answer:<think>Okay, so I've got these two problems to solve about a Portuguese literature professor studying the evolution of archaic words in literary works. Let me try to tackle them one by one.Starting with problem 1: The professor has a dataset of 500 literary works from the 15th to the 21st century, and she's identified 200 key archaic words. For each decade, she calculates the ratio of texts containing at least one of these words to the total number of texts from that decade. This ratio is represented by the function ( f(t) = frac{1}{1 + e^{-(0.2t - 5)}} ), where ( t ) is the number of decades since the 15th century. I need to find the rate of change of this ratio with respect to time at ( t = 20 ).Alright, so rate of change means I need to find the derivative of ( f(t) ) with respect to ( t ) and then evaluate it at ( t = 20 ). Let me recall how to differentiate such a function. The function looks like a logistic function, which has the form ( frac{1}{1 + e^{-kt + c}} ). The derivative of a logistic function is ( f'(t) = f(t)(1 - f(t)) cdot k ). Let me verify that.Yes, if ( f(t) = frac{1}{1 + e^{-kt + c}} ), then ( f'(t) = frac{d}{dt} left( frac{1}{1 + e^{-kt + c}} right) ). Using the chain rule, the derivative is ( frac{e^{-kt + c} cdot k}{(1 + e^{-kt + c})^2} ). Which can be rewritten as ( f(t) cdot (1 - f(t)) cdot k ). So that's correct.In this case, the function is ( f(t) = frac{1}{1 + e^{-(0.2t - 5)}} ). So comparing to the standard logistic function, ( k = 0.2 ) and ( c = -5 ). Therefore, the derivative should be ( f'(t) = f(t)(1 - f(t)) cdot 0.2 ).So, first, I need to compute ( f(20) ), then compute ( 1 - f(20) ), multiply them together, and then multiply by 0.2 to get the rate of change at ( t = 20 ).Let me compute ( f(20) ):( f(20) = frac{1}{1 + e^{-(0.2*20 - 5)}} )Calculating the exponent:0.2*20 = 44 - 5 = -1So exponent is -(-1) = 1Therefore, ( f(20) = frac{1}{1 + e^{1}} )I know that ( e ) is approximately 2.71828, so ( e^1 = 2.71828 )Thus, ( f(20) = frac{1}{1 + 2.71828} = frac{1}{3.71828} approx 0.2689 )Now, ( 1 - f(20) = 1 - 0.2689 = 0.7311 )So, ( f'(20) = 0.2689 * 0.7311 * 0.2 )Let me compute that step by step:First, 0.2689 * 0.7311. Let me approximate this:0.2689 * 0.7 = 0.188230.2689 * 0.0311 ≈ 0.00836Adding them together: 0.18823 + 0.00836 ≈ 0.19659Then, multiply by 0.2: 0.19659 * 0.2 ≈ 0.039318So, approximately 0.0393.But let me check my calculations for precision.Alternatively, maybe I can compute it more accurately.First, compute ( f(20) ):( f(20) = 1 / (1 + e^{1}) )( e^1 ) is about 2.718281828So, denominator is 1 + 2.718281828 = 3.718281828Thus, ( f(20) = 1 / 3.718281828 ≈ 0.268941421 )Then, ( 1 - f(20) = 1 - 0.268941421 ≈ 0.731058579 )Multiply these two: 0.268941421 * 0.731058579Let me compute this more accurately:0.268941421 * 0.731058579First, 0.2 * 0.7 = 0.140.2 * 0.031058579 ≈ 0.0062117160.068941421 * 0.7 ≈ 0.0482589950.068941421 * 0.031058579 ≈ ~0.00214Adding all together:0.14 + 0.006211716 + 0.048258995 + 0.00214 ≈ 0.14 + 0.006211716 = 0.146211716 + 0.048258995 = 0.194470711 + 0.00214 ≈ 0.196610711So, approximately 0.19661Then, multiply by 0.2: 0.19661 * 0.2 = 0.039322So, approximately 0.039322So, the rate of change is approximately 0.0393 per decade.But maybe I should express it more precisely.Alternatively, perhaps I can compute it symbolically first.Given ( f(t) = frac{1}{1 + e^{-(0.2t - 5)}} )Then, ( f'(t) = frac{d}{dt} left( frac{1}{1 + e^{-(0.2t - 5)}} right) )Let me compute the derivative step by step.Let me set ( u = -(0.2t - 5) = -0.2t + 5 )Then, ( f(t) = frac{1}{1 + e^{u}} )So, ( f'(t) = frac{d}{dt} left( frac{1}{1 + e^{u}} right) = frac{-e^{u} cdot u'}{(1 + e^{u})^2} )But ( u = -0.2t + 5 ), so ( u' = -0.2 )Thus, ( f'(t) = frac{-e^{u} cdot (-0.2)}{(1 + e^{u})^2} = frac{0.2 e^{u}}{(1 + e^{u})^2} )But ( e^{u} = e^{-0.2t + 5} ), so:( f'(t) = frac{0.2 e^{-0.2t + 5}}{(1 + e^{-0.2t + 5})^2} )Alternatively, since ( f(t) = frac{1}{1 + e^{-0.2t + 5}} ), then ( 1 - f(t) = frac{e^{-0.2t + 5}}{1 + e^{-0.2t + 5}} )Therefore, ( f'(t) = 0.2 f(t) (1 - f(t)) ), which is consistent with what I had earlier.So, at ( t = 20 ):( u = -0.2*20 + 5 = -4 + 5 = 1 )So, ( e^{u} = e^{1} ≈ 2.71828 )Thus, ( f'(20) = 0.2 * (1 / (1 + e^{1})) * (e^{1} / (1 + e^{1})) )Which is 0.2 * (1 / 3.71828) * (2.71828 / 3.71828)Compute each part:1 / 3.71828 ≈ 0.268942.71828 / 3.71828 ≈ 0.73106Multiply them: 0.26894 * 0.73106 ≈ 0.19661Multiply by 0.2: 0.19661 * 0.2 ≈ 0.03932So, approximately 0.03932.Therefore, the rate of change at ( t = 20 ) is approximately 0.0393 per decade.Wait, but let me check if this is correct. Since the function is a logistic function, it has an S-shape, and the derivative should be maximum at the inflection point, which is at the midpoint of the curve.Given the function ( f(t) = frac{1}{1 + e^{-(0.2t - 5)}} ), the midpoint occurs where the exponent is zero, so when ( 0.2t - 5 = 0 ), which is at ( t = 25 ). So, at ( t = 25 ), the function is at its steepest point.At ( t = 20 ), which is 5 decades before the midpoint, the function is still increasing, but perhaps not as steep as at ( t = 25 ).Wait, but according to our calculation, the derivative at ( t = 20 ) is about 0.0393. Let me see if that makes sense.Alternatively, perhaps I can compute it numerically.Compute ( f(20) = 1 / (1 + e^{1}) ≈ 0.2689 )Compute ( f(21) = 1 / (1 + e^{-(0.2*21 -5)}) = 1 / (1 + e^{-(4.2 -5)}) = 1 / (1 + e^{0.8}) ≈ 1 / (1 + 2.2255) ≈ 1 / 3.2255 ≈ 0.3100So, the change from t=20 to t=21 is approximately 0.3100 - 0.2689 ≈ 0.0411 per decade.Which is close to our derivative value of 0.0393. So, that seems consistent.Therefore, I think my calculation is correct.So, the rate of change is approximately 0.0393 per decade.Moving on to problem 2: The professor models the influence of external languages, such as Arabic and Latin, as a periodic function ( g(t) = A sin(omega t + phi) + B ). Given that ( g(t) ) has a maximum influence of 0.1 at ( t = 25 ) and a minimum influence of -0.1 at ( t = 35 ), I need to determine the constants ( A, omega, phi, ) and ( B ).Alright, let's break this down.First, the function is a sine wave with amplitude ( A ), angular frequency ( omega ), phase shift ( phi ), and vertical shift ( B ).Given that the maximum value is 0.1 and the minimum is -0.1, we can find ( A ) and ( B ).The maximum of ( g(t) ) is ( B + A ), and the minimum is ( B - A ).Given that maximum is 0.1 and minimum is -0.1, so:( B + A = 0.1 )( B - A = -0.1 )Let me solve these two equations.Adding them together:( (B + A) + (B - A) = 0.1 + (-0.1) )Simplify:( 2B = 0 )Thus, ( B = 0 )Substituting back into the first equation:( 0 + A = 0.1 ) => ( A = 0.1 )So, we have ( A = 0.1 ) and ( B = 0 ).Now, we need to find ( omega ) and ( phi ).We know that the maximum occurs at ( t = 25 ) and the minimum at ( t = 35 ).In a sine function ( sin(theta) ), the maximum occurs at ( theta = pi/2 ) and the minimum at ( theta = 3pi/2 ).So, for ( g(t) = 0.1 sin(omega t + phi) ), the maximum occurs when ( omega t + phi = pi/2 ), and the minimum when ( omega t + phi = 3pi/2 ).Given that the maximum is at ( t = 25 ):( omega * 25 + phi = pi/2 ) ...(1)And the minimum is at ( t = 35 ):( omega * 35 + phi = 3pi/2 ) ...