🧠:<think>Okay, so Juan is planning this event to showcase traditional Filipino cuisine, and he needs to figure out some probabilities related to his cooking schedule and food consumption. Let me try to break down the two problems he's facing.Starting with the first problem: He wants to calculate the probability that the total preparation time for Adobo, Sinigang, and Lechon exceeds 8 hours. Each dish has its own preparation time, which follows a normal distribution with given means and standard deviations. The cooking times are independent, so I can use properties of normal distributions to find the total time.First, let me recall that if you have independent normal random variables, their sum is also a normal random variable. The mean of the sum is the sum of the means, and the variance of the sum is the sum of the variances. So, I can calculate the total mean and total variance by adding up the individual means and variances.For Adobo, the mean is 2 hours, and the standard deviation is 0.5 hours. So, variance is (0.5)^2 = 0.25.Sinigang has a mean of 1.5 hours and a standard deviation of 0.3 hours. Variance is (0.3)^2 = 0.09.Lechon has a mean of 4 hours and a standard deviation of 0.8 hours. Variance is (0.8)^2 = 0.64.So, total mean (mu_total) = 2 + 1.5 + 4 = 7.5 hours.Total variance (sigma_total^2) = 0.25 + 0.09 + 0.64 = 0.98.Therefore, the total standard deviation (sigma_total) is the square root of 0.98. Let me calculate that: sqrt(0.98) ≈ 0.9899 hours, approximately 0.99 hours.Now, we need the probability that the total time exceeds 8 hours. So, we can model this as P(X > 8), where X is a normal variable with mean 7.5 and standard deviation ~0.99.To find this probability, I can standardize the variable. The z-score is calculated as (X - mu)/sigma.So, z = (8 - 7.5)/0.99 ≈ 0.5 / 0.99 ≈ 0.5051.Looking up this z-score in the standard normal distribution table, we can find the probability that Z is less than 0.5051, and then subtract that from 1 to get P(Z > 0.5051).From the z-table, a z-score of 0.5 corresponds to about 0.6915, and 0.51 is about 0.6950. Since 0.5051 is halfway between 0.5 and 0.51, maybe around 0.6933? Alternatively, using a calculator, the exact value can be found.Alternatively, using a calculator, the cumulative distribution function (CDF) for z = 0.5051 is approximately 0.6933. So, P(Z > 0.5051) = 1 - 0.6933 = 0.3067.Therefore, the probability that the total preparation time exceeds 8 hours is approximately 30.67%.Wait, let me double-check my calculations. The total variance is 0.25 + 0.09 + 0.64 = 0.98. So, standard deviation is sqrt(0.98) ≈ 0.9899, which is approximately 0.99. Then, z = (8 - 7.5)/0.99 ≈ 0.5051. The CDF at 0.5051 is roughly 0.6933, so 1 - 0.6933 is 0.3067. Yeah, that seems right.Moving on to the second problem: Juan wants to calculate the probability that the total amount of food consumed is between 18 kg and 35 kg. The number of guests is uniformly distributed between 30 and 50, and each guest consumes an average of 0.6 kg with a standard deviation of 0.1 kg.Hmm, okay, so the total food consumed is the number of guests multiplied by the amount each consumes. But since the number of guests is a random variable, and the consumption per guest is also a random variable, we need to model the total consumption.Wait, actually, the total consumption is the sum of each guest's consumption. Since each guest's consumption is normal, the sum will also be normal, but scaled by the number of guests. However, the number of guests is a uniform random variable, so the total consumption is a bit more complicated because it's a mixture