🧠:<think>Alright, so I have this problem about budget allocation for a film grant. The total budget is 50,000, and it's divided into three categories: Equipment (E), Production (P), and Post-Production (PP). I need to figure out two things: 1. The probability that the combined cost of Equipment and Production will exceed 40,000.2. Given that Equipment actually costs 14,000, and Production is still uniformly distributed, find the expected value and variance of the remaining budget for Post-Production.Let me start by understanding the distributions involved.First, Equipment cost follows a normal distribution with a mean of 15,000 and a standard deviation of 3,000. So, E ~ N(15,000; 3,000²). Production cost is uniformly distributed between 20,000 and 25,000. So, P ~ U(20,000; 25,000). Post-Production is whatever is left from the total budget after subtracting Equipment and Production. So, PP = 50,000 - E - P.Starting with the first question: Probability that E + P > 40,000.Hmm, okay. So, I need to find P(E + P > 40,000). Since E is normal and P is uniform, their sum will have a distribution that's a convolution of the two. But calculating that convolution might be complicated. Maybe I can approximate it or find another way.Alternatively, since E and P are independent (I assume they are, unless stated otherwise), the sum E + P will have a mean equal to the sum of their means and variance equal to the sum of their variances.Let me calculate that.Mean of E: μ_E = 15,000Mean of P: Since P is uniform between 20,000 and 25,000, the mean is (20,000 + 25,000)/2 = 22,500.So, mean of E + P: μ_E+P = 15,000 + 22,500 = 37,500.Variance of E: σ²_E = 3,000² = 9,000,000.Variance of P: For a uniform distribution U(a; b), variance is (b - a)² / 12. So, (25,000 - 20,000)² / 12 = (5,000)² / 12 = 25,000,000 / 12 ≈ 2,083,333.33.So, variance of E + P: σ²_E+P = 9,000,000 + 2,083,333.33 ≈ 11,083,333.33.Therefore, standard deviation of E + P: σ_E+P = sqrt(11,083,333.33) ≈ 3,330.33.Now, we need to find P(E + P > 40,000). Since E + P is approximately normally distributed (due to the Central Limit Theorem, even though P is uniform, the sum should be roughly normal), we can standardize this.Let me denote S = E + P.So, P(S > 40,000) = P((S - μ_S)/σ_S > (40,000 - 37,500)/3,330.33) = P(Z > (2,500)/3,330.33) ≈ P(Z > 0.75).Looking up the standard normal distribution table, P(Z > 0.75) is approximately 0.2266. So, about 22.66% chance.Wait, let me verify that calculation. 40,000 - 37,500 is 2,500. Divided by 3,330.33 is approximately 0.75. Yes, that's correct.So, the probability is roughly 22.66%.But wait, is E + P exactly normal? E is normal, P is uniform. The sum of a normal and a uniform variable isn't exactly normal, but for the purposes of this problem, maybe we can approximate it as normal.Alternatively, perhaps we can model it more precisely. Let me think.Since E is normal and P is uniform, the sum S = E + P will have a distribution that is a convolution of the normal and uniform distributions. The convolution of a normal and a uniform is another normal distribution? Wait, no, the convolution of a normal and a uniform is a normal distribution convolved with a uniform, which results in a normal distribution with the same mean but increased variance. Wait, actually, no, that's not correct. The convolution of a normal and a uniform is not a normal distribution. It's a normal distribution convolved with a uniform, which is a kind of "smearing" of the normal distribution over the interval of the uniform.But in any case, the mean and variance calculations I did earlier still hold because variance is additive for independent variables regardless of their distribution. So, the mean is 37,500 and variance is approximately 11,083,333.33.Therefore, even