🧠:<think>Alright, so I need to figure out the total weighted scores for each candidate in this Weighted Preferential Voting System (WPVS). The weighting function is such that if a candidate is ranked k-th by a voter, they get a score of 1/k². First, let's tackle part 1 with the original 3 candidates: A, B, and C, with 100 voters. The rankings are as follows:- 30 voters: A > B > C- 40 voters: B > C > A- 20 voters: C > A > B- 10 voters: C > B > AI need to calculate the total score for each candidate. Let's break it down step by step.Starting with candidate A:1. In the first group of 30 voters, A is ranked 1st. So each of these voters gives A a score of 1/1² = 1. Therefore, total score from this group is 30 * 1 = 30.2. In the second group of 40 voters, A is ranked 3rd. So each of these voters gives A a score of 1/3² = 1/9 ≈ 0.1111. Total score from this group is 40 * (1/9) ≈ 4.4444.3. In the third group of 20 voters, A is ranked 2nd. So each gives A a score of 1/2² = 1/4 = 0.25. Total score here is 20 * 0.25 = 5.4. In the fourth group of 10 voters, A is ranked 3rd. So each gives A 1/9 ≈ 0.1111. Total score is 10 * (1/9) ≈ 1.1111.Adding these up for A: 30 + 4.4444 + 5 + 1.1111 ≈ 40.5555.Now, moving on to candidate B:1. In the first group of 30 voters, B is ranked 2nd. So each gives B 1/4 = 0.25. Total score: 30 * 0.25 = 7.5.2. In the second group of 40 voters, B is ranked 1st. Each gives B 1. Total score: 40 * 1 = 40.3. In the third group of 20 voters, B is ranked 3rd. Each gives B 1/9 ≈ 0.1111. Total score: 20 * (1/9) ≈ 2.2222.4. In the fourth group of 10 voters, B is ranked 2nd. Each gives B 1/4 = 0.25. Total score: 10 * 0.25 = 2.5.Adding these up for B: 7.5 + 40 + 2.2222 + 2.5 ≈ 52.2222.Next, candidate C:1. In the first group of 30 voters, C is ranked 3rd. Each gives C 1/9 ≈ 0.1111. Total score: 30 * (1/9) ≈ 3.3333.2. In the second group of 40 voters, C is ranked 2nd. Each gives C 1/4 = 0.25. Total score: 40 * 0.25 = 10.3. In the third group of 20 voters, C is ranked 1st. Each gives C 1. Total score: 20 * 1 = 20.4. In the fourth group of 10 voters, C is ranked 1st. Each gives C 1. Total score: 10 * 1 = 10.Adding these up for C: 3.3333 + 10 + 20 + 10 ≈ 43.3333.So, summarizing the scores:- A: ≈40.5555- B: ≈52.2222- C: ≈43.3333Therefore, under WPVS, candidate B has the highest score and would win.Now, moving on to part 2 where a new candidate D is introduced. The new rankings are:- 25 voters: A > B > C > D- 25 voters: B > D > C > A- 25 voters: C > A > D > B- 25 voters: D > C > B > AWe need to calculate the total weighted scores for A, B, C, and D.Let's start with each candidate.First, candidate A:1. In the first group of 25 voters, A is 1st. So each gives A 1. Total: 25 * 1 = 25.2. In the second group of 25 voters, A is 4th. So each gives A 1/4² = 1/16 ≈ 0.0625. Total: 25 * 0.0625 ≈ 1.5625.3. In the third group of 25 voters, A is 2nd. Each gives A 1/4 = 0.25. Total: 25 * 0.25 = 6.25.4. In the fourth group of 25 voters, A is 4th. Each gives A 1/16 ≈ 0.0625. Total: 25 * 0.0625 ≈ 1.5625.Adding up for A: 25 + 1.5625 + 6.25 + 1.5625 ≈ 34.375.Next, candidate B:1. In the first group of 25 voters, B is 2nd. Each gives B 1/4 = 0.25. Total: 25 * 0.25 = 6.25.2. In the second group of 25 voters, B is 1st. Each gives B 1. Total: 25 * 1 = 25.3. In the third