distribution.Alternatively, perhaps we can model the total consumption as a random variable where the number of guests N is uniform between 30 and 50, and for each N, the total consumption is N * X, where X is a normal variable with mean 0.6 and standard deviation 0.1.But since N is discrete, but the number of guests is between 30 and 50, which is a range of 21 possible values. That might be a bit tedious to compute exactly, but maybe we can approximate it.Alternatively, since N is uniform, perhaps we can find the expected value and variance of the total consumption, treating N as a random variable.Wait, let me think. Let me denote T as the total consumption, which is T = N * X, where N is uniform over integers from 30 to 50, and X is normal with mean 0.6 and variance (0.1)^2 = 0.01.But actually, T is the sum of N independent normal variables each with mean 0.6 and variance 0.01. So, for a given N, T | N ~ Normal(N*0.6, N*0.01). But since N itself is random, T is a mixture of normals.Calculating the probability that T is between 18 and 35 is challenging because it's a mixture distribution. Maybe we can approximate it by considering the expected value and variance of T.First, let's compute E[T] and Var(T).E[T] = E[N * X] = E[N] * E[X] because N and X are independent.E[N] is the average number of guests, which is (30 + 50)/2 = 40.E[X] is 0.6 kg.So, E[T] = 40 * 0.6 = 24 kg.Now, Var(T) = Var(N * X). Since N and X are independent, Var(T) = E[N]^2 * Var(X) + Var(N) * (E[X])^2.Wait, is that correct? Let me recall the formula for variance of a product of independent variables.Yes, for independent variables, Var(aX + bY) = a^2 Var(X) + b^2 Var(Y). But here, it's N * X, where N is a random variable.So, Var(T) = E[N^2] * Var(X) + Var(N) * (E[X])^2.Yes, that's the formula. So, we need E[N^2] and Var(N).First, N is uniform over integers from 30 to 50 inclusive. So, the number of possible values is 21 (since 50 - 30 + 1 = 21).The variance of N is ((b - a + 1)^2 - 1)/12, where a = 30, b = 50. So, ((50 - 30 + 1)^2 - 1)/12 = (21^2 - 1)/12 = (441 - 1)/12 = 440/12 ≈ 36.6667.Wait, actually, for a discrete uniform distribution over integers from a to b, the variance is ((b - a + 1)^2 - 1)/12. So, that's correct.So, Var(N) ≈ 36.6667.E[N] = 40, as before.E[N^2] can be calculated as Var(N) + (E[N])^2 = 36.6667 + 40^2 = 36.6667 + 1600 = 1636.6667.So, Var(T) = E[N^2] * Var(X) + Var(N) * (E[X])^2.Var(X) is 0.01, and (E[X])^2 is 0.36.So, Var(T) = 1636.6667 * 0.01 + 36.6667 * 0.36.Calculating each term:1636.6667 * 0.01 = 16.36666736.6667 * 0.36 ≈ 13.2So, total Var(T) ≈ 16.366667 + 13.2 ≈ 29.566667.Therefore, the standard deviation of T is sqrt(29.566667) ≈ 5.437 kg.So, T is approximately normally distributed with mean 24 kg and standard deviation ~5.437 kg.Wait, but is T actually normal? Since T is a mixture of normals with different means and variances, it might not be exactly normal, but for practical purposes, especially with a large number of guests, it might be approximately normal.So, assuming T is approximately normal with mu = 24 and sigma ≈ 5.437, we can calculate the probability that T is between 18 and 35.So, we can compute the z-scores for 18 and 35.First, z1 = (18 - 24)/5.437 ≈ (-6)/5.437 ≈ -1.103z2 = (35 - 24)/5.437 ≈ 11/5.437 ≈ 2.023Now, we need to find P(-1.103 < Z < 2.023).Using the standard normal table, P(Z < 2.023) is approximately 0.9783, and P(Z < -1.103) is approximately 0.1357.So, the probability is 0.9783 - 0.1357 ≈ 0.8426, or 84.26%.Wait, but let me verify the z-scores more accurately.For z = -1.103, looking up in the z-table: the value for z = -1.10 is 0.1357, and for z = -1.11 is 