though the distribution isn't exactly normal, using the normal approximation might still give a reasonable estimate for the probability.Alternatively, maybe we can compute the exact probability by integrating over the joint distribution. Since E is normal and P is uniform, the joint distribution is the product of the two individual distributions.So, the probability that E + P > 40,000 is the double integral over E and P where E + P > 40,000.Mathematically, it's:P(E + P > 40,000) = ∫ (from p=20,000 to 25,000) ∫ (from e=40,000 - p to ∞) f_E(e) f_P(p) de dpWhere f_E(e) is the normal density function and f_P(p) is the uniform density function.Since f_P(p) is 1/(25,000 - 20,000) = 1/5,000 for p between 20,000 and 25,000.So, the integral becomes:∫ (p=20,000 to 25,000) [1/5,000 * P(E > 40,000 - p)] dpWhich is:(1/5,000) * ∫ (p=20,000 to 25,000) P(E > 40,000 - p) dpNow, P(E > x) is 1 - Φ((x - μ_E)/σ_E), where Φ is the standard normal CDF.So, substituting:(1/5,000) * ∫ (p=20,000 to 25,000) [1 - Φ((40,000 - p - 15,000)/3,000)] dpSimplify the argument inside Φ:(40,000 - p - 15,000)/3,000 = (25,000 - p)/3,000So, the integral becomes:(1/5,000) * ∫ (p=20,000 to 25,000) [1 - Φ((25,000 - p)/3,000)] dpLet me make a substitution: Let z = (25,000 - p)/3,000. Then, when p = 20,000, z = (25,000 - 20,000)/3,000 = 5,000/3,000 ≈ 1.6667. When p = 25,000, z = 0.Also, dp = -3,000 dz.So, changing the limits accordingly:When p = 20,000, z = 1.6667When p = 25,000, z = 0So, the integral becomes:(1/5,000) * ∫ (z=1.6667 to 0) [1 - Φ(z)] * (-3,000 dz)The negative sign flips the limits:(1/5,000) * 3,000 ∫ (z=0 to 1.6667) [1 - Φ(z)] dzSimplify constants:(3,000 / 5,000) = 0.6So, 0.6 * ∫ (0 to 1.6667) [1 - Φ(z)] dzNow, ∫ [1 - Φ(z)] dz from 0 to a is equal to Φ(a) * a - φ(a) - [Φ(0)*0 - φ(0)] = Φ(a) * a - φ(a) - (0 - φ(0)) = Φ(a) * a - φ(a) + φ(0)Wait, actually, let me recall that ∫ [1 - Φ(z)] dz from 0 to a is equal to aΦ(a) - φ(a) + φ(0). Hmm, not sure. Maybe better to look up the integral of 1 - Φ(z).Alternatively, recall that ∫ [1 - Φ(z)] dz = zΦ(z) - φ(z) + C. Let me differentiate zΦ(z) - φ(z):d/dz [zΦ(z) - φ(z)] = Φ(z) + zφ(z) - (-zφ(z)) = Φ(z) + zφ(z) + zφ(z) = Φ(z) + 2zφ(z)Hmm, not quite 1 - Φ(z). Maybe I need a different approach.Alternatively, perhaps use integration by parts.Let me set u = 1 - Φ(z), dv = dzThen, du = -φ(z) dz, v = zSo, ∫ [1 - Φ(z)] dz = z(1 - Φ(z)) - ∫ z*(-φ(z)) dz = z(1 - Φ(z)) + ∫ zφ(z) dzNow, ∫ zφ(z) dz is known. Since φ(z) is the standard normal PDF, ∫ zφ(z) dz = -φ(z) + C.So, putting it together:∫ [1 - Φ(z)] dz = z(1 - Φ(z)) - φ(z) + CTherefore, the definite integral from 0 to a is:[a(1 - Φ(a)) - φ(a)] - [0(1 - Φ(0)) - φ(0)] = a(1 - Φ(a)) - φ(a) - (-φ(0)) = a(1 - Φ(a)) - φ(a) + φ(0)Since φ(0) is the standard normal PDF at 0, which is 1/√(2π) ≈ 0.3989.So, plugging in a = 1.6667 (which is 5/3 ≈ 1.6667):First, compute Φ(1.6667). Looking at standard normal tables, Φ(1.67) ≈ 0.9525. So, 1 - Φ(1.6667) ≈ 1 - 0.9525 = 0.0475.Then, φ(1.6667) = (1/√(2π)) * e^(- (1.6667)^2 / 2). Let's compute that:(1.6667)^2 ≈ 2.7778So, exponent is -2.7778 / 2 ≈ -1.3889e^(-1.3889) ≈ 0.25So, φ(1.6667) ≈ 0.3989 * 0.25 ≈ 0.0997Similarly, φ(0) ≈ 0.3989Therefore, the integral from 0 to 1.6667 of [1 - Φ(z)] dz ≈ 1.6667 * 0.0475 - 0.0997 + 0.3989Compute each term:1.6667 * 0.0475 ≈ 0.079166So, 0.079166 - 0.0997 + 0.3989 ≈ 0.079166 - 0.0997 = -0.020534 + 0.3989 ≈ 0.378366Therefore, the integral is approximately 0.378366.So, going back to the probability:0.6 * 0.378366 ≈ 0.2270Which is approximately 22.7%, which is very close to our initial approximation of 22.66%. So, that seems consistent.Therefore, the probability that Equipment and Production combined exceed 40,000 is approximately 22.7%.Okay, moving on to the second