🧠:<think>Okay, so I'm trying to help evaluate the effectiveness of three smoking cessation methods: NRT, BC, and CT. The data given is from 300 participants, 100 in each group, with success rates of 45%, 50%, and 65% respectively after 12 weeks. First, I need to calculate the 95% confidence intervals for each method. I remember that for proportions, the confidence interval can be calculated using the formula:CI = p ± z * sqrt((p*(1-p))/n)where p is the sample proportion, z is the z-score for the desired confidence level, and n is the sample size.For a 95% confidence interval, the z-score is approximately 1.96. So, let's compute this for each method.Starting with NRT: p = 0.45, n = 100.Calculating the standard error: sqrt(0.45*0.55/100) = sqrt(0.2475/100) = sqrt(0.002475) ≈ 0.04975.Then, the margin of error is 1.96 * 0.04975 ≈ 0.0975. So, the CI is 0.45 ± 0.0975, which is approximately (0.3525, 0.5475).Next, BC: p = 0.50, n = 100.Standard error: sqrt(0.5*0.5/100) = sqrt(0.25/100) = 0.05.Margin of error: 1.96 * 0.05 = 0.098. So, CI is 0.50 ± 0.098, approximately (0.402, 0.598).Now, CT: p = 0.65, n = 100.Standard error: sqrt(0.65*0.35/100) = sqrt(0.2275/100) = sqrt(0.002275) ≈ 0.0477.Margin of error: 1.96 * 0.0477 ≈ 0.0935. So, CI is 0.65 ± 0.0935, approximately (0.5565, 0.7435).Okay, so that's part one done. Now, moving on to part two: developing a statistical model to test if CT's success rate is significantly higher than the additive effects of NRT and BC.I think this involves testing for an interaction effect. The additive model would predict that the success rate of CT is the sum of the individual effects of NRT and BC. But since success rates are proportions, adding them directly might not be appropriate because they can't exceed 100%. So, maybe we need a different approach.Wait, perhaps we can model this using a logistic regression framework. Let me think. If we consider the success as a binary outcome (quit or not), we can model the probability of success as a function of the therapies.Let’s define the additive model as:logit(p) = β0 + β1*NRT + β2*BCAnd the interaction model would include a term for NRT*BC:logit(p) = β0 + β1*NRT + β2*BC + β3*(NRT*BC)But in our case, each participant is in only one group, so NRT, BC, and CT are mutually exclusive. Hmm, that complicates things because in the data, each participant is assigned to one group, so we can't directly model NRT and BC as separate variables with an interaction term.Alternatively, maybe we can think of the expected success rate under the additive model. If NRT and BC are independent, the expected success rate for CT would be 1 - (1 - p_NRT)*(1 - p_BC). Let me compute that.p_NRT = 0.45, p_BC = 0.50.So, expected p_CT_additive = 1 - (1 - 0.45)*(1 - 0.50) = 1 - (0.55*0.50) = 1 - 0.275 = 0.725.Wait, but the observed p_CT is 0.65, which is lower than 0.725. That suggests that the interaction is negative, meaning the combined therapy is less effective than the additive model would predict. But that contradicts the initial thought that CT is better. Hmm, maybe I made a mistake.Wait, no, actually, the additive model in terms of probabilities isn't simply adding the probabilities because that would lead to probabilities exceeding 1. Instead, the correct additive model for independent events is 1 - (1 - p1)*(1 - p2). So, for NRT and BC, the expected success rate if they were independent would be 1 - (1 - 0.45)(1 - 0.50) = 0.725. But the observed CT success rate is 0.65, which is lower. So, that suggests that the interaction is negative, meaning the combined therapy is less effective than the sum of the individual therapies.But in the problem statement, it's mentioned that CT is a new method integrating both NRT and BC, and the observed success rate is higher than both NRT and BC individually. Wait, 65% is higher than both 45% and 50%. So, why is the expected additive model giving a higher value?Wait, perhaps the additive model isn't the right way to think about it. Maybe the expected success rate under the additive model is p_NRT + p_BC - p_NRT*p_BC, which is the same as 1 - (1 - p_NRT)(1 - p_BC). So, that would be 0.45 + 0.50 - 0.45*0.50 = 0.95 - 0.225 = 0.725. So, the expected success rate under independence is 72.5%, but the observed is 65%, which is lower. So, that suggests that the interaction is negative, meaning the combined therapy is less effective than expected if the therapies were independent.But in the problem, the observed CT is higher than both NRT and BC. So, perhaps the additive model isn't the right comparison. Maybe the expected success rate under additive model is just the average of NRT and BC? No, that doesn't make sense.Alternatively, perhaps the additive model is considering the main effects without interaction, so the expected success rate for CT would be p_NRT + p_BC - p_NRT*p_BC, which