🧠:<think>Okay, so I have this problem where I need to analyze the impact of comedy game shows on my live performance ticket sales. I've been given two functions: one for ticket sales, T(t), and another for viewership, G(t). There are two sub-problems to solve here. Let me take them one by one.Starting with Sub-problem 1: Determine the rate of change in ticket sales for my live performances at t = 3 years. Hmm, rate of change usually means I need to find the derivative of the function T(t) with respect to t. That makes sense because the derivative gives the instantaneous rate of change at a particular point in time.So, T(t) is given as 2t² + 3t + 10. To find the rate of change, I need to compute T'(t). Let me recall how to take derivatives. The derivative of t² is 2t, so multiplying by the coefficient 2, that becomes 4t. Then, the derivative of 3t is just 3. The derivative of a constant, like 10, is zero. So putting it all together, T'(t) = 4t + 3.Now, I need to evaluate this derivative at t = 3. Plugging in 3 for t, we get T'(3) = 4*3 + 3. Let me calculate that: 4*3 is 12, plus 3 is 15. So, the rate of change in ticket sales at t = 3 is 15 thousand tickets per year. That means, at the third year, my ticket sales are increasing at a rate of 15,000 tickets annually. That seems pretty good!Wait, let me double-check my derivative. The original function is quadratic, so the derivative should be linear, which it is. 2t² becomes 4t, 3t becomes 3, and the constant disappears. Yeah, that seems correct. So, T'(3) = 15 is correct. Okay, Sub-problem 1 seems done.Moving on to Sub-problem 2: Find the total viewership of the comedy game show over the 5-year period by evaluating the definite integral of G(t) from t = 0 to t = 5. Alright, so G(t) is given as 5e^(-0.4t) + 2. I need to integrate this function from 0 to 5 and sum up the viewership over those five years.First, let me write down the integral: ∫₀⁵ [5e^(-0.4t) + 2] dt. To solve this, I can split the integral into two parts: the integral of 5e^(-0.4t) dt plus the integral of 2 dt.Starting with the first part: ∫5e^(-0.4t) dt. The integral of e^(kt) dt is (1/k)e^(kt) + C, right? So here, k is -0.4. Therefore, the integral should be 5 * [1/(-0.4)] e^(-0.4t) + C. Simplifying that, 5 divided by (-0.4) is -12.5. So, the integral becomes -12.5 e^(-0.4t).Now, the second part: ∫2 dt is straightforward. The integral of 2 with respect to t is 2t + C.Putting it all together, the indefinite integral is -12.5 e^(-0.4t) + 2t + C. Now, I need to evaluate this from t = 0 to t = 5.So, let's compute the definite integral:First, evaluate at t = 5:-12.5 e^(-0.4*5) + 2*5Simplify:-12.5 e^(-2) + 10Then, evaluate at t = 0:-12.5 e^(0) + 2*0Simplify:-12.5 * 1 + 0 = -12.5Subtracting the lower limit from the upper limit:[-12.5 e^(-2) + 10] - [-12.5] = -12.5 e^(-2) + 10 + 12.5Combine like terms:(-12.5 e^(-2)) + (10 + 12.5) = -12.5 e^(-2) + 22.5Now, let me compute the numerical value. First, e^(-2) is approximately 0.1353. So, multiplying that by -12.5:-12.5 * 0.1353 ≈ -1.69125Then, adding 22.5:-1.69125 + 22.5 ≈ 20.80875So, the total viewership over the 5-year period is approximately 20.80875 million. Since the question asks for the total viewership, I can round this to a reasonable number of decimal places. Maybe two decimal places: 20.81 million.Wait, let me verify my calculations step by step to make sure I didn't make any errors.First, the integral of 5e^(-0.4t) is indeed -12.5 e^(-0.4t). Correct. The integral of 2 is 2t. Correct.Evaluating at t = 5: -12.5 e^(-2) + 10. Correct. e^(-2) is about 0.1353, so -12.5 * 0.1353 ≈ -1.69125. Then, adding 10 gives approximately 8.30875.Wait, hold on. Wait, no. Wait, when I evaluated at t=5, I had -12.5 e^(-2) + 10, which is approximately -1.69125 + 10 = 8.30875.Then, evaluating at t=0: -12.5 e^(0) + 0 = -12.5.Subtracting: 8.30875 - (-12.5) = 8.30875 + 12.5 = 20.80875. Oh, okay, so that part is correct. So, 20.80875 million viewers over 5 years.Expressed as 20.81 million when rounded to two decimal places. Alternatively, if we want to keep it exact, we can write it in terms of e, but since the question says to evaluate the definite integral, I think a numerical approximation is acceptable here.Alternatively, maybe I can express it as 22.5 - 12.5 e^(-2). Let me compute that:22.5 - 12.5 * e^(-2). Since e^(-2) is approximately 0.1353, so 12.5 * 0.1353 ≈ 1.69125. Then, 22.5 - 1.69125 ≈ 20.80875. Yep, same result.So, I think that's correct. So, the total viewership is approximately 20.81 million over the five years.Wait, just to make sure, let me think about the integral again. The function G(t) is 5e^(-0.4t) + 2. So, integrating