question.Assuming the actual cost for Equipment is 14,000, which is less than the mean of 15,000, and the actual cost for Production is uniformly distributed as given (between 20,000 and 25,000), what is the expected value and variance of the remaining budget for Post-Production.So, PP = 50,000 - E - P. Given E = 14,000, so PP = 50,000 - 14,000 - P = 36,000 - P.Since P is uniformly distributed between 20,000 and 25,000, then PP is 36,000 - P, which is a linear transformation of P.Therefore, PP is uniformly distributed between 36,000 - 25,000 = 11,000 and 36,000 - 20,000 = 16,000.So, PP ~ U(11,000; 16,000).Therefore, the expected value of PP is the mean of the uniform distribution:E[PP] = (11,000 + 16,000)/2 = 27,000 / 2 = 13,500.Variance of PP is ((16,000 - 11,000)^2)/12 = (5,000)^2 / 12 = 25,000,000 / 12 ≈ 2,083,333.33.Alternatively, since PP = 36,000 - P, and P ~ U(20,000; 25,000), we can compute E[PP] and Var(PP) directly.E[PP] = E[36,000 - P] = 36,000 - E[P] = 36,000 - 22,500 = 13,500.Var(PP) = Var(36,000 - P) = Var(P) = (25,000 - 20,000)^2 / 12 = 25,000,000 / 12 ≈ 2,083,333.33.So, same result.Therefore, the expected value is 13,500 and the variance is approximately 2,083,333.33.But let me double-check.Given E = 14,000, so PP = 50,000 - 14,000 - P = 36,000 - P.Since P is uniform between 20,000 and 25,000, subtracting P from 36,000 will shift the distribution. So, PP ranges from 36,000 - 25,000 = 11,000 to 36,000 - 20,000 = 16,000, which is correct.Therefore, PP is uniform between 11,000 and 16,000.Mean is (11,000 + 16,000)/2 = 13,500.Variance is (16,000 - 11,000)^2 / 12 = 25,000,000 / 12 ≈ 2,083,333.33.Yes, that's correct.So, summarizing:1. Probability that E + P > 40,000 is approximately 22.7%.2. Given E = 14,000, the expected value of PP is 13,500 and variance is approximately 2,083,333.33.I think that's it.Final Answer1. The probability is boxed{0.227}.2. The expected value is boxed{13500} and the variance is boxed{2083333.33}.</think>

🧠:<think>Okay, so I have this problem about predicting the number of votes in an election five years from now for the Dehra constituency. Let me try to break it down step by step.First, the problem states that the population is growing at a compound annual growth rate (CAGR) of 2.5% annually. The current population is 150,000, and in the most recent election, 60% of the population was eligible to vote. Out of those eligible voters, 70% actually voted. So, the first part is to calculate the expected total number of votes five years from now, assuming the same voting patterns hold and the population continues to grow at the same CAGR.Alright, let me start by understanding what CAGR means. Compound Annual Growth Rate is the mean annual growth rate of an investment over a specified period of time longer than one year. It's calculated by taking the nth root of the total growth rate, where n is the number of years. In this case, it's 2.5% per year for five years.So, to find the population in five years, I can use the formula for compound growth:Future Population = Current Population * (1 + CAGR)^nWhere:- Current Population = 150,000- CAGR = 2.5% = 0.025- n = 5 yearsPlugging in the numbers:Future Population = 150,000 * (1 + 0.025)^5Let me compute (1.025)^5 first. I think I can calculate this step by step.1.025^1 = 1.0251.025^2 = 1.025 * 1.025 = 1.0506251.025^3 = 1.050625 * 1.025 ≈ 1.0768906251.025^4 ≈ 1.076890625 * 1.025 ≈ 1.1038128906251.025^5 ≈ 1.103812890625 * 1.025 ≈ 1.131408212890625So, approximately 1.1314.Therefore, Future Population ≈ 150,000 * 1.1314 ≈ Let me compute that.150,000 * 1.1314 = ?Well, 150,000 * 1 = 150,000150,000 * 0.1314 = ?First, 150,000 * 0.1 = 15,000150,000 * 0.03 = 4,500150,000 * 0.0014 = 210So, adding those together: 15,000 + 4,500 = 19,500; 