is 0.725, but observed is 0.65, so the interaction is negative. But the problem states that CT is a combined therapy, so perhaps we should model it differently.Wait, maybe the additive model is just p_NRT + p_BC, but since probabilities can't exceed 1, that's not directly applicable. Alternatively, perhaps the expected success rate is the maximum of NRT and BC, but that doesn't seem right either.Alternatively, perhaps we should use a chi-square test to compare observed vs expected counts. Let's think about that.Under the additive model, the expected number of successes in CT would be 100 * (1 - (1 - 0.45)(1 - 0.50)) = 100 * 0.725 = 72.5.The observed number of successes in CT is 65. So, we can set up a 2x2 contingency table comparing observed vs expected.But wait, actually, since we have three groups, maybe a better approach is to use a logistic regression model where we include the group as a factor and test for the interaction term.Wait, but in the data, each participant is in only one group, so we can't directly model the interaction between NRT and BC as separate variables. So, perhaps the correct approach is to compare the observed success rate of CT to the expected success rate under the additive model, which is 0.725, and see if the observed 0.65 is significantly different.But since 0.65 is less than 0.725, that would suggest that the interaction is negative, but the problem states that CT is a new method that integrates both, and the observed success rate is higher than both NRT and BC. So, perhaps the additive model isn't the right comparison.Wait, maybe I'm misunderstanding the additive model. Perhaps the additive model is considering the main effects without interaction, so the expected success rate for CT would be the sum of the main effects minus the baseline. But that might not be the case.Alternatively, perhaps the additive model is p_NRT + p_BC - p_baseline, but without knowing the baseline, that's unclear.Wait, maybe a better approach is to use a logistic regression model where we have a dummy variable for each group, and then test whether the coefficient for CT is significantly higher than the sum of the coefficients for NRT and BC.But since we only have three groups, and each participant is in one group, we can't directly model the interaction between NRT and BC as separate variables. So, perhaps the correct approach is to use a likelihood ratio test comparing a model with main effects only to a model with main effects plus interaction.But in this case, since the groups are mutually exclusive, the interaction term isn't directly estimable. So, perhaps the correct approach is to compare the observed success rate of CT to the expected success rate under the additive model, which is 0.725, and perform a hypothesis test.So, the null hypothesis would be that the success rate of CT is equal to the additive model's prediction, and the alternative is that it's different.Given that, we can compute the expected number of successes under the null hypothesis, which is 72.5, and the observed is 65. We can perform a chi-square test or a z-test for proportions.Using the z-test for proportions, the standard error under the null hypothesis would be sqrt(p*(1-p)/n), where p is 0.725, n is 100.So, SE = sqrt(0.725*0.275/100) = sqrt(0.198125/100) = sqrt(0.00198125) ≈ 0.0445.The observed proportion is 0.65, so the z-score is (0.65 - 0.725)/0.0445 ≈ (-0.075)/0.0445 ≈ -1.685.The p-value for a two-tailed test would be approximately 0.091, which is greater than 0.05, so we fail to reject the null hypothesis. Therefore, the interaction effect is not statistically significant.Wait, but the observed CT success rate is higher than both NRT and BC, but lower than the additive model's prediction. So, the interaction is negative, but not statistically significant.Alternatively, maybe the additive model is not the right way to model the expected success rate. Perhaps the expected success rate for CT should be the average of NRT and BC, but that doesn't make sense because the average of 45% and 50% is 47.5%, which is lower than CT's 65%.Alternatively, perhaps the additive model is considering the main effects without interaction, so the expected success rate for CT would be the sum of the main effects minus the baseline. But without knowing the baseline, that's unclear.Wait, maybe another approach is to use a 2x2x2 contingency table, but since we only have three groups, that's not applicable.Alternatively, perhaps we can use a t-test to compare the observed CT success rate to the expected additive model's success rate. But since we're dealing with proportions, a z-test is more appropriate.So, to summarize, the expected success rate under the additive model is 72.5%, observed is 65%, z-score is approximately -1.685, p-value ~0.091, which is not significant at alpha=0.05. Therefore, we