from 0 to 5, we have the integral of 5e^(-0.4t) dt + integral of 2 dt.Yes, that's correct. So, the first part is -12.5 e^(-0.4t) evaluated from 0 to 5, which is -12.5 [e^(-2) - 1], and the second part is 2t evaluated from 0 to 5, which is 10 - 0 = 10.So, putting it together: -12.5 [e^(-2) - 1] + 10.Compute that:-12.5 e^(-2) + 12.5 + 10 = (-12.5 e^(-2)) + 22.5.Which is the same as before. So, 22.5 - 12.5 e^(-2) ≈ 22.5 - 1.69125 ≈ 20.80875. So, that's consistent.Therefore, I can be confident that the total viewership over the five years is approximately 20.81 million.So, summarizing:Sub-problem 1: The rate of change in ticket sales at t=3 is 15 thousand tickets per year.Sub-problem 2: The total viewership over 5 years is approximately 20.81 million.I think that's all. I don't see any mistakes in my calculations, so I feel confident with these answers.Final AnswerSub-problem 1: The rate of change in ticket sales at ( t = 3 ) years is boxed{15} thousand tickets per year.Sub-problem 2: The total viewership over the 5-year period is boxed{20.81} million.</think>

🧠:<think>First, I need to calculate the total monthly expenses and savings by adding up all the given amounts: rent, groceries, utilities, childcare, and emergency savings.Next, I'll subtract the total expenses and savings from the monthly income to determine the disposable income. This will show how much money is left after all obligations are met.To find the percentage of income saved each month, I'll divide the total savings by the monthly income and then multiply by 100 to get the percentage.For the college fund, I'll use the future value of an ordinary annuity formula. This formula accounts for the monthly contributions, the annual interest rate, and the number of years the money will be invested. By plugging in the numbers, I can calculate the total amount that will be in the college fund after 10 years.</think>

🧠:<think>Okay, so I have this problem where a defense attorney is trying to allocate resources effectively for a court case. There are two main parts to this problem: one about the complexity of the case and another about the probability of the jury deciding in favor of the client. Let me try to tackle each part step by step.Starting with the first part: Case Complexity and Time Allocation. The attorney has modeled the complexity of the case with the function ( C(t) = 10e^{-0.1t} + 2sin(t) ), where ( t ) is the number of hours spent preparing. The goal is to find the critical points of ( C(t) ) within the interval ( t in [0, 40] ). Critical points are where the derivative is zero or undefined, right? So, I need to find ( C'(t) ) and solve for ( t ) when ( C'(t) = 0 ).First, let me write down the function again:( C(t) = 10e^{-0.1t} + 2sin(t) )To find the critical points, I need to compute the derivative ( C'(t) ). Let's do that term by term.The derivative of ( 10e^{-0.1t} ) with respect to ( t ) is ( 10 times (-0.1)e^{-0.1t} = -e^{-0.1t} ).The derivative of ( 2sin(t) ) with respect to ( t ) is ( 2cos(t) ).So, putting it together, the derivative ( C'(t) ) is:( C'(t) = -e^{-0.1t} + 2cos(t) )Now, to find the critical points, we set ( C'(t) = 0 ):( -e^{-0.1t} + 2cos(t) = 0 )Which simplifies to:( 2cos(t) = e^{-0.1t} )Or:( cos(t) = frac{e^{-0.1t}}{2} )Hmm, this equation involves both an exponential function and a trigonometric function. I don't think there's an algebraic solution for this, so I might need to solve it numerically or graphically. Maybe I can use some numerical methods or graph both sides to find the points where they intersect.Let me think about the behavior of both sides. The left side is ( cos(t) ), which oscillates between -1 and 1 with a period of ( 2pi ) (approximately 6.28). The right side is ( frac{e^{-0.1t}}{2} ), which is a decaying exponential starting at ( frac{1}{2} ) when ( t = 0 ) and approaching zero as ( t ) increases.So, the right side is always positive and decreasing, while the left side oscillates between -1 and 1. Therefore, the equation ( cos(t) = frac{e^{-0.1t}}{2} ) will have solutions only where ( cos(t) ) is positive, which is in the intervals ( (-pi/2 + 2pi k, pi/2 + 2pi k) ) for integer ( k ).Given that ( t ) is between 0 and 40, let's see how many periods of cosine we have. Since the period is about 6.28, 40 divided by 6.28 is roughly 6.37. So, there are about 6 full periods in the interval [0, 40].But since ( frac{e^{-0.1t}}{2} ) is decreasing, the number of solutions might decrease as ( t ) increases because the right side becomes smaller. Let me try to estimate the number of solutions.At ( t = 0 ): ( cos(0) = 1 ), and ( frac{e^{0}}{2} = 0.5 ). So, 1 > 0.5, so