(2)Subtracting equation (1) from equation (2):( omega * 35 + phi - (omega * 25 + phi) = 3pi/2 - pi/2 )Simplify:( 10 omega = pi )Thus, ( omega = pi / 10 )Now, substitute ( omega = pi / 10 ) back into equation (1):( (pi / 10) * 25 + phi = pi / 2 )Compute ( (pi / 10) * 25 = (25/10) pi = 2.5 pi )So, 2.5 π + φ = π / 2Thus, φ = π / 2 - 2.5 π = π / 2 - 5π / 2 = (-4π)/2 = -2πBut since sine is periodic with period 2π, adding or subtracting multiples of 2π doesn't change the function. So, φ = -2π is equivalent to φ = 0.But let me check:If φ = -2π, then the function becomes ( 0.1 sin(omega t - 2π) ), which is the same as ( 0.1 sin(omega t) ), since sine is 2π periodic.Therefore, φ can be taken as 0.Wait, but let me verify:If φ = -2π, then at t = 25:( omega * 25 + φ = (π / 10)*25 - 2π = 2.5π - 2π = 0.5π = π/2 ), which is correct.Similarly, at t = 35:( (π / 10)*35 - 2π = 3.5π - 2π = 1.5π = 3π/2 ), which is correct.So, φ can be -2π, but since sine is periodic, it's equivalent to φ = 0.But to keep it simple, we can set φ = 0.Therefore, the function is ( g(t) = 0.1 sin(pi t / 10) ).Wait, but let me check the period.The period ( T ) of the sine function is ( 2π / omega ). Here, ( omega = π / 10 ), so period ( T = 2π / (π / 10) ) = 20 ) decades.So, the function has a period of 20 decades.But the maximum occurs at t = 25 and the minimum at t = 35, which are 10 decades apart. Since the period is 20, the distance between a maximum and the next minimum is half a period, which is 10 decades. So that makes sense.Therefore, the function is ( g(t) = 0.1 sin(π t / 10) ).Alternatively, since φ = 0, we can write it as ( g(t) = 0.1 sin(π t / 10) ).So, the constants are:A = 0.1ω = π / 10φ = 0 (or -2π, but 0 is simpler)B = 0Wait, but let me double-check.If I set φ = 0, then at t = 25:( g(25) = 0.1 sin(π *25 /10) = 0.1 sin(2.5π) = 0.1 sin(π/2 + 2π) = 0.1 sin(π/2) = 0.1 * 1 = 0.1 ), which is correct.At t = 35:( g(35) = 0.1 sin(π *35 /10) = 0.1 sin(3.5π) = 0.1 sin(3π/2) = 0.1 * (-1) = -0.1 ), which is correct.So yes, φ = 0 works.Therefore, the values are:A = 0.1ω = π / 10φ = 0B = 0So, summarizing:Problem 1: The rate of change at t = 20 is approximately 0.0393 per decade.Problem 2: The constants are A = 0.1, ω = π/10, φ = 0, and B = 0.I think that's it. Let me just recap to make sure I didn't miss anything.For problem 1, I correctly identified the function as a logistic function, computed its derivative using the chain rule, evaluated it at t = 20, and got approximately 0.0393.For problem 2, I recognized that the maximum and minimum give the amplitude and vertical shift, then used the points of maximum and minimum to find the angular frequency and phase shift, resulting in A = 0.1, ω = π/10, φ = 0, and B = 0.Yes, that seems solid.</think>

answer:<think>Okay, so I have this legal case to analyze, and I need to figure out whether the client should accept a settlement offer or go to trial. The lawyer wants to use game theory and probability to make the best decision. Let me break this down step by step.First, the settlement offer is 500,000. That seems straightforward. If they accept the settlement, they get that amount right away without any further costs or uncertainties. But if they go to trial, there are more variables involved. The probability of winning the trial is 0.6, which is 60%. If they win, they get 1,200,000. However, if they lose, they get nothing. Also, regardless of the outcome, there are legal costs of 150,000. So, I need to calculate the expected monetary value (EMV) for both options.Let me start with the EMV of accepting the settlement. That's simple because it's just the settlement amount minus any costs. Wait, actually, the problem doesn't mention any costs associated with accepting the settlement. So, EMV for settlement is just 500,000.Now, for going to trial, the EMV calculation is a bit more involved. There are two possible outcomes: winning or losing. The probability of winning is 0.6, and if they win, they receive 1,200,000. But they have to subtract the legal costs of 150,000. So, the net gain if they win is 1,200,000 - 150,000 = 1,050,000.If they lose, which has a probability of 0.4, they get nothing but still have to pay the legal costs of 150,000. So, the net loss in that case is -150,000.Therefore, the EMV for going to trial is (0.6 * 1,050,000) + (0.4 * -150,000). Let me compute that:0.6 * 1,050,000 = 630,0000.4 * (-150,000) = -60,000Adding these together: 630,000 - 60,000 = 570,000So, the EMV for going to trial is 570,000, and the EMV for accepting the settlement is 500,000. Comparing the two, 570,000 is higher than 500,000, so based on EMV, the lawyer should recommend going to trial.Wait, but hold on. The problem mentions that if they go to trial and lose, there's a 30% chance of an appeal. Hmm, that's part 2, right? So in part 1, we don't consider the appeal. So, for part 1, the EMV is 570,000 vs. 500,000, so go to trial.Now, moving on to part 2. The legal system introduces a new dimension: a 30% chance of an appeal if they lose at trial. If they appeal, there's a 50% chance of reversing the decision. If reversed, the client gets 800,000 with no additional legal costs. So, I need to adjust the decision tree to include this.Let me think about how this affects the EMV for going to trial. So, if they lose at trial (probability 0.4), then there's a 30% chance to appeal. If they appeal, there's a 50% chance of winning 800,000, and a 50% chance of losing again, which would mean they get nothing. But wait, if they lose the appeal, do they have to pay more costs? The problem says no additional legal costs if reversed, so I assume if they lose the appeal, they still have the original legal costs of 150,000, but no further costs.So, let's break it down. The loss at trial is 0.4 probability. Then, 30% chance to appeal, which is 0.3, and 70% chance not to appeal, which is 0.7.If they don't appeal, they just get nothing and have the legal costs of 150,000. So, that outcome is 0.4 * 0.7 = 0.28 probability, resulting in -150,000.If they do appeal, which is 0.4 * 0.3 = 0.12 probability, then there's a 50% chance of getting 800,000 and a 50% chance of getting nothing. But wait, in the case of getting nothing, do they have any additional costs? The problem says no additional legal costs if reversed, so I think the only cost is the initial 150,000 regardless of the appeal outcome.So, the appeal outcome has two possibilities:1. Win the appeal: 0.5 probability, receive 800,000, net gain is 800,000 - 150,000 = 650,000.2. Lose the appeal: 0.5 probability, receive nothing, net loss is -150,000.Therefore, the EMV for the appeal path is (0.5 * 650,000) + (0.5 * -150,000) = 325,000 - 75,000 = 250,000.But this is conditional on losing at trial and deciding to appeal. So, the overall probability for this path is 0.4 * 0.3 = 0.12, and the EMV contribution is 0.12 * 250,000 = 30,000.Wait, no, actually, I think I need to compute the EMV step by step. Let me structure it properly.First, the trial outcome:- Win: 0.6 probability, payout 1,200,000, net gain 1,050,000.- Lose: 0.4 probability, which leads to two possibilities: - Appeal: 0.3 probability, which has two outcomes: - Win appeal: 0.5 probability, payout 800,000, net gain 650,000. - Lose appeal: 0.5 probability, payout 0, net loss 150,000. - Don't appeal: 0.7 probability, payout 0, net loss 150,000.So, the EMV for the lose at trial branch is:EMV_lose = (0.3 * [0.5 * 650,000 + 0.5 * (-150,000)]) + (0.7 * (-150,000))Let me compute that:First, compute the appeal part:0.5 * 650,000 = 325,0000.5 * (-150,000) = -75,000So, appeal EMV: 325,000 - 75,000 = 250,000Then, the don't appeal part: 0.7 * (-150,000) = -105,000So, EMV_lose = 0.3 * 250,000 + 0.7 * (-150,000)Wait, no. Actually, the 0.3 and 0.7 are within the lose at trial branch.Wait, perhaps another way: the total EMV for losing at trial is:(0.3 * [0.5 * 650,000 + 0.5 * (-150,000)]) + (0.7 * (-150,000))Which is:0.3 * (250,000) + 0.7 * (-150,000) = 75,000 - 105,000 = -30,000So, the EMV for losing at trial is -30,000.Therefore, the overall EMV for going to trial is:EMV_trial = (0.6 * 1,050,000) + (0.4 * (-30,000)) = 630,000 - 12,000 = 618,000Wait, that doesn't seem right. Let me double-check.Wait, no. The EMV for losing at trial is -30,000, which is already considering the appeal and not appealing. So, the overall EMV is:EMV_trial = (0.6 * 1,050,000) + (0.4 * (-30,000)) = 630,000 - 12,000 = 618,000Wait, but earlier without the appeal, the EMV was 570,000. Now, with the appeal, it's 618,000, which is higher. So, the EMV increased because of the possibility of appealing and potentially getting a higher payout.But let me verify the calculations step by step.First, the win at trial: 0.6 * (1,200,000 - 150,000) = 0.6 * 1,050,000 = 630,000.Lose at trial: 0.4 probability.Within lose at trial:- Appeal: 0.3 probability. - Win appeal: 0.5 * (800,000 - 150,000) = 0.5 * 650,000 = 325,000. - Lose appeal: 0.5 * (-150,000) = -75,000. So, EMV for appeal: 325,000 - 75,000 = 250,000.- Don't appeal: 0.7 probability. - EMV: 0.7 * (-150,000) = -105,000.So, total EMV for losing at trial:(0.3 * 250,000) + (0.7 * -150,000) = 75,000 - 105,000 = -30,000.Therefore, overall EMV for going to trial:630,000 (win) + (-30,000) (lose) = 600,000.Wait, that contradicts my earlier calculation. Wait, 630,000 - 30,000 is 600,000.But earlier I thought it was 618,000. Hmm, I must have miscalculated.Wait, no. The EMV for losing at trial is -30,000, so the total EMV is 630,000 + (-30,000) = 600,000.So, the EMV for going to trial is now 600,000, which is higher than the settlement of 500,000. So, the recommendation remains to go to trial.Wait, but let me think again. The EMV for losing at trial is -30,000, which is better than the previous -60,000 without the appeal. So, the overall EMV increased from 570,000 to 600,000.So, the lawyer should still recommend going to trial because 600,000 > 500,000.But let me make sure I didn't make a mistake in the calculations.Alternatively, maybe I should compute the EMV for the entire process step by step.Starting from the trial:- Win: 0.6 * (1,200,000 - 150,000) = 0.6 * 1,050,000 = 630,000.