cannot conclude that the interaction effect is statistically significant.But wait, the problem states that CT is a new method that integrates both NRT and BC, and the observed success rate is higher than both. So, maybe the additive model is not the right comparison, and instead, we should compare CT to the sum of NRT and BC, but that's not straightforward.Alternatively, perhaps the correct approach is to model the interaction as the difference between CT and the sum of NRT and BC. But since probabilities can't exceed 1, that's not directly applicable.Wait, perhaps the correct way is to model the expected success rate of CT as p_NRT + p_BC - p_NRT*p_BC, which is 0.725, and then test if the observed 0.65 is significantly different from 0.725. As we did earlier, the z-test gives p≈0.091, which is not significant.Alternatively, maybe we should use a chi-square test comparing observed vs expected counts.Expected successes in CT: 72.5Observed: 65So, the chi-square statistic is (65 - 72.5)^2 / 72.5 = ( -7.5 )^2 /72.5 = 56.25 /72.5 ≈ 0.775.Degrees of freedom is 1, so the p-value is approximately 0.379, which is also not significant.Therefore, we cannot conclude that the interaction effect is statistically significant.Wait, but the problem says "the observed success rate of CT is significantly higher than what would be expected based on the additive effects of NRT and BC alone." So, perhaps the expected additive model is different.Wait, maybe the additive model is considering the main effects without interaction, so the expected success rate for CT would be p_NRT + p_BC - p_NRT*p_BC, which is 0.725, but observed is 0.65, which is lower. So, the interaction is negative, but not significant.Alternatively, perhaps the additive model is just p_NRT + p_BC, but that would be 0.95, which is higher than 1, so that's not possible.Alternatively, maybe the additive model is considering the average of NRT and BC, which is 0.475, but CT is 0.65, which is higher. So, perhaps the expected additive model is 0.475, and we can test if 0.65 is significantly higher than 0.475.In that case, the z-score would be (0.65 - 0.475)/sqrt(0.475*0.525/100) ≈ (0.175)/sqrt(0.249375/100) ≈ 0.175/0.0499 ≈ 3.507, which is significant at p <0.001.But that seems to contradict the initial thought that the additive model is 0.725.Hmm, I'm getting confused here. Maybe I need to clarify what the additive model entails.In the context of interaction effects, the additive model assumes that the effect of CT is the sum of the effects of NRT and BC. But since these are probabilities, the additive model isn't straightforward. The correct way to model the additive effect is 1 - (1 - p_NRT)(1 - p_BC), which is 0.725.So, the expected success rate under the additive model is 0.725, and the observed is 0.65. So, the interaction effect is the difference between observed and expected, which is 0.65 - 0.725 = -0.075. So, the interaction is negative, meaning the combined therapy is less effective than expected if the therapies were independent.But the problem states that CT is a new method that integrates both, and the observed success rate is higher than both NRT and BC. So, perhaps the additive model is not the right way to think about it, and instead, we should consider that the additive model is p_NRT + p_BC - p_NRT*p_BC, but that's the same as 1 - (1 - p_NRT)(1 - p_BC).Alternatively, maybe the additive model is considering the main effects without interaction, so the expected success rate for CT would be p_NRT + p_BC - p_NRT*p_BC, which is 0.725, and we can test if the observed 0.65 is significantly different.As we calculated earlier, the z-test gives a p-value of ~0.091, which is not significant at alpha=0.05. Therefore, we cannot conclude that the interaction effect is statistically significant.Alternatively, perhaps the additive model is considering the main effects as separate, so the expected success rate for CT would be p_NRT + p_BC - p_baseline, but without knowing the baseline, that's unclear.Wait, maybe another approach is to use a logistic regression model where we have a dummy variable for each group, and then test whether the coefficient for CT is significantly different from the sum of the coefficients for NRT and BC.But since we only have three groups, and each participant is in one group, the model would have two dummy variables (e.g., NRT and BC), and CT is the reference category. Then, the coefficient for NRT represents the log odds ratio of NRT vs CT, and similarly for BC.But to test the interaction, we would need to include an interaction term between NRT and BC, but since participants are in only one group, the interaction term isn't directly applicable.Alternatively, perhaps we can use a likelihood ratio test comparing a model with main effects only to a model that includes the interaction. But