0.1335. So, -1.103 is approximately 0.1357 - (0.1357 - 0.1335)*(0.003/0.01) ≈ 0.1357 - (0.0022)*0.3 ≈ 0.1357 - 0.00066 ≈ 0.1350.Similarly, for z = 2.023, the z-table for 2.02 is 0.9783, and for 2.03 is 0.9798. So, 2.023 is approximately 0.9783 + (0.9798 - 0.9783)*(0.003/0.01) ≈ 0.9783 + (0.0015)*0.3 ≈ 0.9783 + 0.00045 ≈ 0.97875.So, the probability is approximately 0.97875 - 0.1350 ≈ 0.84375, or 84.375%.So, approximately 84.4%.But wait, is this a valid approach? Because T is a mixture of normals, each with different means and variances, the overall distribution might not be exactly normal, but given that the number of guests is large (between 30 and 50), and the consumption per guest is normal, the Central Limit Theorem might make T approximately normal. So, this approximation should be reasonable.Alternatively, another approach is to consider that for each possible N, compute the probability that T is between 18 and 35, and then average over all N from 30 to 50. But that would be more accurate but computationally intensive.Given that Juan is planning, an approximate value is probably sufficient, so 84.4% is a reasonable estimate.Wait, but let me think again. The total consumption is the sum of N independent normal variables, each with mean 0.6 and variance 0.01. So, for a given N, T | N ~ Normal(N*0.6, N*0.01). So, the distribution of T is a mixture of normals with different means and variances.To compute P(18 < T < 35), we would need to integrate over all possible N from 30 to 50, weighted by the probability of each N. Since N is uniform, each integer N from 30 to 50 has probability 1/21.So, P(18 < T < 35) = (1/21) * sum_{N=30}^{50} P(18 < T < 35 | N).Each term P(18 < T < 35 | N) can be computed as the probability that a normal variable with mean 0.6*N and variance 0.01*N is between 18 and 35.So, for each N, compute z1 = (18 - 0.6N)/sqrt(0.01N) and z2 = (35 - 0.6N)/sqrt(0.01N), then find the difference in CDF.This would give a more accurate result, but it's a lot of calculations. Maybe we can approximate it numerically.Alternatively, perhaps we can model T as a normal variable with mean 24 and variance 29.5667 as before, but I wonder how accurate that is.Wait, let's see. The mean is correct, 24 kg. The variance is also correct, as we calculated E[N^2] * Var(X) + Var(N) * (E[X])^2. So, that should be accurate.Therefore, treating T as normal with mean 24 and SD ~5.437 is a valid approximation, so the probability of 84.4% is acceptable.So, summarizing:1. The probability that total preparation time exceeds 8 hours is approximately 30.67%.2. The probability that total food consumed is between 18 kg and 35 kg is approximately 84.4%.These probabilities can help Juan plan his event. For the first part, knowing there's about a 30% chance that cooking will take longer than 8 hours, he might want to schedule some buffer time or consider starting earlier. For the second part, with an 84% chance that food consumption is between 18 and 35 kg, he can plan his food quantities accordingly, ensuring he has enough to cover the likely range and maybe a bit more to account for the remaining probability.Final Answer1. The probability that the total preparation time exceeds 8 hours is boxed{0.3067}.2. The probability that the total food consumed is between 18 kg and 35 kg is boxed{0.8437}.</think>