- Lose: 0.4. - Appeal: 0.3. - Win appeal: 0.5 * (800,000 - 150,000) = 0.5 * 650,000 = 325,000. - Lose appeal: 0.5 * (-150,000) = -75,000. So, EMV for appeal: 325,000 - 75,000 = 250,000. - Don't appeal: 0.7 * (-150,000) = -105,000. So, EMV for lose at trial: (0.3 * 250,000) + (0.7 * -150,000) = 75,000 - 105,000 = -30,000.Therefore, total EMV for trial:630,000 (win) + (-30,000) (lose) = 600,000.Yes, that seems correct. So, the EMV increased from 570,000 to 600,000 due to the appeal possibility. Therefore, the lawyer should still recommend going to trial.Wait, but let me check if I considered all the probabilities correctly. The 30% chance of appeal is conditional on losing at trial, so it's 0.4 * 0.3 = 0.12 probability for the appeal path. Similarly, not appealing is 0.4 * 0.7 = 0.28.So, the EMV contributions are:- Win at trial: 0.6 * 1,050,000 = 630,000.- Appeal and win: 0.12 * 0.5 * 650,000 = 0.06 * 650,000 = 39,000.- Appeal and lose: 0.12 * 0.5 * (-150,000) = 0.06 * (-150,000) = -9,000.- Don't appeal: 0.28 * (-150,000) = -42,000.So, total EMV:630,000 + 39,000 - 9,000 - 42,000 = 630,000 + 39,000 = 669,000; 669,000 - 9,000 = 660,000; 660,000 - 42,000 = 618,000.Wait, now I get 618,000. Hmm, conflicting results.Wait, I think the confusion is in how I'm calculating the EMV. Let me structure it properly.The total EMV is the sum of all possible outcomes multiplied by their respective probabilities.So, the possible outcomes are:1. Win at trial: probability 0.6, payout 1,200,000 - 150,000 = 1,050,000.2. Lose at trial, appeal, win appeal: probability 0.4 * 0.3 * 0.5 = 0.06, payout 800,000 - 150,000 = 650,000.3. Lose at trial, appeal, lose appeal: probability 0.4 * 0.3 * 0.5 = 0.06, payout -150,000.4. Lose at trial, don't appeal: probability 0.4 * 0.7 = 0.28, payout -150,000.So, EMV = (0.6 * 1,050,000) + (0.06 * 650,000) + (0.06 * (-150,000)) + (0.28 * (-150,000)).Calculating each term:0.6 * 1,050,000 = 630,0000.06 * 650,000 = 39,0000.06 * (-150,000) = -9,0000.28 * (-150,000) = -42,000Adding them up:630,000 + 39,000 = 669,000669,000 - 9,000 = 660,000660,000 - 42,000 = 618,000So, the total EMV is 618,000.Wait, so earlier I had two different results: 600,000 and 618,000. Which one is correct?I think the second method is correct because it breaks down all possible outcomes with their exact probabilities. So, the EMV is 618,000.Therefore, with the appeal, the EMV for going to trial is 618,000, which is higher than the settlement of 500,000. So, the lawyer should still recommend going to trial.But wait, in the first method, I calculated the EMV for losing at trial as -30,000, leading to total EMV of 600,000. But the second method gives 618,000. There's a discrepancy here.I think the confusion arises from how the probabilities are being handled. In the first method, I calculated the EMV for losing at trial as -30,000, which is correct because it's the expected value of the lose at trial branch. Then, adding that to the win at trial EMV gives 600,000. However, in the second method, I broke it down into all possible terminal nodes, which should give the same result but I got 618,000.Wait, let me check the second method again.Terminal nodes:1. Win at trial: 0.6 * 1,050,000 = 630,000.2. Appeal and win: 0.06 * 650,000 = 39,000.3. Appeal and lose: 0.06 * (-150,000) = -9,000.4. Don't appeal: 0.28 * (-150,000) = -42,000.Total: 630,000 + 39,000 - 9,000 - 42,000 = 630,000 + 39,000 = 669,000; 669,000 - 9,000 = 660,000; 660,000 - 42,000 = 618,000.Wait, but in the first method, the EMV for losing at trial was -30,000, so total EMV is 630,000 - 30,000 = 600,000.But according to the second method, it's 618,000. So, which is correct?I think the second method is correct because it accounts for all terminal nodes with their exact probabilities. The first method might have an error in how the probabilities are combined.Wait, in the first method, I calculated the EMV for losing at trial as:EMV_lose = (0.3 * [0.5 * 650,000 + 0.5 * (-150,000)]) + (0.7 * (-150,000)) = 75,000 - 105,000 = -30,000.But actually, the 0.3 and 0.7 are conditional probabilities within the lose at trial branch, which has a probability of 0.4. So, the total probability for the appeal path is 0.4 * 0.3 = 0.12, and for not appealing is 0.4 * 0.7 = 0.28.Therefore, the EMV for the lose at trial branch is:(0.12 * [0.5 * 650,000 + 0.5 * (-150,000)]) + (0.28 * (-150,000)).Which is:0.12 * (250,000) + 0.28 * (-150,000) = 30,000 - 42,000 = -12,000.Wait, that's different. So, the EMV for losing at trial is -12,000, not -30,000.Therefore, the total EMV for going to trial is 630,000 (win) + (-12,000) (lose) = 618,000.Ah, that matches the second method. So, my initial first method was incorrect because I didn't properly account for the fact that the 0.3 and 0.7 are within the 0.4 probability of losing at trial. Therefore, the correct EMV for losing at trial is -12,000, leading to a total EMV of 618,000.So, to summarize:Without appeal, EMV for trial is 570,000.With appeal, EMV for trial is 618,000.Therefore, the lawyer should recommend going to trial in both scenarios, but the EMV increased due to the appeal possibility.Wait, but in the first part, without the appeal, the EMV was 570,000, which is higher than the settlement of 500,000. So, the recommendation was to go to trial.In the second part, with the appeal, the EMV increased to 618,000, which is still higher than 500,000. So, the recommendation remains the same.Therefore, the lawyer should recommend going to trial in both cases, but the EMV is higher with the appeal, making the recommendation even stronger.I think that's the conclusion.</think>

answer:<think>Okay, so I have two problems to solve here. The first one is about modeling Alex's yard gains over four high school seasons using a quadratic function, and the second one is about modeling his single-game yardage in his senior year using a sinusoidal function. Let me tackle them one by one.Starting with the first problem. It says that Alex's yard gains over four years follow a quadratic growth model, given by the equation ( y = ax^2 + bx + c ). Here, ( x ) represents the year number, with ( x = 1 ) being freshman year and ( x = 4 ) being senior year. We know that in his freshman year, he gained 800 yards, and by his senior year, he was gaining 2400 yards. Additionally, the total yardage over the four years is 6400 yards.So, I need to find the coefficients ( a ), ( b ), and ( c ). Since it's a quadratic model, it's a second-degree polynomial, which means we need three equations to solve for the three unknowns. The given information provides two points: when ( x = 1 ), ( y = 800 ); and when ( x = 4 ), ( y = 2400 ). The third equation comes from the total yardage over the four years, which is the sum of the yardages for each year from ( x = 1 ) to ( x = 4 ) equaling 6400.Let me write down the equations based on the given information.First, for the freshman year (( x = 1 )):( a(1)^2 + b(1) + c = 800 )Simplifying:( a + b + c = 800 ) ...(1)Second, for the senior year (( x = 4 )):( a(4)^2 + b(4) + c = 2400 )Simplifying:( 16a + 4b + c = 2400 ) ...(2)Third, the sum of the yardages over four years is 6400. So, we need to compute ( y ) for ( x = 1, 2, 3, 4 ) and sum them up.Let me denote the yardage for each year as:- Freshman: ( y_1 = a(1)^2 + b(1) + c = a + b + c = 800 )- Sophomore: ( y_2 = a(2)^2 + b(2) + c = 4a + 2b + c )- Junior: ( y_3 = a(3)^2 + b(3) + c = 9a + 3b + c )- Senior: ( y_4 = a(4)^2 + b(4) + c = 16a + 4b + c = 2400 )So, the total yardage is:( y_1 + y_2 + y_3 + y_4 = 6400 )Substituting the expressions:( (a + b + c) + (4a + 2b + c) + (9a + 3b + c) + (16a + 4b + c) = 6400 )Let me combine like terms:- For ( a ): ( 1a + 4a + 9a + 16a = 30a )- For ( b ): ( 1b + 2b + 3b + 4b = 10b )- For ( c ): ( 1c + 1c + 1c + 1c = 4c )So, the equation becomes:( 30a + 10b + 4c = 6400 ) ...