19,500 + 210 = 19,710Therefore, 150,000 * 1.1314 ≈ 150,000 + 19,710 = 169,710So, approximately 169,710 people will be the population in five years.Wait, let me double-check that multiplication because sometimes when I break it down, I might make a mistake.Alternatively, 1.1314 * 150,000.Let me compute 150,000 * 1.1314:150,000 * 1 = 150,000150,000 * 0.1 = 15,000150,000 * 0.03 = 4,500150,000 * 0.0014 = 210So, adding all these:150,000 + 15,000 = 165,000165,000 + 4,500 = 169,500169,500 + 210 = 169,710Yes, that seems correct.So, the population in five years is approximately 169,710.Now, moving on. In the most recent election, 60% of the population was eligible to vote. So, I need to calculate the number of eligible voters in five years.Eligible Voters = Future Population * 60% = 169,710 * 0.6Let me compute that.169,710 * 0.6 = ?Well, 100,000 * 0.6 = 60,00060,000 * 0.6 = 36,0009,710 * 0.6 = 5,826Wait, maybe a better way:169,710 * 0.6Break it down:169,710 * 0.6 = (170,000 - 290) * 0.6 = 170,000 * 0.6 - 290 * 0.6170,000 * 0.6 = 102,000290 * 0.6 = 174So, 102,000 - 174 = 101,826So, 101,826 eligible voters.Wait, let me verify:169,710 * 0.6Multiply 169,710 by 6, then divide by 10.169,710 * 6 = ?169,710 * 6:100,000 * 6 = 600,00060,000 * 6 = 360,0009,710 * 6 = 58,260So, 600,000 + 360,000 = 960,000960,000 + 58,260 = 1,018,260Divide by 10: 101,826Yes, that's correct. So, 101,826 eligible voters.Now, in the recent election, 70% of eligible voters actually voted. So, the number of votes cast would be 70% of eligible voters.Total Votes = Eligible Voters * 70% = 101,826 * 0.7Let me compute that.101,826 * 0.7Again, break it down:100,000 * 0.7 = 70,0001,826 * 0.7 = 1,278.2So, total votes = 70,000 + 1,278.2 = 71,278.2Since the number of votes should be a whole number, we can round it to 71,278 votes.Wait, let me compute 101,826 * 0.7 directly.101,826 * 0.7:100,000 * 0.7 = 70,0001,826 * 0.7 = 1,278.2So, adding together: 70,000 + 1,278.2 = 71,278.2Yes, so approximately 71,278 votes.Alternatively, 101,826 * 0.7 = 71,278.2, which is 71,278 when rounded down or 71,279 when rounded up. Since we can't have a fraction of a vote, we might need to consider whether to round up or down. But in the context of elections, it's usually rounded to the nearest whole number, so 71,278.2 would be approximately 71,278 votes.So, the expected total number of votes in five years is approximately 71,278.Wait, but let me make sure I didn't make any calculation errors.First, population in five years: 150,000 * (1.025)^5 ≈ 169,710.Eligible voters: 169,710 * 0.6 = 101,826.Voters who actually voted: 101,826 * 0.7 = 71,278.2 ≈ 71,278.Yes, that seems consistent.So, the answer to the first part is approximately 71,278 votes.Now, moving on to the second part. The critical constituent wants to ensure a victory by securing at least 55% of the votes. So, we need to determine the minimum number of votes they need to secure based on the expected total number of votes calculated in part 1.So, 55% of 71,278 votes is the minimum number needed.Let me compute 55% of 71,278.55% is equivalent to 0.55.So, 71,278 * 0.55.Let me compute that.71,278 * 0.5 = 35,63971,278 * 0.05 = 3,563.9Adding them together: 35,639 + 3,563.9 = 39,202.9So, approximately 39,202.9 votes.Since you can't have a fraction of a vote, we need to round up to ensure at least 55%. So, 39,203 votes.Wait, let me verify:71,278 * 0.55Compute 71,278 * 55 / 100First, 71,278 * 55:Compute 71,278 * 50 = 3,563,90071,278 * 5 = 356,390Adding together: 3,563,900 + 356,390 = 3,920,290Now, divide by 100: 3,920,290 / 100 = 39,202.9So, 39,202.9 votes, which we can round up to 39,203 votes to ensure at least 55%.Therefore, the minimum number of votes needed is 39,203.Wait, but let me think again. If you get exactly 