group of 25 voters, B is 4th. Each gives B 1/16 ≈ 0.0625. Total: 25 * 0.0625 ≈ 1.5625.4. In the fourth group of 25 voters, B is 3rd. Each gives B 1/9 ≈ 0.1111. Total: 25 * (1/9) ≈ 2.7778.Adding up for B: 6.25 + 25 + 1.5625 + 2.7778 ≈ 35.5903.Now, candidate C:1. In the first group of 25 voters, C is 3rd. Each gives C 1/9 ≈ 0.1111. Total: 25 * (1/9) ≈ 2.7778.2. In the second group of 25 voters, C is 3rd. Each gives C 1/9 ≈ 0.1111. Total: 25 * (1/9) ≈ 2.7778.3. In the third group of 25 voters, C is 1st. Each gives C 1. Total: 25 * 1 = 25.4. In the fourth group of 25 voters, C is 2nd. Each gives C 1/4 = 0.25. Total: 25 * 0.25 = 6.25.Adding up for C: 2.7778 + 2.7778 + 25 + 6.25 ≈ 36.8056.Lastly, candidate D:1. In the first group of 25 voters, D is 4th. Each gives D 1/16 ≈ 0.0625. Total: 25 * 0.0625 ≈ 1.5625.2. In the second group of 25 voters, D is 2nd. Each gives D 1/4 = 0.25. Total: 25 * 0.25 = 6.25.3. In the third group of 25 voters, D is 3rd. Each gives D 1/9 ≈ 0.1111. Total: 25 * (1/9) ≈ 2.7778.4. In the fourth group of 25 voters, D is 1st. Each gives D 1. Total: 25 * 1 = 25.Adding up for D: 1.5625 + 6.25 + 2.7778 + 25 ≈ 35.5903.So, summarizing the scores:- A: ≈34.375- B: ≈35.5903- C: ≈36.8056- D: ≈35.5903Comparing these, candidate C has the highest score, followed closely by B and D. So, under the WPVS, candidate C would win.Wait, let me double-check the calculations to make sure I didn't make any arithmetic errors.For candidate A:25 (1st) + 1.5625 (4th) + 6.25 (2nd) + 1.5625 (4th) = 25 + 1.5625 = 26.5625; 26.5625 + 6.25 = 32.8125; 32.8125 + 1.5625 = 34.375. Correct.For candidate B:6.25 (2nd) + 25 (1st) + 1.5625 (4th) + 2.7778 (3rd) = 6.25 + 25 = 31.25; 31.25 + 1.5625 = 32.8125; 32.8125 + 2.7778 ≈ 35.5903. Correct.For candidate C:2.7778 (3rd) + 2.7778 (3rd) + 25 (1st) + 6.25 (2nd) = 2.7778 + 2.7778 = 5.5556; 5.5556 + 25 = 30.5556; 30.5556 + 6.25 ≈ 36.8056. Correct.For candidate D:1.5625 (4th) + 6.25 (2nd) + 2.7778 (3rd) + 25 (1st) = 1.5625 + 6.25 = 7.8125; 7.8125 + 2.7778 ≈ 10.5903; 10.5903 + 25 ≈ 35.5903. Correct.Yes, the calculations seem accurate. So, with the introduction of D, the scores have changed, and now C is the winner.</think>

🧠:<think>Okay, so I have this problem where a travel blogger visited five sections of the Great Barrier Reef and recorded coral coverage and the number of species in each section. They want to model the relationship between coral coverage (C) and number of species (S) using a linear regression of the form S = aC + b. Then, they want to predict the number of species for a new section with 75% coral coverage, considering a margin of error of ±2 species. First, I need to recall how linear regression works, specifically the least squares method. From what I remember, linear regression finds the best-fitting line through the data points by minimizing the sum of the squares of the vertical distances between the observed points and the line. The formula for the slope (a) and the intercept (b) can be calculated using some summations.Let me write down the data again to make sure I have it correctly:| Section | Coral Coverage (C) | Number of Species (S) ||---------|--------------------|-----------------------|| A | 60 | 15 || B | 70 | 20 || C | 55 | 18 || D | 65 | 22 || E | 50 | 17 |So, there are five data