🧠:<think>Alright, so I'm trying to help this young car enthusiast optimize their engine's performance by working through these thermodynamic problems. Let me take it step by step because I want to make sure I understand each part correctly.Starting with the first problem: It's about an isochoric process where the pressure increases from 100 kPa to 300 kPa, and the volume remains constant at 2 m³. The initial temperature is 300 K, and I need to find the change in temperature, ΔT.Okay, since it's an isochoric process, that means the volume doesn't change. I remember that for an isochoric process, the ideal gas law can be applied, and since volume is constant, the pressure and temperature are directly proportional. So, the formula should be something like P1/T1 = P2/T2 because n and R are constants here.Let me write that down:P₁ / T₁ = P₂ / T₂We can rearrange this to solve for T₂:T₂ = (P₂ / P₁) * T₁Plugging in the numbers:P₁ = 100 kPa, P₂ = 300 kPa, T₁ = 300 K.So,T₂ = (300 kPa / 100 kPa) * 300 KSimplify the ratio:300/100 = 3So,T₂ = 3 * 300 K = 900 KTherefore, the change in temperature ΔT is T₂ - T₁ = 900 K - 300 K = 600 K.Wait, that seems like a huge jump in temperature. Is that realistic? Well, in an isochoric process, if pressure triples, temperature should also triple if volume is constant, right? So, 300 K to 900 K is a tripling, which makes sense. So, ΔT is 600 K.Moving on to the second problem: It's an adiabatic expansion where the volume goes from 2 m³ to 4 m³. The process follows PV^γ = constant, with γ = 1.4. The initial pressure is P₂ = 300 kPa, and I need to find the final pressure P₃.I recall that for an adiabatic process, the relation is PV^γ = constant. So, we can write:P₂ * V₂^γ = P₃ * V₃^γWe can solve for P₃:P₃ = P₂ * (V₂ / V₃)^γPlugging in the numbers:P₂ = 300 kPa, V₂ = 2 m³, V₃ = 4 m³, γ = 1.4.So,P₃ = 300 kPa * (2 / 4)^1.4Simplify the ratio:2/4 = 0.5So,P₃ = 300 kPa * (0.5)^1.4Now, I need to calculate (0.5)^1.4. Hmm, I don't remember the exact value, but I can approximate it or use logarithms.Let me compute ln(0.5^1.4) = 1.4 * ln(0.5) ≈ 1.4 * (-0.6931) ≈ -0.9703Then, exponentiate that: e^(-0.9703) ≈ 0.378So, (0.5)^1.4 ≈ 0.378Therefore,P₃ ≈ 300 kPa * 0.378 ≈ 113.4 kPaSo, the final pressure P₃ is approximately 113.4 kPa.Wait, let me double-check the exponent calculation because 0.5^1.4 is the same as e^(1.4 * ln 0.5). Let me compute ln(0.5) first, which is approximately -0.6931. Multiply by 1.4: -0.6931 * 1.4 ≈ -0.9703. Then e^(-0.9703) is approximately 0.378. So, yes, that seems correct.Alternatively, I can use logarithms or a calculator, but since I don't have one handy, this approximation should be sufficient.So, summarizing:1. The change in temperature during the isochoric process is 600 K.2. The final pressure after the adiabatic expansion is approximately 113.4 kPa.I think that's it. Let me just make sure I didn't mix up any formulas. For the isochoric process, since volume is constant, the relation is P1/T1 = P2/T2, which is correct. For the adiabatic process, PV^γ = constant, so P₂V₂^γ = P₃V₃^γ, which is also correct. The calculations seem straightforward, so I think I did it right.Final Answer1. The change in temperature is boxed{600 , text{K}}.2. The final pressure after the adiabatic expansion is boxed{113.4 , text{kPa}}.</think>

🧠:<think>Alright, so I have this problem where a philanthropist has given 10 million to an investigative reporter to study two ecosystems: a rainforest and a coral reef. The reporter wants to maximize the total number of weeks spent researching both places without going over the budget. First, let me understand the cost functions for each ecosystem. For the rainforest, the cost is given by ( C_1(x) = 500,000 + 100,000x^2 ), where ( x ) is the number of weeks. For the coral reef, it's ( C_2(y) = 200,000 + 150,000y^3 ), with ( y ) being the weeks there. The total budget is 10 million, which is 10,000,000.So, the