the left side is above the right side.At ( t = pi/2 approx 1.57 ): ( cos(pi/2) = 0 ), and ( frac{e^{-0.1 times 1.57}}{2} approx frac{e^{-0.157}}{2} approx frac{0.855}{2} approx 0.4275 ). So, left side is 0, right side is ~0.4275. So, the left side is below the right side here.Therefore, between t=0 and t=1.57, the two functions cross somewhere. So, that's one critical point.Similarly, in the next interval where cosine is positive, which is from ( t approx 4.71 ) (which is ( 3pi/2 )) to ( t approx 7.85 ) (which is ( 5pi/2 )), but wait, actually, cosine is positive in the first and fourth quadrants, so between ( -pi/2 + 2pi k ) and ( pi/2 + 2pi k ). So, for k=0, it's from -1.57 to 1.57, but since t starts at 0, it's from 0 to 1.57. For k=1, it's from 4.71 to 7.85, for k=2, from 10.99 to 14.14, and so on.Wait, actually, each positive interval is ( 2pi k - pi/2 ) to ( 2pi k + pi/2 ). So, for k=0: -1.57 to 1.57, but since t starts at 0, it's 0 to 1.57.For k=1: 4.71 to 7.85.For k=2: 10.99 to 14.14.k=3: 17.27 to 20.42.k=4: 23.56 to 26.72.k=5: 29.84 to 32.99.k=6: 36.12 to 39.27.So, in each of these intervals, the cosine function is positive, so potentially, each could have a solution where ( cos(t) = frac{e^{-0.1t}}{2} ).But as t increases, the right side ( frac{e^{-0.1t}}{2} ) decreases. So, in the first interval, t=0 to 1.57, the right side starts at 0.5 and decreases to ~0.4275. The left side, cosine, starts at 1 and decreases to 0. So, they cross once in this interval.In the next interval, t=4.71 to 7.85, the right side is ( frac{e^{-0.1 times 4.71}}{2} approx frac{e^{-0.471}}{2} approx frac{0.624}{2} = 0.312 ), and at t=7.85, it's ( frac{e^{-0.785}}{2} approx frac{0.456}{2} = 0.228 ). The left side, cosine, starts at 0 and goes up to 1 at t=6.28 (which is 2π), then back to 0 at t=7.85. So, the right side is decreasing from ~0.312 to ~0.228, while the left side goes from 0 up to 1 and back to 0. So, they might cross twice in this interval? Wait, no. Because the right side is a straight decay, while the left side goes up and then down. So, maybe they cross once on the way up and once on the way down? Or maybe not, because the right side is always decreasing.Wait, actually, at t=4.71, cosine is 0, and the right side is ~0.312. So, cosine is 0, right side is 0.312. Then, as t increases, cosine increases to 1 at t=6.28, while the right side decreases to ~0.228. So, at t=6.28, cosine is 1, right side is ~0.228. So, cosine is above the right side at t=6.28. Therefore, somewhere between t=4.71 and t=6.28, cosine goes from 0 to 1, crossing the right side which is decreasing from 0.312 to 0.228. So, they must cross once in this interval.Similarly, as t increases beyond 6.28, cosine starts decreasing from 1 back to 0 at t=7.85, while the right side continues to decrease. So, at t=7.85, cosine is 0, right side is ~0.228. So, cosine is decreasing from 1 to 0, while the right side is decreasing from ~0.228 to ~0.228? Wait, no, at t=7.85, it's still decreasing. Wait, actually, the right side is always decreasing, but the left side is first increasing then decreasing.So, in the interval t=4.71 to 7.85, we have two crossings: one when cosine is increasing and one when it's decreasing? Or maybe only one crossing because the right side is always decreasing.Wait, let's think about it. At t=4.71, cosine is 0, right side is ~0.312. So, right side is above cosine. As t increases, cosine increases to 1, while right side decreases to ~0.228. So, at some point, cosine will cross the right side from below. Then, as t continues to increase, cosine starts decreasing back to 0, while the right side continues to decrease. So, cosine is now decreasing from 1 to 0, and the right side is decreasing from ~0.228 to ~0.228 (wait, no, it's still decreasing, right? At t=7.85, it's ~0.228, but at t=7.85, it's actually ( frac{e^{-0.785}}{2} approx 0.228 ). So, the right side is decreasing throughout. So, when cosine is decreasing from 1 to 0, it might cross the right side again if the right side is still above 0. But since the right side is decreasing, it's possible that after the first crossing, the right side is still above the decreasing cosine? Hmm, maybe not. Let me plug in t=6.28: cosine is 1, right side is ~0.228. So, cosine is way above. At t=7.85, cosine is 0, right side is ~0.228. So, cosine is below. So, since cosine is decreasing from 1 to 0 and the right side is decreasing from ~0.228 to ~0.228, wait, no, the right side is still decreasing. Wait, actually, at t=7.85, it's ~0.228, but at t=7.85 - let's compute it more accurately.Wait, ( t = 7.85 ) is approximately ( 2.5pi ). So, ( e^{-0.1 times 