55%, that's 39,202.9, which is approximately 39,203. So, to secure at least 55%, you need to have more than 39,202.9, so 39,203 is the minimum whole number.Yes, that makes sense.So, summarizing:1. The expected total number of votes in five years is approximately 71,278.2. To secure at least 55% of the votes, the critical constituent needs to secure a minimum of 39,203 votes.Just to recap the steps:1. Calculate future population using CAGR formula.2. Determine eligible voters as 60% of future population.3. Calculate total votes as 70% of eligible voters.4. For the second part, find 55% of the total votes calculated in step 3.I think that covers everything. I don't see any mistakes in my calculations, but let me just cross-verify the population growth one more time.Current population: 150,000After 1 year: 150,000 * 1.025 = 153,750After 2 years: 153,750 * 1.025 ≈ 157,593.75After 3 years: 157,593.75 * 1.025 ≈ 161,566.72After 4 years: 161,566.72 * 1.025 ≈ 165,684.19After 5 years: 165,684.19 * 1.025 ≈ 169,710.06Yes, so that's consistent with my earlier calculation of approximately 169,710.So, all the steps check out.Final Answer1. The expected total number of votes in five years is boxed{71278}.2. The minimum number of votes needed to secure at least 55% is boxed{39203}.</think>

🧠:<think>Alright, so I have two problems here about Alex's performance in sprinting and archery. Let me tackle them one by one.Starting with the first problem: Sprinting Performance. The function given is ( v(t) = 12 - 0.2t^2 ), and I need to find the total distance Alex covers in the first 5 seconds. Hmm, okay. I remember that distance covered is the integral of the velocity function over time. So, if I integrate ( v(t) ) from 0 to 5 seconds, that should give me the total distance.Let me write that down. The distance ( D ) is:( D = int_{0}^{5} v(t) , dt = int_{0}^{5} (12 - 0.2t^2) , dt )Alright, integrating term by term. The integral of 12 with respect to t is 12t, and the integral of ( -0.2t^2 ) is ( -0.2 times frac{t^3}{3} ), which simplifies to ( -frac{0.2}{3}t^3 ). So putting it together:( D = [12t - frac{0.2}{3}t^3] ) evaluated from 0 to 5.Let me compute this at 5 first:12 * 5 = 60( frac{0.2}{3} * 5^3 = frac{0.2}{3} * 125 = frac{25}{3} approx 8.3333 )So, subtracting that from 60: 60 - 8.3333 ≈ 51.6667 meters.At t=0, both terms are zero, so the total distance is approximately 51.6667 meters. To be precise, it's ( 51 frac{2}{3} ) meters, which is 51.666... meters.Wait, let me double-check my integration. The integral of ( t^2 ) is ( frac{t^3}{3} ), so multiplying by -0.2 gives ( -frac{0.2}{3}t^3 ). That seems right. Then evaluating from 0 to 5, yes, 12*5 is 60, 0.2/3 is approximately 0.066666..., times 125 is 8.3333. So 60 - 8.3333 is indeed 51.6667. So, 51.6667 meters, which is 51 and two-thirds meters. So, I can write that as ( frac{155}{3} ) meters if I want an exact fraction.But maybe the question expects a decimal? Hmm, 51.6667 is approximately 51.67 meters. But since it's a sprint, maybe they want an exact value. Let me see, 0.2 is 1/5, so 0.2/3 is 1/15. So, 12t - (1/15)t^3. At t=5, that's 60 - (1/15)*125 = 60 - 25/3. 25/3 is approximately 8.3333, so 60 - 8.3333 is 51.6667. So, 51.6667 meters is correct.Alright, so that's the first part. Now, moving on to the second problem: Archery Accuracy. The function given is ( A(x) = -0.5x^2 + 5x ). I need to find the maximum accuracy score and the distance x where this maximum occurs.Hmm, okay. Since this is a quadratic function, and the coefficient of ( x^2 ) is negative (-0.5), the parabola opens downward, so the vertex is the maximum point. The vertex of a parabola given by ( ax^2 + bx + c ) is at ( x = -frac{b}{2a} ).So, in this case, a = -0.5 and b = 5. Plugging into the