total cost should be less than or equal to 10,000,000. That gives me the constraint:( 500,000 + 100,000x^2 + 200,000 + 150,000y^3 leq 10,000,000 )Simplifying that, I can combine the constants:500,000 + 200,000 = 700,000So, the inequality becomes:( 700,000 + 100,000x^2 + 150,000y^3 leq 10,000,000 )Subtracting 700,000 from both sides:( 100,000x^2 + 150,000y^3 leq 9,300,000 )I can divide both sides by 10,000 to make the numbers smaller:( 10x^2 + 15y^3 leq 930 )So, the constraint simplifies to ( 10x^2 + 15y^3 leq 930 ).The objective is to maximize the total number of weeks, which is ( x + y ). So, we need to maximize ( x + y ) subject to ( 10x^2 + 15y^3 leq 930 ).This is an optimization problem with two variables. I think I can use calculus to solve this, specifically using Lagrange multipliers because we have a constraint.Let me set up the Lagrangian function. Let’s denote the Lagrangian multiplier as ( lambda ).The Lagrangian ( L ) is:( L = x + y - lambda(10x^2 + 15y^3 - 930) )To find the maximum, we take the partial derivatives of ( L ) with respect to ( x ), ( y ), and ( lambda ), and set them equal to zero.First, partial derivative with respect to ( x ):( frac{partial L}{partial x} = 1 - lambda(20x) = 0 )So, ( 1 = 20lambda x ) --> Equation 1Partial derivative with respect to ( y ):( frac{partial L}{partial y} = 1 - lambda(45y^2) = 0 )So, ( 1 = 45lambda y^2 ) --> Equation 2Partial derivative with respect to ( lambda ):( frac{partial L}{partial lambda} = -(10x^2 + 15y^3 - 930) = 0 )Which gives us the constraint again:( 10x^2 + 15y^3 = 930 ) --> Equation 3Now, from Equation 1: ( 1 = 20lambda x ) --> ( lambda = frac{1}{20x} )From Equation 2: ( 1 = 45lambda y^2 ) --> ( lambda = frac{1}{45y^2} )Since both expressions equal ( lambda ), we can set them equal to each other:( frac{1}{20x} = frac{1}{45y^2} )Cross-multiplying:( 45y^2 = 20x )Simplify:Divide both sides by 5: 9y^2 = 4xSo, ( x = frac{9}{4}y^2 ) --> Equation 4Now, substitute Equation 4 into Equation 3:( 10x^2 + 15y^3 = 930 )Replace ( x ) with ( frac{9}{4}y^2 ):( 10left(frac{9}{4}y^2right)^2 + 15y^3 = 930 )Compute ( left(frac{9}{4}y^2right)^2 ):( left(frac{81}{16}y^4right) )So, plug that back in:( 10 times frac{81}{16}y^4 + 15y^3 = 930 )Calculate ( 10 times frac{81}{16} ):( frac{810}{16} = frac{405}{8} )So, the equation becomes:( frac{405}{8}y^4 + 15y^3 = 930 )Multiply both sides by 8 to eliminate the denominator:( 405y^4 + 120y^3 = 7440 )Bring all terms to one side:( 405y^4 + 120y^3 - 7440 = 0 )This is a quartic equation, which is quite complex. Maybe I can factor out common terms.First, notice that all coefficients are divisible by 15:Divide each term by 15:( 27y^4 + 8y^3 - 496 = 0 )Hmm, still not easy. Maybe try to factor it or find rational roots.Rational Root Theorem suggests possible roots are factors of 496 divided by factors of 27.Factors of 496: ±1, ±2, ±4, ±8, ±16, ±31, ±62, ±124, ±248, ±496Factors of 27: ±1, ±3, ±9, ±27So possible rational roots are ±1, ±2, ±4, ±8, ±16, ±31, etc., divided by 1,3,9,27.Testing y=2:27*(16) + 8*(8) -496 = 432 + 64 -496 = 0. Oh, y=2 is a root!So, (y - 2) is a factor.Let's perform polynomial division or use synthetic division.Divide 27y^4 + 8y^3 -496 by (y - 2).Using synthetic division:Coefficients: 27, 8, 0, 0, -496Wait, actually, the polynomial is 27y^4 + 8y^3 + 0y^2 + 0y -496Using y=2:Bring down 27.Multiply 27 by 2 = 54. Add to 8: 62Multiply 62 by 2 = 124. Add to 0: 124Multiply 124 by 2 = 248. Add to 0: 248Multiply 248 by 2 = 496. Add to -496: 0So, the polynomial factors as (y - 2)(27y^3 + 62y^2 + 124y + 248) = 0Now, we have to factor 27y^3 + 62y^2 + 124y + 248.Let me try rational