(3)Now, I have three equations:1. ( a + b + c = 800 )2. ( 16a + 4b + c = 2400 )3. ( 30a + 10b + 4c = 6400 )I need to solve this system of equations for ( a ), ( b ), and ( c ).Let me label the equations for clarity:Equation (1): ( a + b + c = 800 )Equation (2): ( 16a + 4b + c = 2400 )Equation (3): ( 30a + 10b + 4c = 6400 )First, I can subtract Equation (1) from Equation (2) to eliminate ( c ).Equation (2) - Equation (1):( (16a + 4b + c) - (a + b + c) = 2400 - 800 )Simplify:( 15a + 3b = 1600 )Divide both sides by 3:( 5a + b = 533.overline{3} ) ...(4)Similarly, let's try to eliminate ( c ) from Equations (1) and (3). Multiply Equation (1) by 4 to make the coefficient of ( c ) equal to 4.Equation (1) * 4:( 4a + 4b + 4c = 3200 ) ...(5)Now, subtract Equation (5) from Equation (3):( (30a + 10b + 4c) - (4a + 4b + 4c) = 6400 - 3200 )Simplify:( 26a + 6b = 3200 )Divide both sides by 2:( 13a + 3b = 1600 ) ...(6)Now, we have two equations:Equation (4): ( 5a + b = 533.overline{3} )Equation (6): ( 13a + 3b = 1600 )Let me solve Equation (4) for ( b ):( b = 533.overline{3} - 5a )Now, substitute this into Equation (6):( 13a + 3(533.overline{3} - 5a) = 1600 )Simplify:( 13a + 1600 - 15a = 1600 )Combine like terms:( -2a + 1600 = 1600 )Subtract 1600 from both sides:( -2a = 0 )So, ( a = 0 )Wait, that can't be right. If ( a = 0 ), then the quadratic model becomes linear, but the problem states it's a quadratic growth model. Maybe I made a mistake in my calculations.Let me check my steps.Starting from Equation (4): ( 5a + b = 533.overline{3} )Equation (6): ( 13a + 3b = 1600 )Express ( b ) from Equation (4):( b = 533.overline{3} - 5a )Substitute into Equation (6):( 13a + 3*(533.overline{3} - 5a) = 1600 )Calculate 3*(533.333...):533.333... * 3 = 1600So, 13a + 1600 - 15a = 1600Combine like terms:(13a -15a) + 1600 = 1600-2a + 1600 = 1600Subtract 1600:-2a = 0So, a = 0Hmm, same result. So, this suggests that a = 0, which would make the model linear, but the problem says quadratic. Maybe there's an error in setting up the equations.Wait, let's go back to the total yardage equation. The total yardage is 6400 yards over four years, which is the sum of y1 + y2 + y3 + y4.I had:( y1 = a + b + c = 800 )( y2 = 4a + 2b + c )( y3 = 9a + 3b + c )( y4 = 16a + 4b + c = 2400 )Sum: 30a + 10b + 4c = 6400Wait, is that correct? Let me recalculate the sum.Sum = y1 + y2 + y3 + y4= (a + b + c) + (4a + 2b + c) + (9a + 3b + c) + (16a + 4b + c)= (1 + 4 + 9 + 16)a + (1 + 2 + 3 + 4)b + (1 + 1 + 1 + 1)c= 30a + 10b + 4cYes, that seems correct.So, 30a + 10b + 4c = 6400But when I subtracted 4*(Equation 1) from Equation 3, I got 26a + 6b = 3200, which simplifies to 13a + 3b = 1600.Then, from Equation 2 - Equation 1: 15a + 3b = 1600Wait, hold on. Equation 2 - Equation 1 was:16a + 4b + c - (a + b + c) = 2400 - 800Which is 15a + 3b = 1600So, Equation (4): 15a + 3b = 1600Equation (6): 13a + 3b = 1600Wait, so both Equation (4) and Equation (6) equal 1600? That can't be unless 15a + 3b = 13a + 3b, which would imply 15a = 13a, so 2a = 0, hence a = 0.But that contradicts the quadratic model. So, perhaps my initial setup is wrong.Wait, maybe the quadratic model is cumulative? Or is it the total yardage each year?Wait, the problem says "Alex’s yard gains over the four high school seasons followed a quadratic growth model." So, does that mean that the total yardage each year is quadratic, or the cumulative yardage is quadratic?I think it's the total yardage each year, so y is the yardage in year x, so y1 is freshman, y2 sophomore, etc.So, then, the sum of y1 + y2 + y3 + y4 is 6400.So, that's correct.But according to the equations, a = 0, which is conflicting.Wait, maybe the problem is that the quadratic model is cumulative, meaning that the total yardage up to year x is quadratic. So, in that case, the total after 4 years is 6400, which is y(4) = 6400.But the problem says "Alex’s yard gains over the four high school seasons followed a quadratic growth model." So, it's the yard gains each year, not cumulative.So, that would mean y1 = 800, y4 = 2400, and sum y1+y2+y3+y4 = 6400.So, the equations are correct.But solving them leads to a = 0, which is a problem.Wait, maybe I made a mistake in the total yardage equation.Wait, let me compute the sum again:Sum = y1 + y2 + y3 + y4 = (a + b + c) + (4a + 2b + c) + (9a + 3b + c) + (16a + 4b + c)So, adding up the a terms: 1 + 4 + 9 + 16 = 30ab terms: 1 + 2 + 3 + 4 = 10bc terms: 1 + 1 + 1 + 1 = 4cSo, 30a + 10b + 4c = 6400. That seems correct.So, if I have:Equation 1: a + b + c = 800Equation 2: 16a + 4b + c = 2400Equation 3: 30a + 10b + 4c = 6400Then, subtract Equation 1 from Equation 2:15a + 3b = 1600 ...(4)Multiply Equation 1 by 4: 4a + 4b + 4c = 3200 ...(5)Subtract Equation 5 from Equation 3:(30a + 10b + 4c) - (4a + 4b + 4c) = 6400 - 320026a + 6b = 3200Divide by 2: 13a + 3b = 1600 ...(6)Now, Equation 4: 15a + 3b = 1600Equation 6: 13a + 3b = 1600Subtract Equation 6 from Equation 4:(15a + 3b) - (13a + 3b) = 1600 - 16002a = 0 => a = 0So, a = 0, which would make the model linear. But the problem states it's quadratic. Hmm.Is there a mistake in the problem statement? Or maybe I misinterpreted something.Wait, maybe the quadratic model is for cumulative yardage, not annual yardage. Let me check.The problem says: "Alex’s yard gains over the four high school seasons followed a quadratic growth model." It doesn't specify whether it's cumulative or annual. Hmm.If it's cumulative, then y(x) would represent the total yardage up to year x. So, y(1) = 800, y(4) = 6400. Then, the total yardage in senior year would be y(4) - y(3) = 2400.So, let's try this approach.Let me redefine the problem:If y(x) is the cumulative yardage up to year x, then:y(1) = 800y(4) = 6400And the yardage in senior year is y(4) - y(3) = 2400So, let's set up the equations accordingly.Given y(x) = ax^2 + bx + cSo,For x = 1: a(1)^2 + b(1) + c = 800 => a + b + c = 800 ...(1)For x = 4: a(4)^2 + b(4) + c = 6400 => 16a + 4b + c = 6400 ...(2)Also, the yardage in senior year is 2400, which is y(4) - y(3) = 2400Compute y(3): a(3)^2 + b(3) + c = 9a + 3b + cSo, y(4) - y(3) = (16a + 4b + c) - (9a + 3b + c) = 7a + b = 2400 ...(3)Now, we have three equations:1. a + b + c = 8002. 16a + 4b + c = 64003. 7a + b = 2400Now, let's solve this system.From Equation 3: 7a + b = 2400 => b = 2400 - 7a ...(4)Substitute Equation 4 into Equation 1:a + (2400 - 7a) + c = 800Simplify:a + 2400 - 7a + c = 800-6a + c = 800 - 2400-6a + c = -1600So, c = 6a - 1600 ...(5)Now, substitute Equation 4 and Equation 5 into Equation 2:16a + 4b + c = 6400Substitute b = 2400 - 7a and c = 6a - 1600:16a + 4*(2400 - 7a) + (6a - 1600) = 6400Calculate:16a + 9600 - 28a + 6a - 1600 = 6400Combine like terms:(16a -28a +6a) + (9600 -1600) = 6400(-6a) + 8000 = 6400-6a = 6400 - 8000-6a = -1600Divide both sides by -6:a = (-1600)/(-6) = 1600/6 ≈ 266.666...So, a = 266.666... or 800/3Then, from Equation 4: b = 2400 - 7a = 2400 - 7*(800/3) = 2400 - 5600/3Convert 2400 to thirds: 2400 = 7200/3So, b = 7200/3 - 5600/3 = 1600/3 ≈ 533.333...From Equation 5: c = 6a - 1600 = 6*(800/3) - 1600 = 1600 - 1600 = 0So, c = 0Therefore, the quadratic model is:y(x) = (800/3)x^2 + (1600/3)x + 0Simplify:y(x) = (800/3)x^2 + (1600/3)xWe can factor out 800/3:y(x) = (800/3)(x^2 + 2x)Alternatively, we can write it as:y(x) = (800/3)x(x + 2)But let's check if this makes sense.Compute y(1): (800/3)(1)(3) = 800, which matches.Compute y(4): (800/3)(4)(6) = (800/3)*24 = 800*8 = 6400, which matches.Compute y(3): (800/3)(3)(5) = (800/3)*15 = 800*5 = 4000Then, senior year yardage is y(4) - y(3) = 6400 - 4000 = 2400, which matches.So, this seems correct.Therefore, the coefficients are:a = 800/3 ≈ 266.666...b = 1600/3 ≈ 533.333...c = 0So, in fraction form, a = 800/3, b = 1600/3, c = 0.Alternatively, as decimals, a ≈ 266.67, b ≈ 533.33, c = 0.But since the problem didn't specify the form, I can present them as fractions.So, that's the solution for part 1.Now, moving on to part 2. Coach Martinez noticed that Alex's maximum single-game yardage in his senior year followed a sinusoidal pattern over the 10-game season, peaking at 350 yards in week 6 and reaching its minimum of 150 yards in week 1. We need to model this using a sinusoidal function of the form ( y = A sin(B(t - C)) + D ), and determine A, B, C, D.First, let's recall that a sinusoidal function can be written in the form:( y = A sin(B(t - C)) + D )Where:- A is the amplitude- B affects the period- C is the phase shift- D is the vertical shift (midline)Given that the maximum is 350 yards and the minimum is 150 yards, we can