in this case, since the groups are mutually exclusive, the interaction term isn't estimable, so the likelihood ratio test wouldn't be valid.Therefore, perhaps the correct approach is to compare the observed success rate of CT to the expected success rate under the additive model, which is 0.725, and perform a hypothesis test.As we did earlier, the z-test gives a p-value of ~0.091, which is not significant at alpha=0.05. Therefore, we cannot conclude that the interaction effect is statistically significant.But wait, the problem states that CT is a new method that integrates both NRT and BC, and the observed success rate is higher than both NRT and BC. So, perhaps the additive model is not the right way to think about it, and instead, we should consider that the additive model is p_NRT + p_BC - p_NRT*p_BC, but that's the same as 1 - (1 - p_NRT)(1 - p_BC).Alternatively, maybe the additive model is considering the main effects without interaction, so the expected success rate for CT would be p_NRT + p_BC - p_NRT*p_BC, which is 0.725, and we can test if the observed 0.65 is significantly different.As we calculated earlier, the z-test gives a p-value of ~0.091, which is not significant at alpha=0.05. Therefore, we cannot conclude that the interaction effect is statistically significant.Alternatively, perhaps the additive model is considering the main effects as separate, so the expected success rate for CT would be p_NRT + p_BC - p_baseline, but without knowing the baseline, that's unclear.Wait, maybe another approach is to use a chi-square test for trend or a test for linear trend, but that might not be applicable here.Alternatively, perhaps we can use a t-test to compare the observed CT success rate to the expected additive model's success rate. But since we're dealing with proportions, a z-test is more appropriate.So, to recap, the expected success rate under the additive model is 0.725, observed is 0.65, z-score ≈ -1.685, p≈0.091, not significant.Therefore, the interaction effect is not statistically significant at alpha=0.05.But wait, the problem states that CT is a new method that integrates both NRT and BC, and the observed success rate is higher than both NRT and BC. So, perhaps the additive model is not the right way to think about it, and instead, we should consider that the additive model is p_NRT + p_BC - p_NRT*p_BC, but that's the same as 1 - (1 - p_NRT)(1 - p_BC).Alternatively, maybe the additive model is considering the main effects without interaction, so the expected success rate for CT would be p_NRT + p_BC - p_NRT*p_BC, which is 0.725, and we can test if the observed 0.65 is significantly different.As we calculated earlier, the z-test gives a p-value of ~0.091, which is not significant at alpha=0.05. Therefore, we cannot conclude that the interaction effect is statistically significant.Alternatively, perhaps the additive model is considering the main effects as separate, so the expected success rate for CT would be p_NRT + p_BC - p_baseline, but without knowing the baseline, that's unclear.Wait, maybe another approach is to use a chi-square test comparing observed vs expected counts.Expected successes in CT: 72.5Observed: 65So, the chi-square statistic is (65 - 72.5)^2 / 72.5 = ( -7.5 )^2 /72.5 = 56.25 /72.5 ≈ 0.775.Degrees of freedom is 1, so the p-value is approximately 0.379, which is also not significant.Therefore, we cannot conclude that the interaction effect is statistically significant.So, in conclusion, the 95% confidence intervals are approximately (35.25%, 54.75%) for NRT, (40.2%, 59.8%) for BC, and (55.65%, 74.35%) for CT. The interaction effect between NRT and BC in CT is not statistically significant at alpha=0.05.</think>

🧠:<think>Alright, so I've got two problems here that the auctioneer needs to solve. Let me take them one at a time.Starting with the first problem: The auctioneer wants to maximize their effectiveness function E(x, y) = 3x² + 2xy + 2y². They have constraints on how they allocate their time between live and online auctions. Specifically, they need to spend at least 40% on live auctions (x ≥ 0.4) and at least 30% on online auctions (y ≥ 0.3). Also, the total time must sum up to 100%, so x + y = 1. Hmm, okay, so this is a constrained optimization problem. I remember that for such problems, we can use the method of Lagrange multipliers. But since there are inequality constraints, maybe I should consider the feasible region and check the boundaries as well.First, let's note the constraints:1. x ≥ 0.42. y ≥ 0.33. x + y = 1Since x + y = 1, we can express y as y = 1 - x. That might simplify things.So substituting y into the effectiveness function:E(x) = 3x² + 2x(1 - x) + 2(1 - x)²Let me expand this:E(x) = 3x² + 2x - 2x² + 2(1 - 2x + x²)Simplify term by term:3x² - 2x² = x²2x remains as is.Then, 2*(1 - 2x + x²) = 2 - 4x + 2x²So putting it all together:E(x) = x² + 2x + 2 - 4x + 2x²Combine like terms:x² + 2x² = 3x²2x - 4x = -2xSo E(x) = 3x² - 2x + 2Now, we need to maximize this quadratic function. Since the coefficient of x² is positive (3), the parabola opens upwards, meaning the vertex is a minimum. But we are looking for a maximum. However, since we have constraints on x, the maximum will occur at one of the endpoints of the feasible region.What's the feasible region for x? From the constraints:x ≥ 0.4y = 1 - x ≥ 0.3 ⇒ 1 - x ≥ 0.3 ⇒ x ≤ 0.7So x must be between 0.4 and 0.7.Therefore, the maximum of E(x) will occur either at x = 0.4 or x = 0.7.Let me compute E(x) at both points.First, at x = 0.4:E(0.4) = 3*(0.4)² - 2*(0.4) + 2Calculate each term:3*(0.16) = 0.48-2*(0.4) = -0.8So E(0.4) = 0.48 - 0.8 + 2 = (0.48 - 0.8) + 2 = (-0.32) + 2 = 1.68Now at x = 0.7:E(0.7) = 3*(0.7)² - 2*(0.7) + 2Calculate each term:3*(0.49) = 1.47-2*(0.7) = -1.4So E(0.7) = 1.47 - 1.4 + 2 = (1.47 - 1.4) + 2 = 0.07 + 2 = 2.07Comparing the two, E(0.7) = 2.07 is higher than E(0.4) = 1.68. Therefore, the maximum effectiveness occurs at x = 0.7 and y = 0.3.Wait, but let me double-check. Since the function is quadratic and opens upwards, the minimum is at the vertex, but the maximum on the interval [0.4, 0.7] is indeed at one of the endpoints. So yes, x = 0.7 gives a higher value.So the optimal allocation is 70% live auctions and 30% online auctions.Moving on to the second problem: The competitor's pricing model is P(t) = 100e^{-0.1t}. The auctioneer wants to find the time t when the price drops to 50% of its initial value.First, the initial value is when t = 0, so P(0) = 100e^0 = 100. So 50% of that is 50.We need to solve for t in the equation:100e^{-0.1t} = 50Divide both sides by 100:e^{-0.1t} = 0.5Take the natural logarithm of both sides:ln(e^{-0.1t}) = ln(0.5)Simplify left side:-0.1t = ln(0.5)Solve for t:t = ln(0.5) / (-0.1)Compute ln(0.5). I remember that ln(1/2) is approximately -0.6931.So t = (-0.6931) / (-0.1) = 6.931So approximately 6.931 minutes. If we round to a reasonable decimal place, maybe 6.93 minutes or 6 minutes and 56 seconds.But since the question asks for the time t, I can present it as approximately 6.93 minutes.Wait, let me verify:Compute e^{-0.1*6.931}:First, 0.1*6.931 = 0.6931e^{-0.6931} ≈ 0.5, which is correct because ln(2) ≈ 0.6931, so e^{-ln(2)} = 1/2.Yes, that checks out.So, the time t when the price drops to 50% is approximately 6.93 minutes.Final Answer1. The optimal allocation is ( x = boxed{0.7} ) and ( y = boxed{0.3} ).2. The time when the price drops to 50% is ( t = boxed{6.93} ) minutes.</think>

🧠:<think>Alright, so I'm trying to help this young chef figure out the coefficients for their dough thickness function. The function is quadratic, given by ( t(x) = ax^2 + bx + c ). They've given me three data points: at 5 minutes, the thickness is 2 mm; at 10 minutes, it's 3 mm; and at 15 minutes, it's 4.5 mm. Okay, so I need to set up a system of equations using these points. Each point will give me an equation when I plug in the x and t values into the quadratic function. Let me write those out.First, at 5 minutes:( t(5) = a(5)^2 + b(5) + c = 2 )So, that's ( 25a + 5b + c = 2 ). Let me note that as equation (1).Next, at 10 minutes:( t(10) = a(10)^2 + b(10) + c = 3 )Which simplifies to ( 100a + 10b + c = 3 ). That's equation (2).Then, at 15 minutes:( t(15) = a(15)^2 + b(15) + c = 4.5 )So, ( 225a + 15b + c = 4.5 ). That's equation (3).Now, I have three equations:1. ( 25a + 5b + c = 2 )2. ( 100a + 10b + c = 3 )3. ( 225a + 15b + c = 4.5 )I need to solve this system for a, b, and c. Let me subtract equation (1) from equation (2) to eliminate c.Equation (2) minus equation (1):( (100a - 25a) + (10b - 5b) + (c - c) = 3 - 2 )Which simplifies to:( 75a + 5b = 1 )Let me call this equation (4).Similarly, subtract equation (2) from equation (3):( (225a - 100a) + (15b - 10b) + (c - c) = 4.5 - 3 )Which simplifies to:( 125a + 5b = 1.5 )Let me call this equation (5).Now, I have two equations with two variables:4. ( 75a + 5b = 1 )5. ( 125a + 5b = 1.5 )Subtract equation (4) from equation (5) to eliminate b:( (125a - 75a) + (5b - 5b) = 1.5 - 1 )Which simplifies to:( 50a = 0.5 )So, ( a = 0.5 / 50 = 0.01 ). Now that I have a, I can plug it back into equation (4) to find b.From equation (4):( 75(0.01) + 5b = 1 )Calculates to:( 0.75 + 5b = 1 )Subtract 0.75:( 5b = 0.25 )So, ( b = 0.25 / 5 = 0.05 ).Now, with a and b known, I can find c using equation (1):( 25(0.01) + 5(0.05) + c = 2 )Calculates to:( 0.25 + 0.25 + c = 2 )Which is:( 0.5 + c = 2 )So, ( c = 2 - 0.5 = 1.5 ).Let me double-check these values with equation (3) to make sure.Plugging into equation (3):( 225(0.01) + 15(0.05) + 1.5 = 2.25 + 0.75 + 1.5 = 4.5 )Yes, that adds up correctly.So, the coefficients are:( a = 0.01 ), ( b = 0.05 ), and ( c = 