, I need to take the partial derivatives of this with respect to x, y, z, and λ, and set them equal to zero.First, partial derivative with respect to x:( frac{partial mathcal{L}}{partial x} = 6x + 2y - lambda = 0 ) --> Equation 1Partial derivative with respect to y:( frac{partial mathcal{L}}{partial y} = 2x + 2y - lambda = 0 ) --> Equation 2Partial derivative with respect to z:( frac{partial mathcal{L}}{partial z} = 1 - lambda = 0 ) --> Equation 3Partial derivative with respect to λ:( frac{partial mathcal{L}}{partial lambda} = -(x + y + z - 10) = 0 ) --> Equation 4So, from Equation 3, we can solve for λ:( 1 - lambda = 0 ) --> ( lambda = 1 )Now, plug λ = 1 into Equations 1 and 2.Equation 1: 6x + 2y - 1 = 0 --> 6x + 2y = 1Equation 2: 2x + 2y - 1 = 0 --> 2x + 2y = 1Hmm, so Equations 1 and 2 are:6x + 2y = 12x + 2y = 1Let me subtract Equation 2 from Equation 1:(6x + 2y) - (2x + 2y) = 1 - 14x = 0 --> x = 0Wait, x is zero? That seems odd. If x is zero, then from Equation 2:2(0) + 2y = 1 --> 2y = 1 --> y = 0.5Then, from Equation 4: x + y + z = 10 --> 0 + 0.5 + z = 10 --> z = 9.5So, according to this, the maximum effectiveness is achieved when Alex spends 0 hours running, 0.5 hours on strength training, and 9.5 hours on flexibility. But that doesn't make much sense because he's not running at all. Maybe I made a mistake somewhere.Let me double-check the partial derivatives.Partial derivative with respect to x: 6x + 2y - λ = 0 --> Correct.Partial derivative with respect to y: 2x + 2y - λ = 0 --> Correct.Partial derivative with respect to z: 1 - λ = 0 --> Correct.Partial derivative with respect to λ: -(x + y + z - 10) = 0 --> Correct.So, the equations seem right. Then, solving them gives x = 0, y = 0.5, z = 9.5.But intuitively, if running contributes quadratically to effectiveness, maybe it's better to spend more time on running. But according to the math, it's optimal to spend zero time on running. Maybe the function is such that the cross term 2xy is not enough to compensate for the lack of x^2 term when x is increased.Wait, let me think about the effectiveness function:( E = 3x^2 + 2xy + y^2 + z )If I set x = 0, then E becomes ( y^2 + z ). If I set x to something else, say x = 1, then E would be 3(1)^2 + 2(1)y + y^2 + z = 3 + 2y + y^2 + z. But with x = 1, the total time becomes 1 + y + z = 10 --> y + z = 9. So, z = 9 - y.So, E = 3 + 2y + y^2 + (9 - y) = 3 + 2y + y^2 + 9 - y = 12 + y + y^2.Compare this to when x = 0: E = y^2 + z = y^2 + (10 - y) = y^2 - y + 10.So, for x = 0, E = y^2 - y + 10.For x = 1, E = y^2 + y + 12.Which one is larger? Let's subtract the two:(y^2 + y + 12) - (y^2 - y + 10) = 2y + 2.So, if 2y + 2 > 0, which is always true since y is non-negative, then E is larger when x = 1 than when x = 0.But according to the Lagrangian method, x = 0 is the maximum. That seems contradictory.Wait, maybe the maximum occurs at the boundary of the feasible region. Because in the Lagrangian method, sometimes the maximum is on the boundary rather than the interior.So, perhaps the maximum is at x = 0, but when I tested x = 1, E was higher. Hmm, maybe I need to check if the critical point found by Lagrangian is a minimum or maximum.Alternatively, maybe the function is concave or convex. Let me check the Hessian matrix.The Hessian matrix for the function E is:Second partial derivatives:E_xx = 6E_xy = 2E_xz = 0E_yx = 2E_yy = 2E_yz = 0E_zx = 