find the amplitude and the midline.The amplitude A is half the difference between the maximum and minimum values.So, A = (max - min)/2 = (350 - 150)/2 = 200/2 = 100The midline D is the average of the maximum and minimum.D = (max + min)/2 = (350 + 150)/2 = 500/2 = 250So, A = 100, D = 250Now, we need to find B and C.The function is a sine function, so we need to determine its period and phase shift.First, the period. The season has 10 games, so the function completes one full cycle over 10 weeks. Therefore, the period is 10 weeks.The general formula for the period of a sine function is ( text{Period} = frac{2pi}{B} )So, ( frac{2pi}{B} = 10 )Solving for B:( B = frac{2pi}{10} = frac{pi}{5} )So, B = π/5Now, we need to find the phase shift C.We know that the maximum occurs at week 6. In a sine function, the maximum occurs at ( frac{pi}{2} ) radians from the start of the cycle. So, we can set up the equation:( B(t - C) = frac{pi}{2} ) when t = 6Plugging in B = π/5:( frac{pi}{5}(6 - C) = frac{pi}{2} )Divide both sides by π:( frac{1}{5}(6 - C) = frac{1}{2} )Multiply both sides by 5:( 6 - C = frac{5}{2} )So,( C = 6 - frac{5}{2} = frac{12}{2} - frac{5}{2} = frac{7}{2} = 3.5 )Therefore, C = 3.5So, putting it all together, the function is:( y = 100 sinleft( frac{pi}{5}(t - 3.5) right) + 250 )Let me verify this.At t = 6:( y = 100 sinleft( frac{pi}{5}(6 - 3.5) right) + 250 = 100 sinleft( frac{pi}{5}(2.5) right) + 250 )Calculate the argument:( frac{pi}{5} * 2.5 = frac{pi}{5} * frac{5}{2} = frac{pi}{2} )So, sin(π/2) = 1Thus, y = 100*1 + 250 = 350, which is correct.At t = 1:( y = 100 sinleft( frac{pi}{5}(1 - 3.5) right) + 250 = 100 sinleft( frac{pi}{5}(-2.5) right) + 250 )Calculate the argument:( frac{pi}{5}*(-2.5) = -frac{pi}{2} )Sin(-π/2) = -1Thus, y = 100*(-1) + 250 = -100 + 250 = 150, which is correct.So, the function is correctly modeled.Therefore, the values are:A = 100B = π/5C = 3.5D = 250Alternatively, C can be written as 7/2.So, summarizing:A = 100B = π/5C = 7/2 or 3.5D = 250I think that's all for part 2.Final Answer1. The quadratic model coefficients are ( a = boxed{dfrac{800}{3}} ), ( b = boxed{dfrac{1600}{3}} ), and ( c = boxed{0} ).2. The sinusoidal function parameters are ( A = boxed{100} ), ( B = boxed{dfrac{pi}{5}} ), ( C = boxed{dfrac{7}{2}} ), and ( D = boxed{250} ).</think>

answer:<think>Okay, so I've got this problem about a historical building restoration project. It's divided into two parts. Let me try to tackle each part step by step.Starting with the first part: The floor plan is a convex polygon with n sides. They need to add diagonal braces by choosing 2 non-adjacent vertices. I need to figure out how many ways there are to do this. Hmm, okay, so it's a combinatorial geometry problem.First, let's recall that in a convex polygon with n sides, the number of vertices is also n. To form a diagonal, we need to connect two non-adjacent vertices. So, the total number of diagonals in a convex polygon is given by a formula. I remember it's something like n(n - 3)/2. Let me verify that.Each vertex can be connected to n - 3 other vertices to form a diagonal because you can't connect a vertex to itself or its two adjacent vertices. Since each diagonal is counted twice (once from each end), the total number is n(n - 3)/2. Yeah, that sounds right.But wait, the question is about the number of ways to choose 2 non-adjacent vertices. So, is that the same as the number of diagonals? I think so because each diagonal corresponds to exactly one pair of non-adjacent vertices. So, the number of ways should be n(n - 3)/2.Let me think if there's another way to approach this. The total number of ways to choose any two vertices is C(n, 2), which is n(n - 1)/2. From this, we need to subtract the number of adjacent vertex pairs because those are not diagonals. In a polygon, each vertex has two adjacent vertices, so the number of adjacent pairs is n. Therefore, the number of non-adjacent pairs is C(n, 2) - n = [n(n - 1)/2] - n = [n(n - 1) - 2n]/2 = [n^2 - n - 2n]/2 = [n^2 - 3n]/2 = n(n - 3)/2. Yep, same result. So, that seems solid.Alright, so the answer to part 1 is n(n - 3)/2. Got that.Moving on to part 2: The society's budget is constrained by an exponential decay model. The initial budget is B0, and it decays continuously at a rate k per year. I need to formulate the budget function B(t) and find the time t when the budget is half of its initial value.Okay, exponential decay. The general formula for continuous decay is B(t) = B0 * e^(-kt). That's because continuous decay is modeled by the exponential function with base e. Let me recall: the formula is N(t) = N0 * e^(-λt), where λ is the decay constant. So, in this case, k is the decay rate, so yes, B(t) = B0 * e^(-kt).Now, to find the time t when the budget is half of its initial value, we set B(t) = B0 / 2 and solve for t.So, B0 / 2 = B0 * e^(-kt)Divide both sides by B0: 1/2 = e^(-kt)Take the natural logarithm of both sides: ln(1/2) = ln(e^(-kt)) => ln(1/2) = -ktWe know that ln(1/2) is equal to -ln(2), so:- ln(2) = -ktMultiply both sides by -1: ln(2) = ktTherefore, t = ln(2) / kSo, the time it takes for the budget to halve is the natural logarithm of 2 divided by the decay rate k.Let me double-check that. If I plug t = ln(2)/k into B(t):B(t) = B0 * e^(-k*(ln(2)/k)) = B0 * e^(-ln(2)) = B0 * (1/e^{ln(2)}) = B0 * (1/2) = B0/2. Perfect, that works.So, summarizing part 2: The budget function is B(t) = B0 * e^(-kt), and the time to half the budget is t = ln(2)/k.I think that covers both parts. Let me just recap:1. For the number of non-adjacent vertex pairs in an n-sided convex polygon, it's n(n - 3)/2.2. The budget function is exponential decay, B(t) = B0 * e^(-kt), and the half-life is ln(2)/k.Yeah, that seems right. I don't think I made any mistakes here. The first part was about combinatorics, making sure to subtract the adjacent pairs from the total pairs, and the second part was applying the exponential decay formula and solving for the half-life.Final Answer1. The number of ways is boxed{dfrac{n(n - 3)}{2}}.2. The budget function is ( B(t) = B_0 e^{-kt} ) and the time when the budget is half is boxed{dfrac{ln 2}{k}}.</think>

answer:<think>Alright, so I have this differential equation to solve: [frac{d^2S}{dt^2} + 5frac{dS}{dt} + 6S = e^{-2t}]with initial conditions ( S(0) = 75 ) and ( frac{dS}{dt}(0) = 5 ). Hmm, okay, it's a second-order linear nonhomogeneous differential equation. I remember that to solve such equations, I need to find the general solution to the homogeneous equation and then find a particular solution for the nonhomogeneous part.First, let me write down the homogeneous equation:[frac{d^2S}{dt^2} + 5frac{dS}{dt} + 6S = 0]To solve this, I need the characteristic equation. The characteristic equation is obtained by replacing the derivatives with powers of ( r ):[r^2 + 5r + 6 = 0]Let me solve this quadratic equation. The discriminant is ( 25 - 24 = 1 ), so the roots are:[r = frac{-5 pm sqrt{1}}{2} = frac{-5 pm 1}{2}]So, the roots are ( r = -2 ) and ( r = -3 ). Both are real and distinct. Therefore, the general solution to the homogeneous equation is:[S_h(t) = C_1 e^{-2t} + C_2 e^{-3t}]Okay, now I need to find a particular solution ( S_p(t) ) for the nonhomogeneous equation. The right-hand side is ( e^{-2t} ). Hmm, I notice that ( e^{-2t} ) is already a solution to the homogeneous equation because ( r = -2 ) is a root. So, in this case, I need to use the method of undetermined coefficients with a modification. Specifically, I should multiply by ( t ) to find a particular solution.So, let me assume a particular solution of the form:[S_p(t) = A t e^{-2t}]Now, I need to compute the first and second derivatives of ( S_p(t) ):First derivative:[S_p'(t) = A e^{-2t} + A t (-2) e^{-2t} = A e^{-2t} - 2A t e^{-2t}]Second derivative:[S_p''(t) = -2A e^{-2t} - 2A e^{-2t} + 4A t e^{-2t} = (-4A e^{-2t}) + 4A t e^{-2t}]Now, substitute ( S_p(t) ), ( S_p'(t) ), and ( S_p''(t) ) into the original differential equation:[(-4A e^{-2t} + 4A t e^{-2t}) + 5(A e^{-2t} - 2A t e^{-2t}) + 6(A t e^{-2t}) = e^{-2t}]Let me expand this:First term: ( -4A e^{-2t} + 4A t e^{-2t} )Second term: ( 5A e^{-2t} - 10A t e^{-2t} )Third term: ( 6A t e^{-2t} )Combine all terms:For ( e^{-2t} ) terms: ( -4A + 5A = A )For ( t e^{-2t} ) terms: ( 4A - 10A + 6A = 0 )So, the equation becomes:[A e^{-2t} = e^{-2t}]Therefore, ( A = 1 ).So, the particular solution is:[S_p(t) = t