7.85} = e^{-0.785} approx 0.456 ). Divided by 2 is ~0.228. So, at t=7.85, right side is ~0.228, and cosine is 0. So, cosine is below the right side at t=7.85.So, in the interval t=6.28 to t=7.85, cosine goes from 1 to 0, while the right side goes from ~0.228 to ~0.228? Wait, no, it's still decreasing. Wait, no, t=6.28 is 2π, so t=6.28 to t=7.85 is from 2π to 2.5π. So, the right side at t=6.28 is ( e^{-0.628}/2 approx e^{-0.628} approx 0.533, divided by 2 is ~0.266. At t=7.85, it's ~0.228. So, it's decreasing from ~0.266 to ~0.228.So, cosine starts at 1, goes down to 0, while the right side is decreasing from ~0.266 to ~0.228. So, cosine starts above the right side (1 > 0.266) and ends below (0 < 0.228). Therefore, they must cross exactly once in this interval as cosine decreases through the right side.Wait, but cosine is decreasing from 1 to 0, and the right side is also decreasing, but starting at 0.266 and going to 0.228. So, the right side is always below 1 and above 0.228. So, cosine starts above the right side and ends below. So, they must cross exactly once in this interval.Therefore, in the interval t=4.71 to t=7.85, we have two critical points: one when cosine is increasing and crosses the right side, and another when cosine is decreasing and crosses the right side.Wait, but hold on, at t=4.71, cosine is 0, right side is ~0.312. So, the right side is above cosine. As t increases, cosine increases to 1, while the right side decreases to ~0.266. So, cosine will cross the right side once when increasing. Then, as t continues to increase beyond 6.28, cosine starts decreasing, and the right side continues to decrease. So, cosine is decreasing from 1 to 0, and the right side is decreasing from ~0.266 to ~0.228. So, since cosine is above the right side at t=6.28 (1 > 0.266) and below at t=7.85 (0 < 0.228), they must cross once in this interval as well. So, that's two critical points in this interval.Wait, but actually, in the interval t=4.71 to t=7.85, which is about 3.14 wide, we have two critical points. So, each positive interval of cosine (each ~3.14 wide) could have two critical points? Or is it only one?Wait, maybe not. Let me think again. In the first interval, t=0 to t=1.57, we have one critical point. In the next interval, t=4.71 to t=7.85, we have two critical points. Hmm, that seems inconsistent. Maybe it's because the right side is higher in the first interval, so only one crossing, but in the next interval, the right side is lower, so two crossings?Wait, no, actually, in the first interval, the right side starts at 0.5 and decreases to ~0.4275, while cosine starts at 1 and decreases to 0. So, they cross once. In the next interval, the right side is lower, but the cosine function is going from 0 to 1 and back to 0, so it crosses the right side twice: once on the way up and once on the way down.Wait, but in the first interval, cosine is only going from 1 to 0, so only one crossing. In the subsequent intervals where cosine goes from 0 to 1 and back to 0, we might have two crossings each.So, perhaps in each interval after the first, we have two critical points. Let me check.For k=1: t=4.71 to t=7.85, two critical points.For k=2: t=10.99 to t=14.14, let's see. At t=10.99, which is approximately 3.5π, cosine is 0. The right side is ( e^{-0.1 times 10.99}/2 approx e^{-1.099}/2 approx 0.333/2 = 0.1665 ). At t=14.14 (~4.5π), cosine is 0 again, and the right side is ( e^{-0.1 times 14.14}/2 approx e^{-1.414}/2 approx 0.242/2 = 0.121 ). So, the right side is decreasing from ~0.1665 to ~0.121. Cosine goes from 0 to 1 and back to 0. So, similar to the previous interval, cosine will cross the right side once on the way up and once on the way down. So, two critical points.Similarly, for k=3: t=17.27 to t=20.42. At t=17.27 (~5.5π), cosine is 0, right side is ( e^{-1.727}/2 approx 0.178/2 = 0.089 ). At t=20.42 (~6.5π), cosine is 0, right side is ( e^{-2.042}/2 approx 0.130/2 = 0.065 ). So, right side is decreasing from ~0.089 to ~0.065. Cosine goes from 0 to 1 and back to 0. So, again, two critical points.Similarly, for k=4: t=23.56 to t=26.72. At t=23.56 (~7.5π), cosine is 0, right side is ( e^{-2.356}/2 approx 0.095/2 = 0.0475 ). At t=26.72 (~8.5π), cosine is 0, right side is ( e^{-2.672}/2 approx 0.068/2 = 0.034 ). So, right side is decreasing from ~0.0475 to ~0.034. Cosine goes from 0 to 1 and back to 0. So, two critical points.For k=5: t=29.84 to t=32.99. At t=29.84 (~9.5π), cosine is 0, right side is ( e^{-2.984}/2 approx 0.046/2 = 0.023 ). At t=32.99 (~10.5π), cosine is 0, right side is ( e^{-3.299}/2 approx 0.036/2 = 0.018 ). So, right side is decreasing from ~0.023 to ~0.018. Cosine goes from 0 to 1 and back to 0. So, two critical points.For k=6: t=36.12 to t=39.27. At t=36.12 (~11.5π), cosine is 0, right side is ( e^{-3.612}/2 approx 0.026/2 = 0.013 ). At t=39.27 (~12.5π), cosine is 0, right side is ( e^{-3.927}/2 approx 0.019/2 = 0.0095 ). So, right side is decreasing from ~0.013 to ~0.0095. Cosine goes from 0 to 1 and back to 0. So, two critical points.Wait, but t=39.27 is beyond 40? No, 39.27 is less than 40, so it's still within the interval. So, in this last interval, we might have two critical points as well.Wait, but let's check t=39.27: it's approximately 12.5π, which is ~39.27. So, the next interval would be t=42.44, which is beyond 40, so we don't need to consider that.So, in total, in each interval where cosine is positive (k=0 to k=6), we have:- k=0: t=0 to t=1.57: 1 critical point.- k=1: t=4.71 to t=7.85: 2 critical points.- k=2: t=10.99 to t=14.14: 2 critical points.- k=3: t=17.27 to t=20.42: 2 critical points.- k=4: t=23.56 to t=26.72: 2 critical points.- k=5: t=29.84 to t=32.99: 2 critical points.- k=6: t=36.12 to t=39.27: 2 critical points.So, total critical points: 1 + 2*6 = 13 critical points.Wait, but let me verify this because sometimes the right side might be too low for cosine to cross it twice in an interval.For example, in the last interval, t=36.12 to t=39.27, the right side is between ~0.013 and ~0.0095. Cosine goes from 0 to 1 and back to 0. So, cosine will cross the right side once on the way up (since cosine starts at 0, right side is ~0.013, so cosine will cross it once when increasing to 1, and then again when decreasing back to 0, crossing the right side which is now ~0.0095. So, two crossings.Similarly, in the interval k=5, the right side is ~0.023 to ~0.018, which is still above zero, so cosine will cross twice.Wait, but as t increases, the right side becomes very small, approaching zero. So, in the intervals where the right side is very small, say, less than 0.01, does cosine still cross it twice? Let's see.At t=36.12, right side is ~0.013, which is still above zero. Cosine goes from 0 to 1 and back to 0, so it must cross the right side twice. Similarly, at t=39.27, right side is ~0.0095, still above zero. So, yes, two crossings.Therefore, in each interval from k=1 to k=6, we have two critical points, and in k=0, we have one. So, total critical points: 1 + 6*2 = 13.But wait, let me check if in the last interval, t=36.12 to t=39.27, the right side is so small that maybe cosine doesn't cross it twice? Let's compute at t=36.12, right side is ~0.013, and at t=39.27, it's ~0.0095. So, cosine starts at 0, goes up to 1 at t=37.7 (which is 36.12 + π/2 ≈ 36.12 + 1.57 ≈ 37.69), then back to 0 at t=39.27. So, at t=37.69, cosine is 1, right side is ( e^{-0.1*37.69}/2 ≈ e^{-3.769}/2 ≈ 0.023/2 ≈ 0.0115 ). So, at t=37.69, cosine is 1, which is above the right side (~0.0115). So, cosine crosses the right side once on the way up (from 0 to 1) and once on the way down (from 1 to 0). So, two crossings.Therefore, it seems that in each interval from k=1 to k=6, we have two critical points, and in k=0, one critical point. So, total of 13 critical points.But wait, 13 critical points in the interval [0,40]. That seems a lot, but considering the periodicity of cosine, it's plausible.However, let me check if all these critical points are within [0,40]. The last critical point would be in the interval t=36.12 to t=39.27, which is within 40. So, yes, all 13 critical points are within [0,40].But wait, let me think again. Each interval where cosine is positive (each ~3.14 wide) can have two critical points, except the first interval which only has one. So, with 6 full positive intervals (k=1 to k=6) each contributing two critical points, plus one from k=0, total 13.But actually, let me count the number of positive intervals in [0,40]. The positive intervals are from t=0 to t=1.57, t=4.71 to t=7.85, t=10.99 to t=14.14, t=17.27 to t=20.42, t=23.56 to t=26.72, t=29.84 to t=32.99, t=36.12 to t=39.27. So, that's 7 intervals. The first interval (k=0) has 1 critical point, and the remaining 6 intervals (k=1 to k=6) have 2 each, totaling 1 + 6*2 = 13.So, 13 critical points in total.But wait, let me check if the last interval (k=6) is entirely within [0,40]. The last interval is t=36.12 to t=39.27, which is within 40, so yes.Therefore, the critical points are at 13 different t values in [0,40].But wait, this seems a bit too much. Let me think if there's a way to confirm this.Alternatively, maybe I can graph the functions ( y = 2cos(t) ) and ( y = e^{-0.1t} ) and see how many times they intersect between t=0 and t=40.But since I can't