1.5 ).Now, moving on to part 2. The chef wants the dough thickness to be 3.5 mm. So, we need to solve ( t(x) = 3.5 ) using the function we found.So, set up the equation:( 0.01x^2 + 0.05x + 1.5 = 3.5 )Subtract 3.5 from both sides:( 0.01x^2 + 0.05x + 1.5 - 3.5 = 0 )Simplify:( 0.01x^2 + 0.05x - 2 = 0 )To make it easier, multiply both sides by 100 to eliminate decimals:( x^2 + 5x - 200 = 0 )Now, solve this quadratic equation. I can use the quadratic formula:( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} )Here, a = 1, b = 5, c = -200.Calculate discriminant:( b^2 - 4ac = 25 - 4(1)(-200) = 25 + 800 = 825 )So, square root of 825 is approximately 28.7228.Thus, solutions are:( x = frac{-5 pm 28.7228}{2} )So, two solutions:1. ( x = frac{-5 + 28.7228}{2} = frac{23.7228}{2} ≈ 11.8614 )2. ( x = frac{-5 - 28.7228}{2} = frac{-33.7228}{2} ≈ -16.8614 )Since time can't be negative, we discard the negative solution. So, approximately 11.86 minutes.Let me check if this makes sense. Since at 10 minutes, the thickness was 3 mm, and at 15 minutes, it was 4.5 mm, so 3.5 mm should be somewhere between 10 and 15 minutes. 11.86 is between 10 and 15, so that seems reasonable.Alternatively, maybe I can express the exact value instead of the decimal. Let's see:The quadratic equation was ( x^2 + 5x - 200 = 0 ). The exact roots are:( x = frac{-5 pm sqrt{25 + 800}}{2} = frac{-5 pm sqrt{825}}{2} )Simplify sqrt(825). 825 factors into 25*33, so sqrt(25*33) = 5*sqrt(33). So, exact form is:( x = frac{-5 pm 5sqrt{33}}{2} )Again, taking the positive root:( x = frac{-5 + 5sqrt{33}}{2} )We can factor out 5:( x = frac{5(-1 + sqrt{33})}{2} )But either way, the approximate value is about 11.86 minutes.So, the cooking time needed is approximately 11.86 minutes.Final Answer1. The coefficients are ( a = boxed{0.01} ), ( b = boxed{0.05} ), and ( c = boxed{1.5} ).2. The cooking time to achieve a dough thickness of 3.5 mm is ( boxed{frac{-5 + 5sqrt{33}}{2}} ) minutes, which is approximately ( boxed{11.86} ) minutes.</think>

🧠:<think>Alright, so I have this problem where I need to analyze gender representation in sports media over the last 20 years and its impact on audience engagement. The user has given me two functions: one for the percentage of female sports broadcasters, F(t), and another for the engagement metric, E(t). I need to calculate the average percentage of female broadcasters over 20 years and then find the time t that maximizes the engagement score.Starting with the first part: calculating the average percentage of female sports broadcasters. The function given is F(t) = 5 + 4 sin(πt/10). I remember that to find the average value of a function over an interval [a, b], the formula is (1/(b-a)) times the integral from a to b of F(t) dt. In this case, the interval is from t=0 to t=20, so a=0 and b=20.So, the average percentage, let's call it F_avg, would be (1/20) times the integral from 0 to 20 of [5 + 4 sin(πt/10)] dt. I can split this integral into two parts: the integral of 5 dt and the integral of 4 sin(πt/10) dt.The integral of 5 dt from 0 to 20 is straightforward. It's 5t evaluated from 0 to 20, which is 5*20 - 5*0 = 100.Now, the integral of 4 sin(πt/10) dt. Let me recall the integral of sin(ax) dx is (-1/a) cos(ax) + C. So, applying that here, the integral of sin(πt/10) dt would be (-10/π) cos(πt/10) + C. Multiplying by 4, it becomes (-40/π) cos(πt/10) + C.So, evaluating this from 0 to 20: [(-40/π) cos(π*20/10)] - [(-40/π) cos(π*0/10)]. Simplifying, cos(2π) is 1 and cos(0) is also 1. So, this becomes [(-40/π)(1)] - [(-40/π)(1)] = (-40/π) + 40/π = 0.Wait, that's interesting. The integral of the sine function over a full period is zero. Since the period of sin(πt/10) is 20, which is exactly the interval we're integrating over, that makes sense. So, the integral of the sine part is zero.Therefore, the total integral is 100 + 0 = 100. Then, the average F_avg is (1/20)*100 = 5. So, the average percentage of female sports broadcasters over the 20-year period is 5%.Hmm, that seems low. Let me double-check my calculations. The integral of 5 from 0 to 20 is indeed 100. The integral of 4 sin(πt/10) over 0 to 20 is zero because it's a full period. So, yeah, the average is 5%. That makes sense because the sine function oscillates around zero, so when you add 5, the average is just 5.Moving on to the second part: determining the time t that maximizes the engagement metric E(t). The function given is E(t) = 100 + 10F(t) - (t^2)/2. Since F(t) is 5 + 4 sin(πt/10), substituting that in, E(t) becomes 100 + 10*(5 + 4 sin(πt/10)) - (t^2)/2.Let me simplify that. 10*5 is 50, and 10*4 sin(πt/10) is 40 sin(πt/10). So, E(t) = 100 + 50 + 40 sin(πt/10) - (t^2)/2. Combining the constants, 100 + 50 is 150. So, E(t) = 150 + 40 sin(πt/10) - (t^2)/2.Now, to find the maximum of E(t), I need to take its derivative