points. I need to compute the slope (a) and intercept (b) for the regression line S = aC + b.The formulas for a and b in simple linear regression are:a = (nΣ(C*S) - ΣCΣS) / (nΣC² - (ΣC)²)b = (ΣS - aΣC) / nWhere n is the number of data points, which is 5 here.So, I need to compute several sums:1. ΣC: sum of all coral coverages2. ΣS: sum of all number of species3. Σ(C*S): sum of the product of each C and S4. ΣC²: sum of the squares of each CLet me compute each of these step by step.First, ΣC:C values are 60, 70, 55, 65, 50.Adding them up: 60 + 70 = 130; 130 + 55 = 185; 185 + 65 = 250; 250 + 50 = 300.So, ΣC = 300.Next, ΣS:S values are 15, 20, 18, 22, 17.Adding them up: 15 + 20 = 35; 35 + 18 = 53; 53 + 22 = 75; 75 + 17 = 92.So, ΣS = 92.Now, Σ(C*S):I need to compute each C*S and then sum them.Let me compute each product:- For section A: 60 * 15 = 900- For section B: 70 * 20 = 1400- For section C: 55 * 18 = 990- For section D: 65 * 22 = 1430- For section E: 50 * 17 = 850Now, adding these up:900 + 1400 = 23002300 + 990 = 32903290 + 1430 = 47204720 + 850 = 5570So, Σ(C*S) = 5570.Next, ΣC²:Compute each C squared and sum them.- 60² = 3600- 70² = 4900- 55² = 3025- 65² = 4225- 50² = 2500Adding them up:3600 + 4900 = 85008500 + 3025 = 1152511525 + 4225 = 1575015750 + 2500 = 18250So, ΣC² = 18250.Now, plug these into the formula for a:a = (nΣ(C*S) - ΣCΣS) / (nΣC² - (ΣC)²)n = 5So, numerator = 5*5570 - 300*92Let me compute 5*5570 first: 5*5000=25000, 5*570=2850, so total is 25000 + 2850 = 27850.Next, compute 300*92: 300*90=27000, 300*2=600, so total is 27000 + 600 = 27600.So, numerator = 27850 - 27600 = 250.Denominator = 5*18250 - (300)^2Compute 5*18250: 18250*5. Let's see, 18000*5=90000, 250*5=1250, so total is 90000 + 1250 = 91250.Compute (300)^2 = 90000.So, denominator = 91250 - 90000 = 1250.Therefore, a = 250 / 1250 = 0.2.So, the slope a is 0.2.Now, compute the intercept b.Formula: b = (ΣS - aΣC) / nWe have ΣS = 92, a = 0.2, ΣC = 300, n = 5.Compute aΣC: 0.2 * 300 = 60.So, ΣS - aΣC = 92 - 60 = 32.Then, b = 32 / 5 = 6.4.So, the intercept b is 6.4.Therefore, the linear regression equation is S = 0.2C + 6.4.Let me double-check the calculations to make sure I didn't make any arithmetic errors.First, ΣC = 300, correct.ΣS = 92, correct.Σ(C*S) = 5570, correct.ΣC² = 18250, correct.Numerator for a: 5*5570 = 27850; 300*92 = 27600; 27850 - 27600 = 250, correct.Denominator: 5*18250 = 91250; 300² = 90000; 91250 - 90000 = 1250, correct.a = 250 / 1250 = 0.2, correct.b: (92 - 0.2*300)/5 = (92 - 60)/5 = 32/5 = 6.4, correct.So, the regression equation is S = 0.2C + 6.4.Now, moving on to part 2. The blogger wants to predict the number of species in a new section with 75% coral coverage, with a margin of error of ±2 species.So, first, compute the predicted number of species when C = 75.Using the regression equation: S = 0.2*75 + 6.4.Compute 0.2*75: 0.2*70=14, 0.2*5=1, so total 15.So, S = 15 + 6.4 = 21.4.So, the predicted number of species is 21.4.But since the margin of error is ±2 species, the expected range is from 21.4 - 2 to 21.4 + 2, which is 19.4 to 23.4.However, since the number of species should be a whole number, the range would be approximately 19 to 23 species.Wait, but the question says "Compute the range in which the number of species is expected to lie." It doesn't