formula:( x = -frac{5}{2*(-0.5)} = -frac{5}{-1} = 5 ).So, the maximum occurs at x = 5 meters. Now, to find the maximum accuracy score, plug x=5 back into the function:( A(5) = -0.5*(5)^2 + 5*(5) = -0.5*25 + 25 = -12.5 + 25 = 12.5 ).So, the maximum accuracy score is 12.5 points at x = 5 meters.Wait, let me make sure I did that correctly. So, the vertex formula is correct? Yes, for a quadratic ( ax^2 + bx + c ), the x-coordinate of the vertex is indeed ( -b/(2a) ). So, plugging in a = -0.5 and b = 5, we get x = -5/(2*(-0.5)) = -5/(-1) = 5. That's correct.Calculating A(5): ( -0.5*(25) + 25 = -12.5 + 25 = 12.5 ). Yep, that's right.Alternatively, I could have completed the square to find the maximum. Let me try that method just to verify.Starting with ( A(x) = -0.5x^2 + 5x ). Factor out -0.5 from the first two terms:( A(x) = -0.5(x^2 - 10x) ).Now, to complete the square inside the parentheses, take half of -10, which is -5, square it to get 25. So, add and subtract 25 inside:( A(x) = -0.5[(x^2 - 10x + 25) - 25] )Simplify:( A(x) = -0.5[(x - 5)^2 - 25] )Distribute the -0.5:( A(x) = -0.5(x - 5)^2 + 12.5 )So, in vertex form, it's ( A(x) = -0.5(x - 5)^2 + 12.5 ). Therefore, the maximum value is 12.5 when x = 5. That confirms my earlier result.So, both methods give the same answer, which is reassuring.Just to think a bit more about it, the function models accuracy as a function of distance from the target center. So, the further away you are, the less accurate you are, which makes sense. The maximum accuracy is achieved at 5 meters, which is probably the optimal distance for Alex's archery setup.Wait, but 5 meters seems quite close for archery. In real competitions, archers shoot at 70 meters or more. But maybe this is a different context, or perhaps it's scaled down for training purposes. Anyway, the math checks out regardless of the real-world context.So, summarizing my findings:1. For the sprint, the total distance covered in the first 5 seconds is approximately 51.67 meters, or exactly ( frac{155}{3} ) meters.2. For the archery, the maximum accuracy score is 12.5 points, achieved at a distance of 5 meters from the target center.I think that's all. I don't see any mistakes in my calculations, and both methods for the archery problem gave the same result, so I feel confident about these answers.Final Answer1. The total distance Alex covers in the first 5 seconds is boxed{dfrac{155}{3}} meters.2. The maximum accuracy score Alex can achieve is boxed{12.5} points at a distance of boxed{5} meters from the target center.</think>

🧠:<think>Alright, so I have this problem where I need to advise a billionaire on the risks of investing in a portfolio that includes both fossil fuel stocks and green energy stocks. The main focus is on how global carbon tax policies might affect the portfolio's returns and volatility over ten years. Let me break this down step by step.First, the portfolio is split 60% fossil fuels and 40% green energy. The expected returns and standard deviations are given for each, along with their correlation. I need to calculate the expected return and variance of the portfolio. Then, considering a carbon tax that reduces fossil fuel returns by 2% per annum, I have to recalculate these metrics and discuss the implications.Starting with part 1: calculating the expected return and variance.For the expected return, it's straightforward. It's just the weighted average of the expected returns of each asset. So, 60% of 5% plus 40% of 10%. Let me compute that:Expected Return = (0.6 * 5%) + (0.4 * 10%) = 3% + 4% = 7%. So, the expected return is 7% per annum.Now, for variance. The formula for the variance of a portfolio with two assets is:Var = (w1^2 * σ1^2) + (w2^2 * σ2^2) + 2 * w1 * w2 * ρ * σ1 * σ2Where w1 