with respect to t, set it equal to zero, and solve for t. Let's compute E'(t).The derivative of 150 is 0. The derivative of 40 sin(πt/10) is 40*(π/10) cos(πt/10) = 4π cos(πt/10). The derivative of -(t^2)/2 is -t. So, E'(t) = 4π cos(πt/10) - t.Setting E'(t) = 0: 4π cos(πt/10) - t = 0. So, 4π cos(πt/10) = t.This is a transcendental equation, meaning it can't be solved algebraically. I'll need to use numerical methods or graphing to approximate the solution. Let me think about how to approach this.First, let's consider the domain of t. Since we're looking at a 20-year period, t is between 0 and 20. Let's analyze the behavior of both sides of the equation 4π cos(πt/10) = t.The left side, 4π cos(πt/10), oscillates because of the cosine function. The amplitude is 4π, which is approximately 12.566. The right side is a linear function, t, which increases from 0 to 20.So, initially, at t=0, left side is 4π ≈12.566, right side is 0. So, left > right. As t increases, the right side increases linearly, while the left side oscillates. The first time they might intersect is somewhere before t=12.566, since beyond that, the right side would be larger than the maximum of the left side.Wait, but the left side is 4π cos(πt/10). The maximum value of cos is 1, so maximum left side is 4π ≈12.566. The minimum is -4π ≈-12.566. So, the equation 4π cos(πt/10) = t will have solutions where t is between 0 and 12.566, because beyond that, t >12.566, and the left side can't reach that.So, let's look for t in [0, 12.566]. Let me try plugging in some values.At t=0: 4π cos(0) = 4π ≈12.566, which is greater than 0.At t=10: 4π cos(π*10/10)=4π cos(π)=4π*(-1)= -12.566. So, left side is -12.566, which is less than t=10. So, somewhere between t=0 and t=10, the left side goes from 12.566 to -12.566, crossing t somewhere.Wait, but t is increasing from 0 to 10, while the left side is decreasing from 12.566 to -12.566. So, they must cross somewhere between t=0 and t=10.Let me try t=5: left side is 4π cos(π*5/10)=4π cos(π/2)=4π*0=0. Right side is 5. So, left side is 0 < 5. So, at t=5, left < right.At t=4: left side is 4π cos(4π/10)=4π cos(2π/5). Cos(2π/5) is approximately 0.3090. So, 4π*0.3090≈4*3.1416*0.3090≈12.566*0.3090≈3.875. Right side is 4. So, left≈3.875 <4. So, left < right.At t=3: left side is 4π cos(3π/10). Cos(3π/10)≈0.5878. So, 4π*0.5878≈12.566*0.5878≈7.39. Right side is 3. So, left≈7.39 >3. So, left > right.So, between t=3 and t=4, the left side goes from ~7.39 to ~3.875, while the right side goes from 3 to 4. So, the crossing point is somewhere between t=3 and t=4.Let me try t=3.5: left side is 4π cos(3.5π/10)=4π cos(0.35π). Cos(0.35π)≈cos(63 degrees)≈0.4540. So, 4π*0.4540≈12.566*0.4540≈5.71. Right side is 3.5. So, left≈5.71 >3.5.t=3.75: left side is 4π cos(3.75π/10)=4π cos(0.375π)=4π cos(67.5 degrees). Cos(67.5)≈0.3827. So, 4π*0.3827≈12.566*0.3827≈4.80. Right side is 3.75. So, left≈4.80 >3.75.t=3.9: left side is 4π cos(3.9π/10)=4π cos(0.39π)=4π cos(70.2 degrees). Cos(70.2)≈0.3420. So, 4π*0.3420≈12.566*0.3420≈4.30. Right side is 3.9. So, left≈4.30 >3.9.t=3.95: left≈4π cos(3.95π/10)=4π cos(0.395π)=4π cos(71.1 degrees). Cos(71.1)≈0.3256. So, 4π*0.3256≈12.566*0.3256≈4.10. Right side is 3.95. So, left≈4.10 >3.95.t=3.98: left≈4π cos(3.98π/10)=4π cos(0.398π)=4π cos(71.64 degrees). Cos(71.64)≈0.316. So, 4π*0.316≈12.566*0.316≈3.97. Right side is 3.98. So, left≈3.97 <3.98.So, between t=3.95 and t=3.98, the left side crosses from above to below the right side. So, the solution is approximately t≈3.97.Let me try t=3.97: left side≈4π cos(3.97π/10)=4π cos(0.397π)=4π cos(71.46 degrees). Cos(71.46)≈0.318. So, 4π*0.318≈12.566*0.318≈3.98. Right side is 3.97. So, left≈3.98 >3.97.t=3.975: left≈4π cos(3.975π/10)=4π cos(0.3975π)=4π cos(71.565 degrees). Cos(71.565)≈0.315. So, 4π*0.315≈12.566*0.315≈3.95. Right side is 3.975. So, left≈3.95 <3.975.So, the root is between t=3.97 and t=3.975. Let's use linear approximation.At t=3.97: left≈3.98, right=3.97. So, difference=0.01.At t=3.975: left≈3.95, right=3.975. Difference= -0.025.We can set up a linear equation between these two points.Let’s denote f(t) = 4π cos(πt/10) - t.At t1=3.97, f(t1)=3.98 -3.97=0.01.At t2=3.975, f(t2)=3.95 -3.975= -0.025.We want to find t where f(t)=0.The change in t is 0.005, and the change in f(t) is -0.035.We need to find Δt such that 0.01 + (-0.035/0.005)*Δt =0.Wait, actually, the slope is (f(t2)-f(t1))/(t2-t1)= (-0.025 -0.01)/0.005= (-0.035)/0.005= -7.So, the linear approximation near t=3.97 is f(t) ≈0.01 -7*(t -3.97).Set this equal to zero: 0.01 -7*(t -3.97)=0.So, 7*(t -3.97)=0.01 => t -3.97=0.01/7≈0.00142857.Thus, t≈3.97 +0.00142857≈3.97142857.So, approximately t≈3.9714 years.To check, let's compute f(3.9714)=4π cos(3.9714π/10) -3.9714.First, compute 3.9714π/10≈0.39714π≈1.247 radians.cos(1.247)≈0.319.So, 4π*0.319≈12.566*0.319≈3.994.Then, subtract 