graph it right now, I have to rely on my reasoning.Alternatively, perhaps I can use the Intermediate Value Theorem on each interval to confirm the number of solutions.But given the periodicity and the decay of the exponential, it's reasonable to assume that in each positive interval of cosine, there are two crossings except the first one, which has one.So, 13 critical points in total.But wait, 13 seems a lot, but considering the function is oscillating and decaying, it's possible.However, let me think about the behavior of the derivative ( C'(t) = -e^{-0.1t} + 2cos(t) ).At t=0: ( C'(0) = -1 + 2*1 = 1 ). So, positive.At t=1.57 (~π/2): ( C'(1.57) = -e^{-0.157} + 2*0 ≈ -0.855 + 0 = -0.855 ). So, negative.Therefore, by Intermediate Value Theorem, there's a critical point between t=0 and t=1.57.Similarly, at t=4.71 (~3π/2): ( C'(4.71) = -e^{-0.471} + 2*0 ≈ -0.624 + 0 = -0.624 ). Negative.At t=6.28 (~2π): ( C'(6.28) = -e^{-0.628} + 2*1 ≈ -0.533 + 2 = 1.467 ). Positive.So, between t=4.71 and t=6.28, the derivative goes from negative to positive, so there's a critical point.Then, at t=7.85 (~5π/2): ( C'(7.85) = -e^{-0.785} + 2*0 ≈ -0.456 + 0 = -0.456 ). Negative.So, between t=6.28 and t=7.85, the derivative goes from positive to negative, so another critical point.Therefore, in the interval t=4.71 to t=7.85, we have two critical points: one at t≈5.x and another at t≈7.x.Similarly, in the next intervals, the same pattern repeats: derivative goes from negative to positive and then back to negative, creating two critical points per interval.Therefore, the number of critical points is indeed 13.But wait, 13 critical points in [0,40]. Let me think about how many maxima and minima that would be. Each critical point is either a maximum or a minimum.But since the function ( C(t) ) is a combination of an exponential decay and a sine wave, it's oscillating with decreasing amplitude.Wait, actually, the exponential term is 10e^{-0.1t}, which is decaying, and the sine term is 2sin(t), which is oscillating with fixed amplitude. So, the overall function is a decaying exponential plus an oscillating term.Therefore, the critical points correspond to the peaks and troughs of the oscillating part, modulated by the exponential decay.But regardless, the critical points are where the derivative is zero, which we've established occurs 13 times in [0,40].But now, the question is: what do these critical points suggest about the preparation strategy?Well, critical points are where the function changes from increasing to decreasing or vice versa. So, in terms of case complexity, which is given by ( C(t) ), the critical points would indicate points where the complexity is either at a local maximum or minimum.Since the attorney wants to minimize the complexity, which is decreasing over time, the critical points would help identify the optimal times to allocate resources.But since the function ( C(t) ) is a combination of a decaying exponential and an oscillating sine function, the complexity will have oscillations with decreasing amplitude around the decaying exponential.Therefore, the critical points would indicate the peaks and troughs of these oscillations.So, the attorney should be aware that the complexity will have fluctuations, and the critical points can help identify when the complexity is at a local high or low.Therefore, the optimal time allocation might involve focusing on periods where the complexity is decreasing and avoiding periods where it's increasing.But since the overall trend is decreasing (due to the exponential term), the complexity will trend downward, but with oscillations.Therefore, the critical points suggest that the attorney should monitor the complexity and adjust preparation time around these critical points to take advantage of the decreasing trend and minimize the complexity.But perhaps more specifically, the critical points can help identify when the complexity is at a local minimum, which would be the best times to present the case, as the complexity is lowest.Alternatively, the attorney might want to prepare more when complexity is increasing (before a critical point) and less when it's decreasing (after a critical point).But since the overall trend is decreasing, the attorney should continue preparing up to a certain point where the marginal decrease in complexity is outweighed by the time invested.But perhaps the critical points indicate the optimal times to stop or increase preparation.Wait, actually, since the complexity function is ( C(t) = 10e^{-0.1t} + 2sin(t) ), the overall trend is decreasing because of the exponential term, but with oscillations.So, the critical