and w2 are the weights, σ1 and σ2 are the standard deviations, and ρ is the correlation coefficient.Plugging in the numbers:w1 = 0.6, σ1 = 12% = 0.12w2 = 0.4, σ2 = 18% = 0.18ρ = -0.3So,Var = (0.6^2 * 0.12^2) + (0.4^2 * 0.18^2) + 2 * 0.6 * 0.4 * (-0.3) * 0.12 * 0.18Let me compute each term step by step.First term: 0.6^2 * 0.12^2 = 0.36 * 0.0144 = 0.005184Second term: 0.4^2 * 0.18^2 = 0.16 * 0.0324 = 0.005184Third term: 2 * 0.6 * 0.4 * (-0.3) * 0.12 * 0.18Let me compute this:2 * 0.6 = 1.21.2 * 0.4 = 0.480.48 * (-0.3) = -0.144-0.144 * 0.12 = -0.01728-0.01728 * 0.18 = -0.0031104So, the third term is -0.0031104Adding all three terms together:0.005184 + 0.005184 = 0.0103680.010368 - 0.0031104 = 0.0072576So, the variance is 0.0072576. To get the standard deviation, I'd take the square root, but since the question asks for variance, I can leave it at that.But wait, the question mentions a 10-year period. Does that affect anything? Hmm, expected return over 10 years would just be 7% per annum, so 70% total, but variance scales with time. Since variance is annual, over 10 years, it would be 10 times the annual variance.Wait, actually, for variance, if returns are independent, the variance over n years is n times the annual variance. So, the 10-year variance would be 10 * 0.0072576 = 0.072576.But let me confirm. The question says "over the 10-year period," but it's unclear whether they want the annualized variance or the total variance over 10 years. Since variance is usually expressed annually, but when considering total variance over 10 years, it would indeed be 10 times the annual variance. However, sometimes people express total variance as (σ * sqrt(n))^2, which is the same as n * σ^2. So, yes, 10 times the annual variance.So, the annual variance is 0.0072576, and over 10 years, it's 0.072576.But let me check the question again. It says "calculate the expected return and variance of the portfolio over the 10-year period." So, it's possible they want the total expected return and total variance over 10 years, not annualized.So, expected return over 10 years would be 7% * 10 = 70%.Variance over 10 years would be 10 * 0.0072576 = 0.072576.But sometimes, people express variance in terms of annualized terms. So, if we consider the variance per year, it's 0.0072576, but over 10 years, it's additive.Alternatively, if we think in terms of standard deviation, the annual standard deviation is sqrt(0.0072576) ≈ 0.0852 or 8.52%. Over 10 years, the standard deviation would be 8.52% * sqrt(10) ≈ 26.97%. But the question asks for variance, so 0.072576.Wait, but actually, the variance of the total return over 10 years is 10 times the annual variance because each year's return is independent. So, yes, 10 * 0.0072576 = 0.072576.But let me make sure. The formula for variance over n periods is n * σ^2 if returns are independent and identically distributed, which they are in this case since each year is a normal distribution with the same mean and variance.So, yes, the variance over 10 years is 10 * 0.0072576 = 0.072576.Alternatively, if they want the variance per year, it's 0.0072576, but since the question says "over the 10-year period," I think they want the total variance, which is 0.072576.But to be safe, maybe I should present both. But I think the question expects the variance over the 10-year period, so 0.072576.Wait, but actually, the variance of the total return is 10 * annual variance. So, yes, 0.072576.So, summarizing part 1:Expected Return: 70% over 10 years.Variance: 0.072576 (which is 7.2576% in decimal terms, but usually variance is expressed in decimal, so 0.072576).Moving on to part 2: analyzing the impact of a global carbon tax that reduces fossil fuel returns by 2% per annum.So, the expected return of