e^{-2t}]Therefore, the general solution to the nonhomogeneous equation is:[S(t) = S_h(t) + S_p(t) = C_1 e^{-2t} + C_2 e^{-3t} + t e^{-2t}]Now, I need to apply the initial conditions to find ( C_1 ) and ( C_2 ).First, compute ( S(0) = 75 ):[S(0) = C_1 e^{0} + C_2 e^{0} + 0 times e^{0} = C_1 + C_2 = 75]So, equation (1): ( C_1 + C_2 = 75 )Next, compute the first derivative ( S'(t) ):First, let's find ( S'(t) ):[S'(t) = -2 C_1 e^{-2t} - 3 C_2 e^{-3t} + e^{-2t} - 2 t e^{-2t}]Simplify:[S'(t) = (-2 C_1 + 1) e^{-2t} - 3 C_2 e^{-3t} - 2 t e^{-2t}]Now, evaluate ( S'(0) = 5 ):[S'(0) = (-2 C_1 + 1) e^{0} - 3 C_2 e^{0} - 0 = (-2 C_1 + 1) - 3 C_2 = 5]So, equation (2): ( -2 C_1 - 3 C_2 + 1 = 5 )Simplify equation (2):[-2 C_1 - 3 C_2 = 4]Now, we have a system of two equations:1. ( C_1 + C_2 = 75 )2. ( -2 C_1 - 3 C_2 = 4 )Let me solve this system.From equation (1): ( C_1 = 75 - C_2 )Substitute into equation (2):[-2 (75 - C_2) - 3 C_2 = 4]Compute:[-150 + 2 C_2 - 3 C_2 = 4]Simplify:[-150 - C_2 = 4]So,[- C_2 = 154 implies C_2 = -154]Then, from equation (1):[C_1 = 75 - (-154) = 75 + 154 = 229]So, ( C_1 = 229 ) and ( C_2 = -154 )Therefore, the particular solution is:[S(t) = 229 e^{-2t} - 154 e^{-3t} + t e^{-2t}]Hmm, let me check if I did everything correctly.Wait, when I substituted into equation (2):Original equation (2): ( -2 C_1 - 3 C_2 + 1 = 5 )So, ( -2 C_1 - 3 C_2 = 4 )Yes, that's correct.Then, substituting ( C_1 = 75 - C_2 ):( -2(75 - C_2) - 3 C_2 = 4 )Which is ( -150 + 2 C_2 - 3 C_2 = 4 )Simplify: ( -150 - C_2 = 4 implies -C_2 = 154 implies C_2 = -154 )Then, ( C_1 = 75 - (-154) = 229 ). That seems correct.So, the solution is:[S(t) = 229 e^{-2t} - 154 e^{-3t} + t e^{-2t}]Wait, let me check the derivative again.Compute ( S'(t) ):[S(t) = 229 e^{-2t} - 154 e^{-3t} + t e^{-2t}]So,First derivative:[S'(t) = -458 e^{-2t} + 462 e^{-3t} + e^{-2t} - 2 t e^{-2t}]Simplify:Combine ( e^{-2t} ) terms: ( (-458 + 1) e^{-2t} = -457 e^{-2t} )Then, ( 462 e^{-3t} ) and ( -2 t e^{-2t} )So,[S'(t) = -457 e^{-2t} + 462 e^{-3t} - 2 t e^{-2t}]At ( t = 0 ):[S'(0) = -457 + 462 + 0 = 5]Which matches the initial condition. So that's correct.Therefore, the solution is correct.Now, moving on to part 2: Calculate the limit of ( S(t) ) as ( t ) approaches infinity.So,[lim_{t to infty} S(t) = lim_{t to infty} left( 229 e^{-2t} - 154 e^{-3t} + t e^{-2t} right)]Let me analyze each term:1. ( 229 e^{-2t} ): As ( t to infty ), ( e^{-2t} to 0 ), so this term goes to 0.2. ( -154 e^{-3t} ): Similarly, as ( t to infty ), ( e^{-3t} to 0 ), so this term also goes to 0.3. ( t e^{-2t} ): Here, we have ( t ) multiplied by ( e^{-2t} ). As ( t to infty ), ( e^{-2t} ) decays exponentially, which dominates the linear growth of ( t ). Therefore, this term also approaches 0.Hence, the limit is 0 + 0 + 0 = 0.Wait, but that seems counterintuitive. If all terms go to zero, then the scores stabilize at zero? But in the context, scores are average scores, which are likely to stabilize at some positive value.Wait, maybe I made a mistake. Let me think again.Wait, the differential equation is:[frac{d^2S}{dt^2} + 5frac{dS}{dt} + 6S = e^{-2t}]As ( t to infty ), the right-hand side ( e^{-2t} ) tends to zero. So, the nonhomogeneous term disappears, and the solution should approach the solution of the homogeneous equation.But in our particular solution, we have ( t e^{-2t} ), which also tends to zero.Wait, but in the homogeneous solution, both ( e^{-2t} ) and ( e^{-3t} ) tend to zero. So, the entire solution tends to zero. But in reality, scores don't go to zero. Maybe the model is such that the scores decay to zero, but that seems odd.Alternatively, perhaps I made a mistake in the particular solution. Let me double-check.Wait, the particular solution was ( t e^{-2t} ). When I substituted back, I got ( A = 1 ). Let me verify that substitution again.Original substitution:After plugging ( S_p(t) = A t e^{-2t} ) into the differential equation, we had:[(-4A e^{-2t} + 4A t e^{-2t}) + 5(A e^{-2t} - 2A t e^{-2t}) + 6(A t e^{-2t}) = e^{-2t}]Simplify term by term:First term: ( -4A e^{-2t} + 4A t e^{-2t} )Second term: ( 5A e^{-2t} - 10A t e^{-2t} )Third term: ( 6A t e^{-2t} )Combine like terms:For ( e^{-2t} ): ( (-4A + 5A) = A )For ( t e^{-2t} ): ( (4A - 10A + 6A) = 0 )So, the equation reduces to ( A e^{-2t} = e^{-2t} implies A = 1 ). That seems correct.Therefore, the particular solution is correct.So, the general solution is correct, and as ( t to infty ), all terms go to zero.But in the context, this would mean that Marieke’s scores stabilize at zero, which doesn't make sense because scores are positive. Maybe the model is not appropriate, or perhaps I misinterpreted the question.Wait, the question says that the scores oscillate around a certain value before stabilizing. But according to the solution, they stabilize at zero. Maybe I made a mistake in interpreting the differential equation.Wait, let me check the differential equation again:[frac{d^2S}{dt^2} + 5frac{dS}{dt} + 6S = e^{-2t}]Yes, that's correct. So, the forcing function is ( e^{-2t} ), which decays to zero. So, the steady-state solution would be the particular solution, but in this case, the particular solution also tends to zero.Wait, but in some cases, when the particular solution is of the form ( t e^{-2t} ), the transient term might have a different behavior. Let me think about the behavior as ( t to infty ).Each term:- ( 229 e^{-2t} ): decays exponentially- ( -154 e^{-3t} ): decays exponentially faster- ( t e^{-2t} ): decays to zero because exponential decay dominates polynomial growthSo, yes, all terms go to zero. Therefore, the limit is zero.But the question says that the scores oscillate around a certain value before stabilizing. Hmm, in our solution, there is no oscillation because the roots are both real and negative, so the solution doesn't oscillate. It just decays to zero.Wait, but the question mentions oscillation. Maybe I made a mistake in solving the differential equation.Wait, let me double-check the characteristic equation.The homogeneous equation is:[r^2 + 5r + 6 = 0]Which factors as ( (r + 2)(r + 3) = 0 ), so roots at ( r = -2 ) and ( r = -3 ). Both are real and negative, so the solution doesn't oscillate. So, why does the question mention oscillation?Wait, perhaps the particular solution causes some oscillation? But in our case, the particular solution is ( t e^{-2t} ), which is a decaying exponential multiplied by t, so it doesn't oscillate either.Hmm, maybe I misread the question. Let me check again.The function ( S(t) ) is modeled by the differential equation:[frac{d^2S}{dt^2} + 5frac{dS}{dt} + 6S = e^{-2t}]with initial conditions ( S(0) = 75 ) and ( frac{dS}{dt}(0) = 5 ).So, the equation is correct. The homogeneous solution has real roots, so no oscillation. The particular solution is also non-oscillatory. Therefore, the overall solution doesn't oscillate. So, the question's statement about oscillation might be a red herring, or perhaps it's referring to something else.Alternatively, maybe the question is expecting a different kind of solution, but I don't see how. The differential equation is linear with constant coefficients, so the solution is as I found.Therefore, despite the question mentioning oscillation, the solution doesn't oscillate because the characteristic roots are real. So, the limit as ( t to infty ) is zero.But in the context of scores, it's odd. Maybe the model is incorrect, but given the problem, I have to go with the mathematics.So, summarizing:1. The solution to the differential equation is:[S(t) = 229 e^{-2t} - 154 e^{-3t} + t e^{-2t}]2. The limit as ( t to infty ) is 0.But wait, let me think again. Maybe I should consider the steady-state solution. The particular solution is ( t e^{-2t} ), which tends to zero. So, the steady-state is zero. But in some cases, if the forcing function were, say, a constant, the steady-state would be a constant. But here, the forcing function decays to zero, so the steady-state is zero.Alternatively, perhaps the question is expecting the limit of the homogeneous solution? But no, the particular solution also tends to zero.Alternatively, maybe I need to consider the behavior of the homogeneous solution as ( t to infty ), which is zero, and the particular solution also tends to zero, so the overall limit is zero.Therefore, despite the mention of oscillation, the limit is zero.Wait, but in the solution, the homogeneous part is ( C_1 e^{-2t} + C_2 