points are where the instantaneous rate of change is zero, i.e., where the function momentarily stops decreasing or increasing.Therefore, the attorney can use these critical points to determine when the complexity is at a local minimum or maximum.Local minima would be the best times to present the case, as the complexity is lowest, while local maxima would be the worst times.Therefore, the attorney should aim to present the case near the local minima of ( C(t) ), which occur at certain critical points.But since the critical points are numerous, the attorney might need to identify the global minimum in the interval [0,40], but since the function is oscillating with decreasing amplitude, the global minimum would be as t approaches infinity, but within [0,40], the minimum complexity would be at t=40, but with oscillations.Wait, actually, the exponential term dominates as t increases, so the complexity tends to zero as t approaches infinity, but within [0,40], the complexity will have oscillations around a decaying exponential.Therefore, the overall minimum complexity in [0,40] would be at t=40, but with possible local minima along the way.But the critical points are where the function changes direction, so the attorney can use these to schedule preparation times to coincide with periods of decreasing complexity.Alternatively, perhaps the attorney should focus on the intervals where the complexity is decreasing and allocate more resources during those periods.But I think the key takeaway is that the critical points indicate where the complexity is at local extrema, so the attorney should be aware of these points to optimize the preparation strategy.Now, moving on to the second part: Jury Decision Probability.The attorney models the probability ( P(x) ) that the jury will decide in favor of the client as a function of the number of arguments ( x ) presented, where ( P(x) = frac{x^2}{x^2 + 25} ).Part (a) asks for the range of ( P(x) ) as ( x ) varies from 0 to infinity.So, ( P(x) = frac{x^2}{x^2 + 25} ).Let me analyze this function.First, as ( x ) approaches 0, ( P(x) ) approaches ( 0/25 = 0 ).As ( x ) approaches infinity, ( P(x) ) approaches ( x^2 / x^2 = 1 ).Therefore, the range of ( P(x) ) is from 0 to 1.But let me confirm.Since ( x^2 ) is always non-negative, the numerator and denominator are both positive for all ( x ).So, ( P(x) ) is always between 0 and 1.Moreover, ( P(x) ) is an increasing function because as ( x ) increases, ( P(x) ) approaches 1 asymptotically.Therefore, the range of ( P(x) ) is ( [0, 1) ). Wait, but as ( x ) approaches infinity, ( P(x) ) approaches 1, but never actually reaches 1. So, the range is ( [0, 1) ).But let me check for ( x = 5 ): ( P(5) = 25 / (25 + 25) = 25/50 = 0.5 ). So, at x=5, P(x)=0.5.As x increases beyond 5, P(x) increases towards 1.Similarly, as x decreases below 5, P(x) decreases towards 0.Therefore, the range is indeed ( [0, 1) ).But wait, when x=0, P(0)=0, and as x increases, P(x) increases without bound? No, it approaches 1.So, the range is ( [0, 1) ).But wait, in the problem statement, it's specified that ( x ) varies from 0 to infinity, so the range is from 0 up to, but not including, 1.Therefore, the range is ( [0, 1) ).But sometimes, depending on the context, the range might be considered as ( [0, 1] ) if we consider the limit as x approaches infinity. But strictly speaking, since x is finite, P(x) never actually reaches 1.So, the range is ( [0, 1) ).Part (b) asks: If the attorney must ensure at least a 75% probability of winning, find the minimum number of arguments that must be presented.So, we need to find the smallest integer x such that ( P(x) geq 0.75 ).Given ( P(x) = frac{x^2}{x^2 + 25} geq 0.75 ).Let me solve for x.( frac{x^2}{x^2 + 25} geq 0.75 )Multiply both sides by ( x^2 + 25 ) (which is positive, so inequality sign doesn't change):( x^2 geq 0.75(x^2 + 25) )Expand the right side:( x^2 geq 0.75x^2 + 18.75 )Subtract ( 0.75x^2 ) from both sides:( x^2 - 0.75x^2 geq 18.75 )Simplify:( 0.25x^2 geq 18.75 )Multiply both sides by 4:( x^2 geq 75 )Take square roots:( x geq sqrt{75} )Simplify sqrt(75):( sqrt{75} = sqrt{25*3} = 5sqrt{3} ≈ 5*1.732 ≈ 8.66 )Since x must be an integer (number of arguments presented), we round up to the next integer, which is 9.Therefore, the minimum number of arguments that must be presented is 9.But let me verify:For x=8:( P(8) = 64 / (64 + 25) = 64/89 ≈ 0.719, which is less than 0.75.For x=9:( P(9) = 81 / (81 + 25) = 81/106 ≈ 0.764, which is greater than 0.75.Therefore, the minimum number of arguments is 9.So, summarizing:1. The critical points of ( C(t) ) in [0,40] are 13 in total, indicating local maxima and minima where the complexity changes direction. The attorney should consider these points to optimize preparation time, focusing on periods where complexity is decreasing.2. (a) The range of ( P(x) ) is ( [0, 1) ). (b) The minimum number of arguments needed for at least a 75% probability is 9.</think>