fossil fuels was 5%, now it becomes 5% - 2% = 3% per annum.We need to recalculate the expected return and variance of the portfolio.First, new expected return:(0.6 * 3%) + (0.4 * 10%) = 1.8% + 4% = 5.8% per annum.Over 10 years, that's 58%.Now, variance. The standard deviations remain the same, only the expected returns change. The correlation is still -0.3.So, using the same formula:Var = (0.6^2 * 0.12^2) + (0.4^2 * 0.18^2) + 2 * 0.6 * 0.4 * (-0.3) * 0.12 * 0.18Wait, but the expected returns changed, but the variances and standard deviations are based on the returns' volatility, not their expected returns. So, the standard deviations remain 12% and 18%, so the variance calculation remains the same as before.Wait, is that correct? Let me think. The standard deviation is a measure of volatility, which is the risk, not the expected return. So, even if the expected return changes, the standard deviation remains the same unless the volatility changes. The problem doesn't mention any change in volatility, only a reduction in expected returns. So, the variance calculation remains the same as in part 1.Therefore, the variance over 10 years is still 0.072576.But wait, let me double-check. The variance depends on the standard deviations and correlation, which haven't changed. So yes, the variance remains the same.So, the new expected return is 5.8% per annum, or 58% over 10 years, and the variance is still 0.072576.Now, discussing the implications. The expected return has decreased from 7% to 5.8% per annum, which is a significant drop. The variance remains the same, so the risk hasn't changed, but the return has decreased. This makes the portfolio less attractive in terms of risk-adjusted returns.Moreover, from a sustainable investment perspective, investing in fossil fuels is becoming riskier as carbon taxes are implemented. The portfolio still has a significant allocation to fossil fuels (60%), which is exposed to policy risks. Sustainable investing typically favors reducing exposure to such assets and increasing in green energy, which is expected to benefit from the transition to a low-carbon economy.Therefore, the portfolio's risk-return profile has worsened, and it may not align well with sustainable investment principles, which would advocate for a higher allocation to green energy and lower to fossil fuels.But let me think again. The variance didn't change because the standard deviations and correlation remained the same. So, the risk is the same, but the return is lower. So, the Sharpe ratio (return per unit of risk) has decreased, making the portfolio less efficient.In terms of sustainable investing, the portfolio still holds a majority in fossil fuels, which is against the principles of sustainability. Sustainable investing would prefer a portfolio with higher green energy allocation to support the transition and avoid the risks associated with fossil fuels.Therefore, the advice would be to reconsider the portfolio composition, possibly reducing fossil fuel exposure and increasing green energy to align with sustainable goals and improve the risk-return profile.Wait, but in part 2, the variance remained the same, but the expected return decreased. So, the portfolio is now offering lower returns for the same level of risk, which is worse. Additionally, the reliance on fossil fuels makes it vulnerable to future policy changes and regulatory risks beyond just the carbon tax, such as stricter emission standards, divestment movements, etc.So, in conclusion, the portfolio's performance has deteriorated, and it doesn't align with sustainable investing principles. The billionaire should consider reallocating to more sustainable assets to mitigate risks and potentially enhance returns in the long run.</think>