roots again for this cubic.Possible roots: factors of 248 over factors of 27.Factors of 248: ±1, ±2, ±4, ±8, ±31, ±62, ±124, ±248Testing y=-2:27*(-8) + 62*(4) + 124*(-2) + 248 = -216 + 248 -248 +248 = (-216 + 248) + (-248 +248) = 32 + 0 = 32 ≠0Testing y=-4:27*(-64) + 62*(16) + 124*(-4) +248 = -1728 + 992 -496 +248 = (-1728 + 992) + (-496 +248) = (-736) + (-248) = -984 ≠0Testing y=-1:27*(-1) +62*(1) +124*(-1) +248 = -27 +62 -124 +248 = (-27 +62) + (-124 +248) = 35 +124=159≠0Testing y=-8:27*(-512) +62*(64) +124*(-8) +248 = -13824 + 3968 -992 +248Compute step by step:-13824 + 3968 = -9856-9856 -992 = -10848-10848 +248 = -10600 ≠0Testing y= -31/3: Probably messy, maybe not.Alternatively, maybe factor by grouping.27y^3 +62y^2 +124y +248Group as (27y^3 +62y^2) + (124y +248)Factor out y^2 from first group: y^2(27y +62)Factor out 124 from second group: 124(y + 2)Hmm, 27y +62 and 124(y +2). Doesn't seem to have a common factor.Alternatively, maybe another grouping.Alternatively, perhaps it's better to use numerical methods or realize that perhaps y=2 is the only real root, and the rest are complex or irrational.Given that, perhaps y=2 is the only feasible solution because other roots might be negative or not integers, which wouldn't make sense for weeks.So, if y=2, then from Equation 4: x = (9/4)y^2 = (9/4)*(4) = 9 weeks.So, x=9, y=2.Let me check if this satisfies the original constraint.Compute total cost:C1(9) = 500,000 +100,000*(81) = 500,000 +8,100,000 =8,600,000C2(2) =200,000 +150,000*(8) =200,000 +1,200,000 =1,400,000Total cost:8,600,000 +1,400,000=10,000,000, which is exactly the budget.So, that works.Therefore, the optimal allocation is 9 weeks in the rainforest and 2 weeks in the coral reef.Now, moving on to part 2.The reporter found that biodiversity index ( B ) and pollution level ( P ) are related by ( B = kP^2 ). For the rainforest, when ( P=3 ), ( B=27 ). So, we can find ( k ).Plugging in P=3, B=27:27 = k*(3)^227 = 9kSo, k=27/9=3.So, k=3.Now, for the coral reef, pollution level changes from P=4 to P=5. We need to find the percentage change in biodiversity index.First, compute B when P=4:B1 = 3*(4)^2 =3*16=48Then, when P=5:B2=3*(5)^2=3*25=75The change in B is 75 -48=27.Percentage change is (27 /48)*100%.Compute 27/48= 9/16=0.5625So, 0.5625*100=56.25%Therefore, the biodiversity index increases by 56.25%.Wait, but let me make sure. Since the relationship is ( B = 3P^2 ), when P increases, B increases quadratically. So, from 4 to5, which is an increase of 25%, but the B increases by (5/4)^2=25/16=1.5625, which is 56.25% increase. So, that's correct.So, the percentage change is 56.25%.Final Answer1. The optimal number of weeks is boxed{9} weeks in the rainforest and boxed{2} weeks in the coral reef.2. The value of ( k ) is boxed{3}, and the percentage change in the biodiversity index is boxed{56.25%}.</think>

🧠:<think>Okay, so I have two problems to solve here related to crime rates and police efficiency. Let me take them one at a time.Starting with the first problem: Crime Rate Analysis. The function given is (C(t) = 1000 cdot e^{-0.05t} + 300 sin(0.2pi t)), where (t) is the number of years since 2001. I need to find the average crime rate over the 15-year period from 2001 to 2015. Hmm, average value of a function over an interval. I remember that the average value of a function (f(t)) over the interval ([a, b]) is given by (frac{1}{b - a} int_{a}^{b} f(t) dt). So in this case, (a = 0) and (b = 15), so the average crime rate (C_{avg}) would be (frac{1}{15} int_{0}^{15} C(t) dt).Let me write that out:(C_{avg} = frac{1}{15} int_{0}^{15} [1000 e^{-0.05t} + 300 