3.9714: 3.994 -3.9714≈0.0226. Hmm, that's still positive. Maybe my linear approximation needs more accurate data.Alternatively, perhaps using a better method like Newton-Raphson.Let me try Newton-Raphson.We have f(t)=4π cos(πt/10) - t.f'(t)= -4π*(π/10) sin(πt/10) -1= - (4π²/10) sin(πt/10) -1.Starting with t0=3.97.Compute f(t0)=4π cos(3.97π/10) -3.97≈4π*0.318 -3.97≈3.98 -3.97=0.01.f'(t0)= - (4π²/10) sin(3.97π/10) -1.Compute sin(3.97π/10)=sin(0.397π)=sin(71.46 degrees)≈0.947.So, f'(t0)= - (4*(9.8696)/10)*0.947 -1≈ - (39.4784/10)*0.947 -1≈ -3.9478*0.947 -1≈-3.737 -1≈-4.737.So, Newton-Raphson update: t1= t0 - f(t0)/f'(t0)=3.97 - (0.01)/(-4.737)=3.97 +0.00211≈3.97211.Compute f(t1)=4π cos(3.97211π/10) -3.97211.Compute 3.97211π/10≈0.397211π≈1.247 radians.cos(1.247)≈0.319.So, 4π*0.319≈3.994.Then, 3.994 -3.97211≈0.02189. Still positive.Wait, that's not decreasing as expected. Maybe my approximation of cos(1.247) is too rough.Alternatively, perhaps using a calculator for more precise values.Alternatively, maybe it's better to accept that the root is approximately t≈3.97 years.But let's check at t=3.9714:Compute 3.9714π/10≈0.39714π≈1.247 radians.cos(1.247)= approximately, using Taylor series around π/3 (≈1.047), but maybe better to use calculator-like approximation.Alternatively, use a calculator:cos(1.247)=cos(71.46 degrees)= approximately 0.319.So, 4π*0.319≈3.994.3.994 -3.9714≈0.0226.So, still positive. So, need to go a bit higher.Wait, perhaps my initial assumption is wrong. Maybe the root is around t≈3.97.Alternatively, perhaps using a better approach.Alternatively, maybe using a graphing calculator or software, but since I'm doing this manually, let's try t=3.975.Compute f(3.975)=4π cos(3.975π/10) -3.975.3.975π/10≈0.3975π≈1.248 radians.cos(1.248)= approximately 0.315.So, 4π*0.315≈3.96.3.96 -3.975≈-0.015.So, f(3.975)= -0.015.So, between t=3.97 and t=3.975, f(t) goes from +0.01 to -0.015.So, using linear approximation:At t=3.97, f=0.01.At t=3.975, f=-0.015.The difference in t is 0.005, and the difference in f is -0.025.We need to find t where f=0.So, the fraction is 0.01 / 0.025=0.4.So, t=3.97 +0.4*0.005=3.97 +0.002=3.972.So, t≈3.972.Testing t=3.972:Compute f(t)=4π cos(3.972π/10) -3.972.3.972π/10≈0.3972π≈1.247 radians.cos(1.247)=≈0.319.So, 4π*0.319≈3.994.3.994 -3.972≈0.022.Hmm, still positive. Maybe my approximations are too rough.Alternatively, perhaps the exact solution is around t≈3.972.But considering the oscillation of the sine function, maybe there's another maximum beyond t=10? Wait, no, because beyond t≈12.566, the right side t would be larger than the maximum of the left side, which is 12.566. So, the only solution is around t≈3.97.Wait, but let me check t=10: f(t)=4π cos(π) -10= -4π -10≈-12.566 -10≈-22.566, which is negative.t=5: f(t)=4π cos(π/2) -5=0 -5=-5.t=0: f(t)=4π -0≈12.566.So, the function f(t)=4π cos(πt/10) -t starts at 12.566, decreases, crosses zero around t≈3.97, then continues to decrease to negative values.Therefore, the only critical point is at t≈3.97. Since E(t) is a function that starts at E(0)=150 +40*1 -0=190, then increases or decreases?Wait, let's think about the behavior of E(t). E(t)=150 +40 sin(πt/10) - (t^2)/2.At t=0: E=150 +0 -0=150.Wait, no, wait: E(t)=100 +10F(t) -t²/2, and F(t)=5 +4 sin(πt/10). So, E(t)=100 +10*(5 +4 sin(πt/10)) -t²/2=100 +50 +40 sin(πt/10) -t²/2=150 +40 sin(πt/10) -t²/2.So, at t=0: E=150 +0 -0=150.At t=10: E=150 +40 sin(π) -100=150 +0 -100=50.At t=20: E=150 +40 sin(2π) -200=150 +0 -200=-50.So, E(t) starts at 150, goes up and down due to the sine term, but overall is decreasing because of the -t²/2 term.So, the maximum of E(t) is likely at t≈3.97, as that's where the derivative is zero, and since the function is concave down there (second derivative negative), it's a maximum.Let me check the second derivative to confirm.E''(t)= derivative of E'(t)= derivative of (4π cos(πt/10) -t)= -4π*(π/10) sin(πt/10) -1= - (4π²/10) sin(πt/10) -1.At t≈3.97, sin(πt/10)=sin(0.397π)=sin(71.46 degrees)≈0.947.So, E''(t)= - (4π²/10)*0.947 -1≈ - (39.4784/10)*0.947 -1≈-3.9478*0.947 -1≈-3.737 -1≈-4.737 <0.So, concave down, which means it's a local maximum. Since it's the only critical point in the domain where E(t) is increasing before and decreasing after, this is the global maximum.Therefore, the time t that maximizes E(t) is approximately 3.97 years, which is roughly 3 years and 11.6 months, or about 3.97 years.But since the problem asks for the time t within the 20-year period, I can present it as approximately 3.97 years, or more precisely, around 3.97 years.Alternatively, if more decimal places are needed, but I think 3.97 is sufficient.So, summarizing:1. The average percentage of female sports broadcasters over 20 years is 5%.2. The time t that maximizes the engagement score E(t) is approximately 3.97 years.</think>