e^{-3t} ), which decays to zero, and the particular solution is ( t e^{-2t} ), which also decays to zero. So, yes, the limit is zero.Therefore, the answer is zero.But just to be thorough, let me compute the limit step by step.Compute:[lim_{t to infty} 229 e^{-2t} = 0][lim_{t to infty} -154 e^{-3t} = 0][lim_{t to infty} t e^{-2t} = 0]Because exponential decay dominates polynomial growth. So, indeed, each term tends to zero, so the sum tends to zero.Therefore, the limit is zero.But again, in the context, it's strange. Maybe the question is expecting a different interpretation, but based on the mathematics, the limit is zero.Alternatively, perhaps I made a mistake in the particular solution. Let me check again.Wait, the particular solution was ( t e^{-2t} ). Let me plug it back into the equation to verify.Compute ( S_p(t) = t e^{-2t} )First derivative: ( S_p' = e^{-2t} - 2 t e^{-2t} )Second derivative: ( S_p'' = -2 e^{-2t} - 2 e^{-2t} + 4 t e^{-2t} = (-4 e^{-2t}) + 4 t e^{-2t} )Now, plug into the equation:( S_p'' + 5 S_p' + 6 S_p = (-4 e^{-2t} + 4 t e^{-2t}) + 5(e^{-2t} - 2 t e^{-2t}) + 6(t e^{-2t}) )Compute term by term:-4 e^{-2t} + 4 t e^{-2t} + 5 e^{-2t} - 10 t e^{-2t} + 6 t e^{-2t}Combine like terms:For ( e^{-2t} ): (-4 + 5) = 1For ( t e^{-2t} ): (4 -10 +6) = 0So, total: ( e^{-2t} ), which is equal to the RHS. So, correct.Therefore, the particular solution is correct.Hence, the conclusion is that the limit is zero.But the question says that the scores oscillate around a certain value before stabilizing. Maybe the question is referring to oscillation in the transient response, but in our case, the transient response is decaying exponentials, not oscillations. So, perhaps the question is incorrect in stating that there is oscillation.Alternatively, maybe I made a mistake in the characteristic equation.Wait, characteristic equation is ( r^2 + 5r + 6 = 0 ), which factors as ( (r + 2)(r + 3) = 0 ). So, roots at -2 and -3, which are real and negative. So, no oscillation. So, the solution doesn't oscillate.Therefore, the mention of oscillation in the question is perhaps a mistake, or maybe it's referring to something else.But in any case, based on the mathematics, the solution is as I found, and the limit is zero.Therefore, the answers are:1. ( S(t) = 229 e^{-2t} - 154 e^{-3t} + t e^{-2t} )2. The limit is 0.But just to make sure, let me write the solution again:[S(t) = 229 e^{-2t} - 154 e^{-3t} + t e^{-2t}]Yes, that's correct.So, final answers:1. The solution is ( S(t) = 229 e^{-2t} - 154 e^{-3t} + t e^{-2t} )2. The limit as ( t to infty ) is 0.But wait, the question says "Marieke’s scores oscillate around a certain value before stabilizing." Since our solution doesn't oscillate, perhaps I made a mistake in the differential equation.Wait, let me check the differential equation again.It's:[frac{d^2S}{dt^2} + 5frac{dS}{dt} + 6S = e^{-2t}]Yes, that's correct. So, with real roots, no oscillation.Alternatively, maybe the question intended complex roots, but the characteristic equation has real roots.Wait, unless I made a mistake in solving the characteristic equation.Wait, discriminant is ( 25 - 24 = 1 ), so roots are real.Yes, so no oscillation.Therefore, the mention of oscillation is perhaps a mistake in the question.Alternatively, maybe the particular solution introduces some oscillation? But in this case, the particular solution is ( t e^{-2t} ), which is a decaying function without oscillation.Therefore, the conclusion is that the solution doesn't oscillate, and the limit is zero.So, I think that's the answer.</think>

answer:<think>Okay, so I have this problem about optimizing resource allocation for an Apache Beam pipeline written in Go. It's divided into two parts. Let me try to understand each part step by step.Starting with the first part: The pipeline processes N transactions, where N is 10^6. Each transaction T_i has a processing time P_i that follows a normal distribution with mean μ = 0.5 seconds and standard deviation σ = 0.1 seconds. I need to calculate the expected total processing time for all transactions and determine the standard deviation of the total processing time.Alright, so let's break this down. Each transaction's processing time is a normal random variable. Since all the transactions are processed, the total processing time is the sum of all individual processing times.Let me denote the total processing time as T. So,T = P₁ + P₂ + ... + P_NSince each P_i is normally distributed, the sum of normally distributed variables is also normally distributed. That makes sense because the normal distribution is closed under addition.Now, to find the expected total processing time, E[T], I can use the linearity of expectation. The expectation of a sum is the sum of expectations.E[T] = E[P₁ + P₂ + ... + P_N] = E[P₁] + E[P₂] + ... + E[P_N]Since each E[P_i] is μ = 0.5 seconds, and there are N = 10^6 transactions,E[T] = N * μ = 10^6 * 0.5 = 500,000 seconds.Okay, that seems straightforward. Now, for the standard deviation of T.The variance of the sum of independent random variables is the sum of their variances. Since each P_i is independent (I assume they are independent because there's no indication otherwise), the variance of T is N times the variance of each P_i.Var(T) = Var(P₁) + Var(P₂) + ... + Var(P_N) = N * Var(P_i)Var(P_i) is σ² = (0.1)^2 = 0.01.So,Var(T) = 10^6 * 0.01 = 10,000.Therefore, the standard deviation of T is the square root of Var(T):SD(T) = sqrt(10,000) = 100 seconds.Wait, that seems a bit high, but considering we're summing a million variables each with a standard deviation of 0.1, it makes sense. The standard deviation scales with the square root of N, so sqrt(10^6) is 1000, but wait, no, hold on.Wait, Var(T) is N * σ², so sqrt(N * σ²) = sqrt(N) * σ.So, sqrt(10^6) is 1000, and 1000 * 0.1 is 100. Yeah, that's correct. So the standard deviation is 100 seconds.Alright, so that's part 1 done. The expected total processing time is 500,000 seconds with a standard deviation of 100 seconds.Moving on to part 2: The cost function C is given by C = k * t², where k is a constant and t is the total processing time from part 1. I need to determine the expected cost and its variance.Hmm, okay. So, C is a function of t, which is a random variable. So, I need to find E[C] and Var(C).Given that t is normally distributed with mean μ_t = 500,000 and standard deviation σ_t = 100, so t ~ N(500,000, 100²).But wait, actually, t is a sum of a million normal variables, so it's also normal. So, t is N(500,000, 10,000). Wait, no, variance is 10,000, so standard deviation is 100.So, t ~ N(500,000, 100²).Now, C = k * t². So, we need to find E[C] and Var(C).First, let's find E[C] = E[k * t²] = k * E[t²].We know that for any random variable, E[t²] = Var(t) + (E[t])².So,E[t²] = Var(t) + (E[t])² = 10,000 + (500,000)².Calculating that:(500,000)^2 = 250,000,000,000So,E[t²] = 250,000,000,000 + 10,000 = 250,000,010,000.Therefore,E[C] = k * 250,000,010,000.But since k is a constant, we can just leave it as k multiplied by that value. So, E[C] = k * (250,000,010,000).Now, for the variance of C, Var(C) = Var(k * t²) = k² * Var(t²).So, we need to find Var(t²).To find Var(t²), we can use the formula:Var(t²) = E[t⁴] - (E[t²])².So, we need to compute E[t⁴].But t is a normal random variable. For a normal distribution, there are known formulas for moments. Specifically, for a normal variable X ~ N(μ, σ²), the fourth moment E[X⁴] is given by:E[X⁴] = μ⁴ + 6μ²σ² + 3σ⁴.Let me verify that formula. Yes, for a normal distribution, the fourth moment is μ⁴ + 6μ²σ² + 3σ⁴.So, applying that to t:E[t⁴] = (500,000)^4 + 6*(500,000)^2*(100)^2 + 3*(100)^4.Let me compute each term step by step.First term: (500,000)^4.500,000 is 5*10^5, so (5*10^5)^4 = 5^4 * (10^5)^4 = 625 * 10^20 = 6.25 * 10^22.Wait, let me compute it properly:(500,000)^4 = (5*10^5)^4 = 5^4 * 10^(5*4) = 625 * 10^20 = 6.25 * 10^22.Second term: 6*(500,000)^2*(100)^2.First, compute (500,000)^2 = 250,000,000,000.(100)^2 = 10,000.So, 6 * 250,000,000,000 * 10,000 = 6 * 2.5 * 10^11 * 10^4 = 6 * 2.5 * 10^15 = 15 * 10^15 = 1.5 * 10^16.Third term: 3*(100)^4.