🧠:<think>Okay, so I have this problem where I need to calculate two probabilities related to a web application's request handling and Redis caching. Let me try to break it down step by step.First, the problem is divided into two parts. The first part is about the number of requests per second following a Poisson distribution, and I need to find the probability that more than 320 requests are received in a second. The second part is about the time taken to retrieve data from Redis, which follows an exponential distribution, and I need to find the probability that it takes more than 8 milliseconds.Starting with the first part: Poisson distribution. I remember that the Poisson distribution is used to model the number of events happening in a fixed interval of time or space. The formula for the Poisson probability mass function is P(X = k) = (λ^k * e^-λ) / k!, where λ is the average rate (in this case, 300 requests per second), and k is the number of occurrences.But wait, the question is asking for the probability that the server receives more than 320 requests in a second. So, I need to calculate P(X > 320). Since the Poisson distribution can be approximated by a normal distribution when λ is large, maybe I can use that approximation here because 300 is a pretty large number.So, for the Poisson distribution, the mean μ is equal to λ, which is 300. The variance σ² is also equal to λ, so σ is the square root of 300. Let me calculate that: sqrt(300) is approximately 17.3205.Now, to approximate the Poisson distribution with a normal distribution, I can use the continuity correction. Since we're looking for P(X > 320), we'll adjust it to P(X > 320.5) in the normal distribution. That's because the Poisson is discrete, and the normal is continuous.So, converting 320.5 to a z-score: z = (320.5 - μ) / σ. Plugging in the numbers: (320.5 - 300) / 17.3205 ≈ 20.5 / 17.3205 ≈ 1.183.Now, I need to find the probability that Z > 1.183. Looking at the standard normal distribution table, the area to the left of Z=1.18 is about 0.8810, and for Z=1.19 it's about 0.8830. Since 1.183 is closer to 1.18, maybe I can approximate it as 0.8810. Therefore, the area to the right (which is what we need) is 1 - 0.8810 = 0.1190.But wait, let me check if the normal approximation is appropriate here. The rule of thumb is that both λ and λ(1 - p) should be greater than 5. Here, λ is 300, which is way more than 5, so the approximation should be fine.Alternatively, I could use the Poisson formula directly, but calculating P(X > 320) would require summing from 321 to infinity, which isn't practical. So, the normal approximation is the way to go.Moving on to the second part: exponential distribution. The time taken to retrieve data from Redis follows an exponential distribution with a mean of 5 milliseconds. I need to find the probability that it takes more than 8 milliseconds.The exponential distribution is memoryless, and its probability density function is f(t) = (1/β) * e^(-t/β), where β is the mean. So, β here is 5 milliseconds.The cumulative distribution function (CDF) for the exponential distribution is P(X ≤ t) = 1 - e^(-t/β). Therefore, the probability that X > t is 1 - CDF, which is e^(-t/β).Plugging in t = 8 milliseconds: P(X > 8) = e^(-8/5) = e^(-1.6). Calculating that, e^1.6 is approximately 4.953, so 1/4.953 ≈ 0.202. Therefore, P(X > 8) ≈ 0.202 or 20.2%.Let me double-check that. The formula is correct, right? For exponential distribution, P(X > t) = e^(-λt), where λ is the rate parameter. Since the mean is 5, λ = 1/5 = 0.2. So, P(X > 8) = e^(-0.2*8) = e^(-1.6), which is the same as before. So, that's consistent.Wait, another way to think about it: since the mean is 5, the rate λ is 1/5. So, the CDF is 1 - e^(-λt), so P(X > t) = e^(-λt). Yep, that's correct.So, summarizing:1. For the Poisson distribution, using normal approximation, P(X > 320) ≈ 0.119 or 11.9%.2. For the exponential distribution, P(X > 8) ≈ 0.202 or 20.2%.I think that's it. I should make sure I didn't mix up any formulas, especially between Poisson and exponential. But I believe I applied the correct methods for each distribution.Final Answer1. The probability of receiving more than 320 requests in a second is boxed{0.119}.2. The probability that it takes more than 8 milliseconds to retrieve data is boxed{0.202}.</think>