0E_zy = 0E_zz = 0So, the Hessian is:[6, 2, 0][2, 2, 0][0, 0, 0]This matrix is not positive definite because the last eigenvalue is zero. So, the function is not strictly concave, which means the critical point found might be a saddle point or a minimum.Therefore, the maximum might occur on the boundary of the feasible region.So, the feasible region is x + y + z <= 10, with x, y, z >= 0.So, to find the maximum, I might need to check the boundaries.Let me consider different cases.Case 1: z = 0. Then, x + y = 10.So, E = 3x^2 + 2xy + y^2.We can express y = 10 - x, so E = 3x^2 + 2x(10 - x) + (10 - x)^2.Simplify:3x^2 + 20x - 2x^2 + 100 - 20x + x^2Combine like terms:(3x^2 - 2x^2 + x^2) + (20x - 20x) + 100= 2x^2 + 0 + 100So, E = 2x^2 + 100.To maximize this, since it's a quadratic in x opening upwards, the maximum occurs at the endpoints.So, when x = 0, E = 100.When x = 10, E = 2(100) + 100 = 300.So, in this case, maximum E is 300 when x = 10, y = 0, z = 0.But wait, earlier with x = 0, y = 0.5, z = 9.5, E was:E = 0 + 0 + (0.5)^2 + 9.5 = 0.25 + 9.5 = 9.75.Which is way less than 300. So, clearly, the maximum is at x = 10, y = 0, z = 0.But wait, let's check another case.Case 2: y = 0.Then, x + z = 10.E = 3x^2 + 0 + 0 + z = 3x^2 + (10 - x).So, E = 3x^2 - x + 10.This is a quadratic in x, opening upwards. So, the minimum is at the vertex, but since it's opening upwards, the maximum occurs at the endpoints.At x = 0, E = 10.At x = 10, E = 300 - 10 + 10 = 300.So, again, maximum at x = 10, y = 0, z = 0.Case 3: x = 0.Then, y + z = 10.E = 0 + 0 + y^2 + z = y^2 + (10 - y).So, E = y^2 - y + 10.This is a quadratic in y, opening upwards. The minimum is at y = 0.5, but the maximum occurs at the endpoints.At y = 0, E = 0 + 10 = 10.At y = 10, E = 100 - 10 + 10 = 100.So, maximum at y = 10, x = 0, z = 0.But earlier, when I set x = 10, y = 0, z = 0, E was 300, which is higher.So, seems like the maximum is at x = 10, y = 0, z = 0.But let me check another case where two variables are at their maximum.Wait, but if I set x = 10, y = 0, z = 0, that's 10 hours running, 0 strength, 0 flexibility.But maybe there's a combination where E is higher.Wait, let me think about the effectiveness function again.E = 3x^2 + 2xy + y^2 + z.If I set z = 10 - x - y, then E becomes:3x^2 + 2xy + y^2 + (10 - x - y).So, E = 3x^2 + 2xy + y^2 - x - y + 10.Now, to find the critical points, take partial derivatives with respect to x and y.Partial derivative with respect to x:6x + 2y - 1 = 0 --> 6x + 2y = 1 --> 3x + y = 0.5 --> Equation APartial derivative with respect to y:2x + 2y - 1 = 0 --> 2x + 2y = 1 --> x + y = 0.5 --> Equation BNow, subtract Equation B from Equation A:(3x + y) - (x + y) = 0.5 - 0.52x = 0 --> x = 0Then, from Equation B: 0 + y = 0.5 --> y = 0.5So, critical point at x = 0, y = 0.5, z = 9.5.But earlier, we saw that this gives E = 0.25 + 9.5 = 9.75, which is much less than 300.So, the critical point is a minimum, not a maximum.Therefore, the maximum must occur on the boundary.So, as we saw earlier, the maximum occurs when x = 10, y = 0, z = 0, giving E = 300.But wait, let me check another case where z is not zero.Suppose z is something else, but x and y are positive.Wait, but if I set z = 0, I get E = 3x^2 + 2xy + y^2, which we saw can go up to 300.If I set z positive, then E = 3x^2 + 2xy + y^2 + z.But since z is positive, E will be higher than when z is zero. Wait, no, because