specify whether to round or keep it as decimals. Since the original data has whole numbers, maybe it's better to present the range as 19.4 to 23.4, but if we need to interpret it practically, species counts are whole numbers, so perhaps 19 to 23.But let me check if the margin of error is applied to the predicted value. So, if the predicted value is 21.4, then adding and subtracting 2 gives 19.4 and 23.4. So, the range is 19.4 to 23.4. If we need to present it as a range, we can write it as approximately 19.4 to 23.4 species.Alternatively, if we consider that the number of species must be an integer, the range would be from 19 to 23 species. But the problem doesn't specify, so I think it's safer to present the exact range as 19.4 to 23.4.But let me think again. The margin of error is ±2 species, so it's an interval around the predicted value. So, the lower bound is 21.4 - 2 = 19.4, and the upper bound is 21.4 + 2 = 23.4. So, the range is [19.4, 23.4].Therefore, the number of species is expected to lie between 19.4 and 23.4.But since species can't be a fraction, maybe the blogger would interpret this as between 19 and 23 species. However, the question doesn't specify rounding, so perhaps we should keep it as decimals.Alternatively, maybe the margin of error is given as ±2, so the range is 21.4 ± 2, which is 19.4 to 23.4.I think the answer should be presented as 19.4 to 23.4, but I'll check if the question expects integers. The original data has integer species counts, but the regression can result in a non-integer prediction. So, probably, it's acceptable to have decimal values in the range.So, summarizing:1. The linear regression equation is S = 0.2C + 6.4.2. For C = 75%, S = 21.4, with a margin of error of ±2, so the range is 19.4 to 23.4.Therefore, the final answers are:a = 0.2, b = 6.4Predicted range: 19.4 to 23.4.Wait, but let me double-check the calculation for S when C=75.0.2 * 75 = 15, plus 6.4 is 21.4, correct.So, yes, the calculations seem correct.I think that's it. I don't see any mistakes in the steps. I computed all the necessary sums, applied the formulas correctly, and double-checked the calculations.Final Answer1. The linear regression parameters are ( a = boxed{0.2} ) and ( b = boxed{6.4} ).2. The expected range for the number of species is from ( boxed{19.4} ) to ( boxed{23.4} ).</think>

🧠:<think>Okay, so I have this problem about a professional pumpkin carver who wants to maximize the surface area covered by designs while minimizing the amount of pumpkin flesh removed. There are two parts to this problem. Let me tackle them one by one.Starting with part 1: The carver uses a pattern that covers 75% of the pumpkin's surface area. I need to derive the expression for the surface area covered by the design and then calculate the exact area for a pumpkin with a radius of 12 inches.First, I remember that the surface area of a sphere is given by the formula (4pi r^2). Since the pattern covers 75% of this surface area, I can find the covered area by multiplying 75% (which is 0.75) by the total surface area.So, the expression for the surface area covered by the design should be (0.75 times 4pi r^2). Let me write that out:Surface area covered = (0.75 times 4pi r^2)Simplifying that, 0.75 times 4 is 3, so it becomes (3pi r^2). That