sin(0.2pi t)] dt)I can split this integral into two parts:(C_{avg} = frac{1}{15} left[ int_{0}^{15} 1000 e^{-0.05t} dt + int_{0}^{15} 300 sin(0.2pi t) dt right])Alright, let's compute each integral separately.First integral: (I_1 = int 1000 e^{-0.05t} dt)The integral of (e^{kt}) is (frac{1}{k} e^{kt}), so here (k = -0.05). So,(I_1 = 1000 cdot frac{1}{-0.05} e^{-0.05t} + C = -20000 e^{-0.05t} + C)Now, evaluating from 0 to 15:(I_1 = [-20000 e^{-0.05 cdot 15}] - [-20000 e^{0}])Calculating each term:(e^{-0.75}) is approximately... let me compute that. (e^{-0.75}) is about 0.4723665527. So,First term: (-20000 * 0.4723665527 ≈ -9447.331054)Second term: (-20000 * 1 = -20000), so subtracting that gives:(-9447.331054 - (-20000) = 10552.66895)So, (I_1 ≈ 10552.66895)Now, the second integral: (I_2 = int 300 sin(0.2pi t) dt)The integral of (sin(kt)) is (-frac{1}{k} cos(kt)). So here, (k = 0.2pi), so:(I_2 = 300 cdot left( -frac{1}{0.2pi} cos(0.2pi t) right) + C = -frac{300}{0.2pi} cos(0.2pi t) + C)Simplify (frac{300}{0.2pi}):0.2 is 1/5, so 300 divided by (1/5) is 1500. So,(I_2 = -frac{1500}{pi} cos(0.2pi t) + C)Now, evaluate from 0 to 15:(I_2 = left[ -frac{1500}{pi} cos(0.2pi cdot 15) right] - left[ -frac{1500}{pi} cos(0) right])Compute each term:First, (0.2pi cdot 15 = 3pi). So, (cos(3pi)) is (-1).Second, (cos(0) = 1).So,First term: (-frac{1500}{pi} * (-1) = frac{1500}{pi})Second term: (-frac{1500}{pi} * 1 = -frac{1500}{pi})So, subtracting:(frac{1500}{pi} - (-frac{1500}{pi}) = frac{1500}{pi} + frac{1500}{pi} = frac{3000}{pi})Calculating that numerically, since (pi ≈ 3.1415926536), so:(frac{3000}{3.1415926536} ≈ 954.9296586)So, (I_2 ≈ 954.9296586)Now, putting it all together:Total integral (I = I_1 + I_2 ≈ 10552.66895 + 954.9296586 ≈ 11507.59861)Then, average crime rate (C_{avg} = frac{1}{15} * 11507.59861 ≈ 767.1732407)So, approximately 767.17 crimes per year on average.Wait, let me double-check my calculations because I might have messed up somewhere.First integral: (I_1 = 1000 int e^{-0.05t} dt = -20000 e^{-0.05t}) evaluated from 0 to 15.At t=15: (e^{-0.75} ≈ 0.4723665527), so -20000 * 0.4723665527 ≈ -9447.331054At t=0: (e^{0} = 1), so -20000 * 1 = -20000Subtracting: -9447.331054 - (-20000) = 10552.66895. That seems correct.Second integral: 300 sin(0.2πt). Integral is -1500/π cos(0.2πt). Evaluated from 0 to 15.At t=15: cos(3π) = -1, so -1500/π * (-1) = 1500/πAt t=0: cos(0) = 1, so -1500/π * 1 = -1500/πSubtracting: 1500/π - (-1500/π) = 3000/π ≈ 954.9296586. That also seems correct.Adding both integrals: 10552.66895 + 954.9296586 ≈ 11507.59861Divide by 15: 11507.59861 / 15 ≈ 767.1732407So, approximately 767.17 crimes per year on average. That seems reasonable.Moving on to the second problem: Police Efficiency. The function given is (E(t) = 200 cdot ln(t + 1) + 50), where (t) is the number of years since 2001. We need to find the point in time (t) when the efficiency (E(t)) was increasing at the fastest rate.Hmm, increasing at the fastest rate. That sounds like we need to find the maximum of the derivative of (E(t)). Because the rate of increase is given by the first derivative, and the fastest rate would be where the derivative is maximized, i.e., where the second derivative is zero (critical point).So, let's compute the first derivative (E'(t)) and then find where its derivative, (E''(t)), is zero.First, compute (E'(t)):(E(t) = 200 ln(t + 1) + 50)Derivative of (ln(t + 1)) is (frac{1}{t + 1}), so:(E'(t) = 200 cdot frac{1}{t + 1} + 0 = frac{200}{t + 1})Now, to find when (E'(t)) is increasing the fastest, we need to find the maximum of (E'(t)). Wait, actually, the rate of increase of (E(t)) is (E'(t)), and the fastest rate would be the maximum value of (E'(t)). But (E'(t)) is a decreasing function because as (t) increases, (t + 1) increases, so (1/(t + 1)) decreases. Therefore, (E'(t)) is decreasing over (t). So, its maximum occurs at the smallest (t), which is (t = 0). But that's at the starting point, 2001.Wait, but the problem says "within the range from 2001 to 2015", so (t) ranges from 0 to 15. So, if (E'(t)) is decreasing, its maximum is at (t = 0), and it decreases thereafter.But maybe I misinterpreted. The problem says "when the efficiency (E(t)) was increasing at the fastest rate." So, the rate of increase is (E'(t)), which is (frac{200}{t + 1}). So, to find when this rate is the highest, we need to find the maximum of (E'(t)). Since (E'(t)) is decreasing, its maximum is at the smallest (t), which is (t = 0). So, the efficiency was increasing the fastest at (t = 0), which is 2001.But that seems a bit odd because usually, when we talk about the fastest rate of increase, we might be referring to the point where the function is curving the most, but in this case, since (E'(t)) is strictly decreasing, its maximum is at the beginning.Alternatively, maybe the question is asking for when the efficiency was increasing at the fastest rate, which could be interpreted as when the second derivative is zero, but in this case, the second derivative is negative, so the function is concave down everywhere, meaning the rate of increase is always decreasing.Wait, let's compute the second derivative to check.Compute (E''(t)):(E'(t) = frac{200}{t + 1})Derivative of that is:(E''(t) = -frac{200}{(t + 1)^2})Which is always negative for all (t > -1). So, the function (E(t)) is concave down everywhere in the domain (t geq 0). Therefore, the rate of increase (E'(t)) is always decreasing. So, the maximum rate of increase occurs at the smallest (t), which is (t = 0).Therefore, the efficiency was increasing at the fastest rate at (t = 0), which is the year 2001.But let me think again. Maybe the question is asking for when the efficiency was increasing the fastest in terms of the slope, but if the slope is always decreasing, then the maximum slope is at the start. So, yes, t=0.Alternatively, if we consider the rate of change of the efficiency, which is (E'(t)), and we want to find when this rate is the highest, which is indeed at t=0.So, the answer is t=0, which is 2001.Wait, but let me check if I interpreted the question correctly. It says "the point in time t (within the range from 2001 to 2015) when the efficiency E(t) was increasing at the fastest rate." So, yes, the fastest rate of increase is the maximum of (E'(t)), which is at t=0.Alternatively, sometimes people might confuse the maximum rate with the point where the function is growing the fastest in terms of absolute change, but in this case, since the derivative is always decreasing, the maximum is at the start.So, I think the answer is t=0, which is 2001.But just to be thorough, let's plot or think about the behavior of (E(t)). (E(t)) is a logarithmic function, which grows slower as t increases. So, its slope decreases over time, meaning the rate of increase is highest at the beginning.Yes, that makes sense. So, the efficiency was increasing the fastest at t=0, 2001.So, summarizing:1. The average crime rate over 15 years is approximately 767.17 crimes per year.2. The efficiency was increasing at the fastest rate at t=0, which is the year 2001.Final Answer1. The average crime rate over the 15-year period is boxed{767.17}.2. The police efficiency was increasing at the fastest rate in the year boxed{2001}.</think>