z is added, so higher z would mean higher E, but z is constrained by x + y + z =10.Wait, but if I set z = 10 - x - y, then E = 3x^2 + 2xy + y^2 + (10 - x - y).So, E is a function of x and y, and we saw that the critical point is a minimum.Therefore, to maximize E, we need to set z as small as possible, which is z = 0, and then maximize E with z = 0.Which, as we saw, occurs at x =10, y=0, z=0.Alternatively, if we set z to be something else, say z =1, then x + y =9.E = 3x^2 + 2xy + y^2 +1.But since E is 3x^2 + 2xy + y^2 +1, which is similar to the case when z=0, just shifted up by 1. So, the maximum would still be at x=9, y=0, z=1, giving E=3(81) +0 +0 +1=243 +1=244, which is less than 300.Similarly, if z=2, x+y=8, E=3x^2 +2xy + y^2 +2.Maximum at x=8, y=0, E=3(64) +0 +0 +2=192 +2=194 <300.So, indeed, the maximum occurs when z=0, x=10, y=0.Therefore, the optimal allocation is x=10, y=0, z=0.But wait, that seems counterintuitive because the problem mentions that Alex does running, strength training, and flexibility exercises. If he spends all his time running, he's neglecting the other two, which might be important for overall performance.But according to the effectiveness function given, which is ( E = 3x^2 + 2xy + y^2 + z ), the coefficients for x^2 is higher than y^2, and there's a cross term 2xy, which might suggest that combining x and y could be beneficial.But when we tried to include y, the effectiveness didn't go up as much as when we set y=0 and x=10.Wait, let me compute E when x=10, y=0, z=0: E=3(100) +0 +0 +0=300.If I set x=9, y=1, z=0: E=3(81) +2(9)(1) +1 +0=243 +18 +1=262 <300.x=8, y=2, z=0: E=3(64) +2(8)(2) +4 +0=192 +32 +4=228 <300.x=7, y=3, z=0: E=3(49) +2(7)(3) +9 +0=147 +42 +9=198 <300.So, indeed, the effectiveness decreases as we increase y and decrease x.Therefore, the maximum effectiveness is achieved when x=10, y=0, z=0.But wait, the problem says "each week, he can allocate up to 10 hours for training". So, he doesn't have to use all 10 hours. Maybe he could use less?But if he uses less, say 9 hours, then E would be less than 300.So, to maximize E, he should use all 10 hours.Therefore, the optimal allocation is x=10, y=0, z=0.But that seems strange because the problem mentions all three types of exercises. Maybe the function is not realistic, or perhaps I made a mistake in interpreting the problem.Wait, let me check the function again: ( E(x, y, z) = 3x^2 + 2xy + y^2 + z ).So, z only adds linearly, while x and y have quadratic terms. So, z is less effective per hour than x or y.But even so, if I set z=0, I can get a higher E by increasing x.Alternatively, if I set z=10, then E=0 +0 +0 +10=10, which is much less than 300.So, indeed, the maximum is at x=10, y=0, z=0.Therefore, the answer to part 1 is x=10, y=0, z=0.Now, moving on to part 2.Alex wants to maintain a ratio of 3:2:1 for running to strength training to flexibility.So, the ratio is x:y:z = 3:2:1.That means x = 3k, y=2k, z=k, for some k>0.Since x + y + z <=10, we have 3k + 2k +k =6k <=10 --> k <=10/6≈1.6667.So, k=10/6=5/3≈1.6667.Therefore, x=3*(5/3)=5, y=2*(5/3)=10/3≈3.333, z=5/3≈1.6667.But wait, does this allocation maximize E?Wait, in part 2, it's not necessarily the same as part 1. It's a different constraint.So, in part 2, Alex wants to maintain the ratio 3:2:1, so x=3k, y=2k, z=k.We need to find k such that x + y + z =10, so 6k=10 --> k=10/6=5/3.Therefore, x=5, y=10/3, z=5/3.But does this allocation maximize E under the ratio constraint?Wait, the problem says "using this ratio, determine the number of hours Alex should spend on each type of exercise to maximize the effectiveness function E(x, y, z)."So, it's not about maximizing E without constraints, but under the ratio constraint.So, we can express E in terms of k.E = 3x^2 + 2xy + y^2 + z.Substitute x=3k, y=2k, z=k:E = 3*(9k^2) + 2*(3k)*(2k) + (4k^2) + kSimplify:=27k^2 + 12k^2 +4k^2 +k=43k^2 +kNow, we need to maximize E with respect to k, given that 6k <=10 --> k <=5/3.But E is a quadratic in k, opening upwards (since coefficient of k^2 is positive). Therefore, E is minimized at the vertex and maximized at the endpoints.So, the maximum occurs at k=5/3.Therefore, the allocation is x=5, y=10/3, z=5/3.So, in part 2, the optimal allocation is x=5, y≈3.333, z≈1.6667.But let me compute E at this point:E=43*(25/9) +5/3≈43*(2.7778)+1.6667≈119.444 +1.6667≈121.111.Compare this to the maximum in part 1, which was 300.So, under the ratio constraint, the maximum E is about 121.11, which is much less than 300.Therefore, the answer to part 2 is x=5, y=10/3, z=5/3.But let me write it as fractions:x=5, y=10/3, z=5/3.Alternatively, in decimal form, x=5, y≈3.333, z≈1.6667.But since the problem might prefer fractions, I'll stick with x=5, y=10/3, z=5/3.So, summarizing:1. Without any ratio constraint, the maximum effectiveness is achieved when x=10, y=0, z=0.2. With the ratio constraint 3:2:1, the maximum effectiveness is achieved when x=5, y=10/3, z=5/3.But wait, in part 1, the maximum was at x=10, y=0, z=0, but in part 2, the ratio constraint forces a different allocation.So, the answers are:1. x=10, y=0, z=0.2. x=5, y=10/3, z=5/3.I think that's it.</think>

🧠:<think>Alright, so I have this problem about the Katy High School football team's performance statistics. There are two parts to it. Let me try to tackle them one by one.First, the team played 12 games, and in each game, the number of touchdowns (T) scored follows a Poisson distribution with a mean of 3.5 touchdowns per game. I need to find the probability that they scored exactly 5 touchdowns in a given game.Okay, Poisson distribution. I remember that the Poisson probability formula is:P(T = k) = (λ^k * e^(-λ)) / k!Where λ is the average rate (mean), which is 3.5 here, and k is the number of occurrences, which is 5 in this case.So plugging in the numbers:P(T = 5) = (3.5^5 * e^(-3.5)) / 5!Let me compute this step by step.First, calculate 3.5^5. Let me see, 3.5 squared is 12.25, then 3.5 cubed is 3.5 * 12.25 = 42.875. Then 3.5^4 is 3.5 * 42.875 = 150.0625. Then 3.5^5 is 3.5 * 150.0625 = 525.21875.Next, e^(-3.5). I know that e is approximately 2.71828. So e^(-3.5) is 1 / e^(3.5). Let me compute e^3.5. e^3 is about 20.0855, and e^0.5 is about 1.6487. So e^3.5 is roughly 20.0855 * 1.6487 ≈ 33.115. Therefore, e^(-3.5) ≈ 1 / 33.115 ≈ 0.0302.Now, 5! is 5 factorial, which is 5*4*3*2*1 = 120.Putting it all together:P(T = 5) = (525.21875 * 0.0302) / 120First, multiply 525.21875 by 0.0302. Let me do that:525.21875 * 0.03 = 15.7565625525.21875 * 0.0002 = 0.10504375Adding them together: 15.7565625 + 0.10504375 ≈ 15.86160625Now divide that by 120:15.86160625 / 120 ≈ 0.13218So approximately 0.1322, or 13.22%.Wait, let me double-check my calculations because sometimes when I compute