seems straightforward.Now, for a pumpkin with a radius of 12 inches, I can plug that value into the expression. Let me compute that:Surface area covered = (3pi (12)^2)Calculating (12^2) gives 144, so:Surface area covered = (3pi times 144)Multiplying 3 by 144, that's 432. So, the exact area is (432pi) square inches. That makes sense.Moving on to part 2: The carving process removes a layer of pumpkin flesh, creating a shell of uniform thickness (t). The volume of the carved pumpkin's shell is to be no more than 20% of the original pumpkin's volume. I need to derive the maximum allowable thickness (t) in terms of (r) and then calculate (t) for a pumpkin with a radius of 12 inches.Alright, so the original pumpkin is a sphere with radius (r). When the carver removes a layer of thickness (t), the remaining pumpkin is a smaller sphere with radius (r - t). The volume of the shell (the part that's removed) is the difference between the original volume and the volume of the smaller sphere.The volume of a sphere is given by (frac{4}{3}pi r^3). So, the original volume is (frac{4}{3}pi r^3), and the volume after carving is (frac{4}{3}pi (r - t)^3). Therefore, the volume of the shell is:Volume of shell = (frac{4}{3}pi r^3 - frac{4}{3}pi (r - t)^3)We can factor out (frac{4}{3}pi) to simplify:Volume of shell = (frac{4}{3}pi [r^3 - (r - t)^3])The problem states that this volume should be no more than 20% of the original volume. 20% is 0.2, so:(frac{4}{3}pi [r^3 - (r - t)^3] leq 0.2 times frac{4}{3}pi r^3)I can divide both sides by (frac{4}{3}pi) to simplify:(r^3 - (r - t)^3 leq 0.2 r^3)Let me compute (r^3 - (r - t)^3). Expanding ((r - t)^3) using the binomial theorem:((r - t)^3 = r^3 - 3r^2 t + 3r t^2 - t^3)So, subtracting this from (r^3):(r^3 - (r^3 - 3r^2 t + 3r t^2 - t^3) = 3r^2 t - 3r t^2 + t^3)Therefore, the inequality becomes:(3r^2 t - 3r t^2 + t^3 leq 0.2 r^3)Hmm, this is a cubic equation in terms of (t). I need to solve for (t) in terms of (r). Since (t) is much smaller than (r) (as it's a thin shell), perhaps I can approximate the equation by neglecting the higher-order terms (3r t^2) and (t^3). Let me see if that's a valid assumption.If (t) is small compared to (r), then (t^2) and (t^3) would be much smaller than (r^2 t). So, maybe I can approximate:(3r^2 t approx 0.2 r^3)Solving for (t):(t approx frac{0.2 r^3}{3 r^2} = frac{0.2 r}{3} = frac{r}{15})But wait, is this approximation valid? Let me check. If (t) is approximately (r/15), then (t^2) would be (r^2/225), and (t^3) would be (r^3/3375). Compared to (3r^2 t = 3r^2 (r/15) = r^3/5), which is 0.2 r^3, the neglected terms are much smaller. So, the approximation is reasonable.Therefore, the maximum allowable thickness (t) is approximately (r/15).But let me also consider solving the cubic equation without approximation to see if the exact solution is significantly different.Starting from:(3r^2 t - 3r t^2 + t^3 = 0.2 r^3)Let me divide both sides by (r^3) to make it dimensionless:(3 frac{t}{r} - 3 left(frac{t}{r}right)^2 + left(frac{t}{r}right)^3 = 0.2)Let me set (x = frac{t}{r}), so the equation becomes:(3x - 3x^2 + x^3 = 0.2)Rearranging:(x^3 - 3x^2 + 3x - 0.2 = 0)This is a cubic equation in (x). Let me see if I can find a real root between 0 and 1, since (t) must be less than (r).Let me try plugging