(100)^4 = 100,000,000.So, 3 * 100,000,000 = 300,000,000 = 3 * 10^8.Now, adding all three terms together:E[t⁴] = 6.25 * 10^22 + 1.5 * 10^16 + 3 * 10^8.But wait, these terms are of vastly different magnitudes. The first term is 10^22, the second is 10^16, and the third is 10^8. So, when adding them, the smaller terms are negligible compared to the largest term.But for precision, let's write them all out:6.25 * 10^22 + 1.5 * 10^16 = 6.25 * 10^22 + 0.000000015 * 10^22 = approximately 6.250000015 * 10^22.Adding 3 * 10^8 is even smaller, so it's negligible. So, for all practical purposes, E[t⁴] ≈ 6.25 * 10^22.But let's be precise. Let me compute each term in the same exponent:First term: 6.25 * 10^22.Second term: 1.5 * 10^16 = 0.000000015 * 10^22.Third term: 3 * 10^8 = 0.0000000000003 * 10^22.So, adding them:6.25 * 10^22 + 0.000000015 * 10^22 + 0.0000000000003 * 10^22 ≈ 6.250000015 * 10^22.So, E[t⁴] ≈ 6.250000015 * 10^22.Now, Var(t²) = E[t⁴] - (E[t²])².We already have E[t²] = 250,000,010,000.So, (E[t²])² = (250,000,010,000)^2.Let me compute that:250,000,010,000 is 2.5000001 * 10^11.So, (2.5000001 * 10^11)^2 = (2.5)^2 * (10^11)^2 + 2 * 2.5 * 0.0000001 * (10^11)^2 + (0.0000001)^2 * (10^11)^2.Wait, that might be complicated. Alternatively, note that 250,000,010,000 = 250,000,000,000 + 10,000.So, (a + b)^2 = a² + 2ab + b², where a = 250,000,000,000 and b = 10,000.So,(250,000,000,000 + 10,000)^2 = (250,000,000,000)^2 + 2 * 250,000,000,000 * 10,000 + (10,000)^2.Compute each term:First term: (250,000,000,000)^2 = (2.5 * 10^11)^2 = 6.25 * 10^22.Second term: 2 * 250,000,000,000 * 10,000 = 2 * 2.5 * 10^11 * 10^4 = 5 * 10^15.Third term: (10,000)^2 = 100,000,000 = 10^8.So, adding them:6.25 * 10^22 + 5 * 10^15 + 10^8.Again, similar to before, the first term is dominant, the second is much smaller, and the third is negligible.So, (E[t²])² ≈ 6.25 * 10^22 + 5 * 10^15 ≈ 6.25 * 10^22 (since 5*10^15 is 0.000000005 * 10^22).Wait, actually, 5*10^15 is 0.000000005 * 10^22? Wait, 10^22 / 10^15 = 10^7, so 5*10^15 = 5 * 10^(-7) * 10^22 = 0.0000005 * 10^22.So, 6.25 * 10^22 + 0.0000005 * 10^22 = 6.2500005 * 10^22.Therefore, (E[t²])² ≈ 6.2500005 * 10^22.Now, Var(t²) = E[t⁴] - (E[t²])² ≈ 6.250000015 * 10^22 - 6.2500005 * 10^22.Wait, that would be negative, which can't be right. Hmm, that suggests I made a mistake in my calculations.Wait, let me double-check. E[t⁴] was approximately 6.250000015 * 10^22, and (E[t²])² was approximately 6.2500005 * 10^22.So, subtracting, 6.250000015 * 10^22 - 6.2500005 * 10^22 = (6.250000015 - 6.2500005) * 10^22 = (-0.000000485) * 10^22 = -4.85 * 10^16.But variance can't be negative. So, I must have made a mistake in calculating E[t⁴] or (E[t²])².Wait, let's go back. Maybe my approximation was too rough. Let's compute E[t⁴] and (E[t²])² more accurately.E[t⁴] = μ⁴ + 6μ²σ² + 3σ⁴.Given μ = 500,000, σ = 100.Compute each term:μ⁴ = (500,000)^4 = 6.25 * 10^22.6μ²σ² = 6*(500,000)^2*(100)^2 = 6*(250,000,000,000)*(10,000) = 6*2.5*10^11*10^4 = 6*2.5*10^15 = 15*10^15 = 1.5*10^16.3σ⁴ = 3*(100)^4 = 3*(100,000,000) = 300,000,000 = 3*10^8.So, E[t⁴] = 6.25*10^22 + 1.5*10^16 + 3*10^8.Similarly, (E[t²])² = (250,000,010,000)^2.Let me compute this more accurately.250,000,010,000 = 250,000,000,000 + 10,000.So, (a + b)^2 = a² + 2ab + b².a = 250,000,000,000, b = 10,000.a² = (250,000,000,000)^2 = 6.25*10^22.2ab = 2*250,000,000,000*10,000 = 5*10^15.b² = (10,000)^2 = 10^8.So, (E[t²])² = 6.25*10^22 + 5*10^15 + 10^8.Therefore, Var(t²) = E[t⁴] - (E[t²])² = [6.25*10^22 + 1.5*10^16 + 3*10^8] - [6.25*10^22 + 5*10^15 + 10^8].Subtracting term by term:6.25*10^22 - 6.25*10^22 = 0.1.5*10^16 - 5*10^15 = (1.5 - 0.5)*10^16 = 1*10^16.3*10^8 - 10^8 = 2*10^8.So, Var(t²) = 1*10^16 + 2*10^8.Again, 10^16 is much larger than 10^8, so Var(t²) ≈ 1*10^16.But let's write it as 10,000,000,000,000,000 + 200,000,000 = 10,000,000,200,000,000.But in terms of powers of 10, it's approximately 1.00000002 * 10^16.But for variance, we can consider it as 10^16 + 2*10^8, but since 10^16 is dominant, we can approximate Var(t²) ≈ 10^16.Wait, but let's see:Var(t²) = 1.5*10^16 + 3*10^8 - 5*10^15 - 10^8 = (1.5*10^16 - 5*10^15) + (3*10^8 - 10^8) = (1*10^16) + (2*10^8) = 1.00000002*10^16.So, approximately 1.00000002*10^16.But for the purposes of variance, we can consider it as 1.00000002*10^16, but since the question is about the variance, we can express it as 10^16 + 2*10^8.However, in terms of significant figures, given that the original variance was 10,000 and the mean was 500,000, the higher order terms are negligible. So, perhaps we can approximate Var(t²) ≈ 2*μ²σ² + 4μ⁴.Wait, no, that's not the case. Wait, actually, for a normal variable, Var(t²) = E[t⁴] - (E[t²])².But we have already computed E[t⁴] and (E[t²])², so Var(t²) is 1.00000002*10^16.But let's see, 1.00000002*10^16 is approximately 10^16, so for practical purposes, we can say Var(t²) ≈ 10^16.But let me check the exact value:E[t⁴] = 6.25*10^22 + 1.5*10^16 + 3*10^8.(E[t²])² = 6.25*10^22 + 5*10^15 + 10^8.So, subtracting:E[t⁴] - (E[t²])² = (6.25*10^22 - 6.25*10^22) + (1.5*10^16 - 5*10^15) + (3*10^8 - 10^8) = 0 + 1*10^16 + 2*10^8.So, Var(t²) = 1*10^16 + 2*10^8.Therefore, Var(t²) = 10,000,000,000,000,000 + 200,000,000 = 10,000,000,200,000,000.But in terms of powers of 10, it's 1.00000002*10^16.So, for the variance of C, which is Var(C) = k² * Var(t²) = k² * (1.00000002*10^16).But since 1.00000002 is very close to 1, we can approximate Var(C) ≈ k² * 10^16.However, to be precise, it's k² * (10^16 + 2*10^8).But given that 10^16 is much larger than 2*10^8, the 2*10^8 is negligible. So, we can say Var(C) ≈ k² * 10^16.But let me check if there's another way to compute Var(t²).Alternatively, since t is normal, t² is a scaled chi-squared distribution. But I think the method I used is correct.So, to summarize:E[C] = k * E[t²] = k * (250,000,010,000).Var(C) = k² * Var(t²) ≈ k² * 10^16.But let me write the exact value:Var(C) = k² * (10^16 + 2*10^8).But since 10^16 is 100,000,000 times larger than 10^8, the 2*10^8 is negligible. So, we can approximate Var(C) ≈ k² * 10^16.Therefore, the expected cost is k multiplied by approximately 250,000,010,000, and the variance of the cost is approximately k squared multiplied by 10^16.But let me write the exact values:E[C] = k * (250,000,010,000).Var(C) = k² * (10,000,000,200,000,000).But to express this in terms of powers of 10:E[C] = k * (2.5000001 * 10^11).Var(C) = k² * (1.00000002 * 10^16).Alternatively, if we want to write it as:E[C] = 2.5000001 * 10^11 * k.Var(C) = 1.00000002 * 10^16 * k².But since the problem doesn't specify the value of k, we can leave it in terms of k.So, to recap:1. Expected total processing time: 500,000 seconds, standard deviation 100 seconds.2. Expected cost: k * (250,000,010,000) ≈ 2.5000001 * 10^11 * k.Variance of cost: k² * (1.00000002 * 10^16) ≈ 1.00000002 * 10^16 * k².But perhaps we can write it more neatly.Alternatively, since the variance is dominated by the 10^16 term, we can write Var(C) ≈ k² * 10^16.But to be precise, it's 1.00000002 * 10^16 * k².But in the context of the problem, maybe we can just write Var(C) = k² * (10^16 + 2*10^8).But given that 10^16 is 100 million times larger than 10^8, the 2*10^8 is negligible, so it's approximately 10^16.Therefore, the expected cost is approximately 2.5 * 10^11 * k, and the variance is approximately 10^16 * k².But let me check the exact value of E[t²]:E[t²] = Var(t) + (E[t])² = 10,000 + (500,000)^2 = 10,000 + 250,000,000,000 = 250,000,010,000.So, E[C] = k * 250,000,010,000.Similarly, Var(t²) = E[t⁴] - (E[t²])² = (6.25*10^22 + 1.5*10^16 + 3*10^8) - (6.25*10^22 + 5*10^15 + 10^8) = 1*10^16 + 2*10^8.So, Var(C) = k² * (1*10^16 + 2*10^8) = k² * 10,000,000,200,000,000.But to express this in a more compact form, we can write it as k² * (10^16 + 2*10^8).Alternatively, factor out 10^8:Var(C) = k² * 10^8 * (10^8 + 2) = k² * 10^8 * 100,000,002.But that might not be necessary.In conclusion, the expected cost is k multiplied by 250,000,010,000, and the variance is k squared multiplied by approximately 10^16.But to be precise, it's k² * (10^16 + 2*10^8).So, summarizing:1. Expected total processing time: 500,000 seconds, standard deviation 100 seconds.2. Expected cost: 250,000,010,000 * k.Variance of cost: (10^16 + 2*10^8) * k².But since 10^16 is much larger, we can approximate the variance as 10^16 * k².Alternatively, if we want to write the exact value, it's 10,000,000,200,000,000 * k².But in terms of powers of 10, it's 1.00000002 * 10^16 * k².So, I think that's the answer.Final Answer1. The expected total processing time is boxed{500000} seconds with a standard deviation of boxed{100} seconds.2. The expected cost is boxed{250000010000k} and the variance of the cost is boxed{10000000200000000k^2}.</think>