e^(-3.5), I might have approximated too much.Alternatively, I can use a calculator for e^(-3.5). Let me recall that e^(-3) is about 0.0498, and e^(-0.5) is about 0.6065. So e^(-3.5) is e^(-3) * e^(-0.5) ≈ 0.0498 * 0.6065 ≈ 0.03016. So that's consistent with my earlier approximation.So 525.21875 * 0.03016 ≈ Let me compute this more accurately.525.21875 * 0.03 = 15.7565625525.21875 * 0.00016 = approximately 525.21875 * 0.0001 = 0.052521875, and 525.21875 * 0.00006 = 0.031513125. So total is 0.052521875 + 0.031513125 ≈ 0.084035.So total is 15.7565625 + 0.084035 ≈ 15.8406.Divide by 120: 15.8406 / 120 ≈ 0.132005.So approximately 0.132, or 13.2%.Hmm, that seems a bit high? Wait, the mean is 3.5, so the probability of 5 touchdowns is 13.2%? Let me check with another method.Alternatively, maybe using a calculator or Poisson table. But since I don't have that, I can compute it step by step.Alternatively, use the formula:P(5) = (3.5^5 * e^-3.5) / 5!We have 3.5^5 = 525.21875e^-3.5 ≈ 0.03016Multiply them: 525.21875 * 0.03016 ≈ 15.8406Divide by 120: 15.8406 / 120 ≈ 0.132005So yes, approximately 13.2%.Alternatively, maybe I can compute it more accurately.Wait, 3.5^5: 3.5^2 = 12.25, 3.5^3 = 42.875, 3.5^4 = 150.0625, 3.5^5 = 525.21875. Correct.e^-3.5: Let me compute it more accurately.We know that ln(2) ≈ 0.6931, ln(3) ≈ 1.0986, ln(5) ≈ 1.6094, etc.But maybe using Taylor series for e^x.e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...But since x is negative here, x = -3.5.So e^-3.5 = 1 - 3.5 + (3.5)^2 / 2 - (3.5)^3 / 6 + (3.5)^4 / 24 - (3.5)^5 / 120 + ...Compute up to a certain term to get an approximation.Let me compute up to the 5th term.Term 0: 1Term 1: -3.5Term 2: (12.25)/2 = 6.125Term 3: (-42.875)/6 ≈ -7.1458333Term 4: (150.0625)/24 ≈ 6.252604167Term 5: (-525.21875)/120 ≈ -4.376822917Adding up:1 - 3.5 = -2.5-2.5 + 6.125 = 3.6253.625 - 7.1458333 ≈ -3.5208333-3.5208333 + 6.252604167 ≈ 2.7317708672.731770867 - 4.376822917 ≈ -1.64505205Hmm, that's oscillating and not converging yet. Maybe I need more terms.Term 6: (3.5)^6 / 7203.5^6 = 3.5 * 525.21875 ≈ 1838.2656251838.265625 / 720 ≈ 2.553146736So term 6: +2.553146736Adding to previous total: -1.64505205 + 2.553146736 ≈ 0.908094686Term 7: -(3.5)^7 / 50403.5^7 = 3.5 * 1838.265625 ≈ 6433.92968756433.9296875 / 5040 ≈ 1.2765535So term 7: -1.2765535Adding: 0.908094686 - 1.2765535 ≈ -0.368458814Term 8: (3.5)^8 / 403203.5^8 = 3.5 * 6433.9296875 ≈ 22518.7539062522518.75390625 / 40320 ≈ 0.5584Term 8: +0.5584Adding: -0.368458814 + 0.5584 ≈ 0.189941186Term 9: -(3.5)^9 / 3628803.5^9 = 3.5 * 22518.75390625 ≈ 78815.63867187578815.638671875 / 362880 ≈ 0.2172Term 9: -0.2172Adding: 0.189941186 - 0.2172 ≈ -0.027258814Term 10: (3.5)^10 / 36288003.5^10 = 3.5 * 78815.638671875 ≈ 275854.7353515625275854.7353515625 / 3628800 ≈ 0.076Term 10: +0.076Adding: -0.027258814 + 0.076 ≈ 0.048741186Term 11: -(3.5)^11 / 399168003.5^11 = 3.5 * 275854.7353515625 ≈ 965491.5737303125965491.5737303125 / 39916800 ≈ 0.02417Term 11: -0.02417Adding: 0.048741186 - 0.02417 ≈ 0.024571186Term 12: (3.5)^12 / 4790016003.5^12 = 3.5 * 965491.5737303125 ≈ 3380220.5070560943380220.507056094 / 479001600 ≈ 0.007036Term 12: +0.007036Adding: 0.024571186 + 0.007036 ≈ 0.031607186Term 13: -(3.5)^13 / 62270208003.5^13 = 3.5 * 3380220.507056094 ≈ 11830771.7746963311830771.77469633 / 6227020800 ≈ 0.0019Term 13: -0.0019Adding: 0.031607186 - 0.0019 ≈ 0.029707186Term 14: (3.5)^14 / 871782912003.5^14 = 3.5 * 11830771.77469633 ≈ 41407699.1114371641407699.11143716 / 87178291200 ≈ 0.000475Term 14: +0.000475Adding: 0.029707186 + 0.000475 ≈ 0.030182186Term 15: -(3.5)^15 / 13076743680003.5^15 = 3.5 * 41407699.11143716 ≈ 144926946.89003144926946.89003 / 1307674368000 ≈ 0.0001108Term 15: -0.0001108Adding: 0.030182186 - 0.0001108 ≈ 0.030071386So after 15 terms, we get approximately 0.03007, which is close to our initial approximation of 0.03016. So that seems consistent.Therefore, e^-3.5 ≈ 0.03007.So going back, 3.5^5 = 525.21875, e^-3.5 ≈ 0.03007.Multiply them: 525.21875 * 0.03007 ≈ Let me compute 525.21875 * 0.03 = 15.7565625525.21875 * 0.00007 ≈ 0.0367653125So total ≈ 15.7565625 + 0.0367653125 ≈ 15.7933278125Divide by 5! = 120:15.7933278125 / 120 ≈ 0.1316110651So approximately 0.1316, or 13.16%.So rounding to four decimal places, 0.1316, which is about 13.16%.So the probability is approximately 13.16%.Wait, but earlier I had 0.132, which is 13.2%. So that's consistent.Therefore, the probability that the team scored exactly 5 touchdowns in a given game is approximately 13.2%.Okay, that's part 1.Now, part 2: The commentator wants to understand the relationship between total yards gained (Y) and touchdowns scored (T). Yards gained in a game follow a normal distribution with a mean of 350 yards and a standard deviation of 50 yards. Assuming T and Y are independent, compute the expected total yards gained over the season.So the team played 12 games. Each game, Y is normally distributed with mean 350 and standard deviation 50. Since Y and T are independent, but we are only asked about the expected total yards over the season.Wait, but the relationship between Y and T is not directly needed here because we are just asked for the expected total yards, regardless of touchdowns.Since expectation is linear, the expected total yards over 12 games is just 12 times the expected yards per game.So E[Total Y] = 12 * E[Y per game] = 12 * 350 = 4200 yards.Wait, that seems straightforward. But let me think again.Is there any trick here? The problem mentions that Y and T are independent, but since we are only asked about the expectation of Y, and expectation is linear regardless of independence, so even if they were dependent, the expectation would still be 12 * 350.But since they are independent, it's just reinforcing that the variance might be additive, but since we are only asked for expectation, it doesn't matter.Therefore, the expected total yards gained over the season is 4200 yards.Wait, but just to make sure, is there any other interpretation? Like, maybe the total yards is related to touchdowns? But the problem says to compute the expected total yards gained over the season, assuming Y and T are independent. So I think my initial thought is correct.Alternatively, if they wanted the covariance or something else, but no, it's just expectation.So yes, 12 games, each with mean 350 yards, so total expectation is 12*350=4200.Therefore, the expected total yards gained over the season is 4200 yards.So summarizing:1. Probability of exactly 5 touchdowns in a game: approximately 13.2%.2. Expected total yards over 12 games: 4200 yards.Final Answer1. The probability of scoring exactly 5 touchdowns in a game is boxed{0.132}.2. The expected total yards gained over the season is boxed{4200} yards.</think>