in (x = 0.1):(0.001 - 0.03 + 0.3 - 0.2 = 0.071), which is positive.At (x = 0.1), the left side is 0.071.At (x = 0.2):(0.008 - 0.12 + 0.6 - 0.2 = 0.288), still positive.Wait, that can't be. Wait, let me compute it correctly.Wait, (x = 0.1):(x^3 = 0.001)(-3x^2 = -0.03)(3x = 0.3)So, 0.001 - 0.03 + 0.3 - 0.2 = 0.001 - 0.03 is -0.029, plus 0.3 is 0.271, minus 0.2 is 0.071. So, 0.071.At (x = 0.15):(x^3 = 0.003375)(-3x^2 = -0.0675)(3x = 0.45)So, 0.003375 - 0.0675 + 0.45 - 0.2 = 0.003375 - 0.0675 = -0.064125 + 0.45 = 0.385875 - 0.2 = 0.185875.Still positive.At (x = 0.2):(x^3 = 0.008)(-3x^2 = -0.12)(3x = 0.6)So, 0.008 - 0.12 + 0.6 - 0.2 = 0.008 - 0.12 = -0.112 + 0.6 = 0.488 - 0.2 = 0.288.Still positive.Wait, so the function is positive at x=0.1, 0.15, 0.2. Let me try a higher x.Wait, maybe I made a mistake in the equation.Wait, the original equation is:(3x - 3x^2 + x^3 = 0.2)So, rearranged:(x^3 - 3x^2 + 3x - 0.2 = 0)Wait, perhaps I should compute the value at x=0.25:(x^3 = 0.015625)(-3x^2 = -0.1875)(3x = 0.75)So, 0.015625 - 0.1875 + 0.75 - 0.2 = 0.015625 - 0.1875 = -0.171875 + 0.75 = 0.578125 - 0.2 = 0.378125.Still positive.Wait, maybe I need to go higher. Let's try x=0.3:(x^3 = 0.027)(-3x^2 = -0.27)(3x = 0.9)So, 0.027 - 0.27 + 0.9 - 0.2 = 0.027 - 0.27 = -0.243 + 0.9 = 0.657 - 0.2 = 0.457.Still positive.Wait, maybe I need to go even higher. Let's try x=0.4:(x^3 = 0.064)(-3x^2 = -0.48)(3x = 1.2)So, 0.064 - 0.48 + 1.2 - 0.2 = 0.064 - 0.48 = -0.416 + 1.2 = 0.784 - 0.2 = 0.584.Still positive.Wait, this is confusing. Maybe I made a mistake in the equation.Wait, let's double-check the equation.Original volume of shell: (frac{4}{3}pi [r^3 - (r - t)^3])Set this equal to 0.2 times original volume: (0.2 times frac{4}{3}pi r^3)So, we have:(frac{4}{3}pi [r^3 - (r - t)^3] = 0.2 times frac{4}{3}pi r^3)Divide both sides by (frac{4}{3}pi):(r^3 - (r - t)^3 = 0.2 r^3)So, (r^3 - (r - t)^3 = 0.2 r^3)Which simplifies to:((r - t)^3 = r^3 - 0.2 r^3 = 0.8 r^3)Therefore:(r - t = sqrt[3]{0.8} r)So, (t = r - sqrt[3]{0.8} r = r (1 - sqrt[3]{0.8}))Ah, that's a much simpler way to solve it! I was complicating it earlier by expanding the cube. Let me do that.So, (t = r (1 - sqrt[3]{0.8}))Compute (sqrt[3]{0.8}). Let me calculate that.0.8 is equal to 4/5, so the cube root of 4/5.I know that cube root of 1 is 1, cube root of 0.729 is 0.9 because 0.9^3=0.729. 0.8 is a bit higher than 0.729, so cube root of 0.8 is a bit higher than 0.9.Let me compute it more accurately.Let me denote (x = sqrt[3]{0.8}). So, (x^3 = 0.8).Let me use linear approximation or Newton-Raphson method.Let me use Newton-Raphson.Let me start with an initial guess. Let me take x0 = 0.95.Compute f(x) = x^3 - 0.8.f(0.95) = 0.857375 - 0.8 = 0.057375f'(x) = 3x^2.f'(0.95) = 3*(0.9025) = 2.7075Next approximation: x1 = x0 - f(x0)/f'(x0) = 0.95 - 0.057375 / 2.7075 ≈ 0.95 - 0.0212 ≈ 0.9288Compute f(0.9288):0.9288^3 ≈ Let's compute 0.9^3 = 0.7290.0288^3 is negligible, but let's compute more accurately.Compute 0.9288 * 0.9288 = approx 0.8627Then, 0.8627 * 0.9288 ≈ 0.8627*0.9 = 0.77643, plus 0.8627*0.0288 ≈ 0.0248, total ≈ 0.77643 + 0.0248 ≈ 0.80123So, f(0.9288) ≈ 0.80123 - 0.8 = 0.00123That's pretty close. Compute f'(0.9288) = 3*(0.9288)^2 ≈ 3*(0.8627) ≈ 2.5881Next approximation: x2 = x1 - f(x1)/f'(x1) ≈ 0.9288 - 0.00123 / 2.5881 ≈ 0.9288 - 0.000475 ≈ 0.9283Compute f(0.9283):0.9283^3 ≈ Let's compute 0.9^3 = 0.7290.0283^3 is negligible, but let's compute more accurately.0.9283 * 0.9283 ≈ 0.86180.8618 * 0.9283 ≈ 0.8618*0.9 = 0.77562, plus 0.8618*0.0283 ≈ 0.02438, total ≈ 0.77562 + 0.02438 ≈ 0.8000So, f(0.9283) ≈ 0.8000 - 0.8 = 0.0000So, cube root of 0.8 is approximately 0.9283.Therefore, (t = r (1 - 0.9283) = r (0.0717))So, (t ≈ 0.0717 r)Which is approximately 7.17% of the radius.Earlier, my approximation was (t ≈ r/15 ≈ 0.0667 r), which is about 6.67%. So, the exact value is a bit higher, around 7.17%.So, the exact expression is (t = r (1 - sqrt[3]{0.8})), which is approximately 0.0717 r.Therefore, for a pumpkin with radius 12 inches, (t ≈ 0.0717 times 12 ≈ 0.8604) inches.Let me compute that more accurately.0.0717 * 12:0.07 * 12 = 0.840.0017 * 12 = 0.0204So, total ≈ 0.84 + 0.0204 ≈ 0.8604 inches.So, approximately 0.86 inches.But let me compute it more precisely using the exact cube root.Since we found that (sqrt[3]{0.8} ≈ 0.9283), so (t = 12 (1 - 0.9283) = 12 * 0.0717 ≈ 0.8604) inches.So, approximately 0.86 inches.But let me check if I can express this more precisely.Alternatively, since (sqrt[3]{0.8} = (4/5)^{1/3}), we can write:(t = r left(1 - left(frac{4}{5}right)^{1/3}right))But for the purposes of the answer, it's probably acceptable to write it in terms of the cube root or approximate it numerically.So, summarizing:1. The surface area covered is (3pi r^2), and for r=12 inches, it's (432pi) square inches.2. The maximum allowable thickness (t) is (r (1 - sqrt[3]{0.8})), which is approximately 0.0717 r. For r=12 inches, (t ≈ 0.86) inches.Wait, but let me double-check the exact value of the cube root of 0.8.Using a calculator, cube root of 0.8 is approximately 0.928315.So, 1 - 0.928315 ≈ 0.071685.Therefore, (t ≈ 0.071685 r).For r=12 inches:t ≈ 0.071685 * 12 ≈ 0.86022 inches.So, approximately 0.8602 inches, which we can round to 0.86 inches.Alternatively, if we want to express it exactly, it's (12(1 - sqrt[3]{0.8})) inches.But since the problem asks for the exact expression in terms of r and then to calculate t for r=12, I think it's acceptable to write the exact expression as (t = r (1 - sqrt[3]{0.8})) and then compute the numerical value as approximately 0.86 inches.Alternatively, we can write (sqrt[3]{0.8}) as (sqrt[3]{4/5}), so (t = r (1 - sqrt[3]{4/5})).But perhaps the problem expects the exact form, so I'll present both the exact expression and the numerical value.So, to recap:1. Surface area covered = (3pi r^2). For r=12, it's (432pi) square inches.2. Maximum thickness (t = r (1 - sqrt[3]{0.8})). For r=12, (t ≈ 0.86) inches.I think that's it. Let me just make sure I didn't make any calculation errors.For part 1:Surface area of sphere: (4pi r^2). 75% is 0.75, so 0.75 * 4πr² = 3πr². Correct.For r=12: 3π*(12)^2 = 3π*144 = 432π. Correct.For part 2:Volume of shell: original volume minus carved volume. Original volume is (4/3)πr³. Carved volume is (4/3)π(r - t)³. So, shell volume is (4/3)π[r³ - (r - t)³]. Set this equal to 0.2*(4/3)πr³, so r³ - (r - t)³ = 0.2 r³. Therefore, (r - t)³ = 0.8 r³, so r - t = (0.8)^(1/3) r, so t = r - (0.8)^(1/3) r = r(1 - (0.8)^(1/3)). Correct.Calculating (0.8)^(1/3) ≈ 0.9283, so t ≈ 0.0717 r. For r=12, t ≈ 0.86 inches. Correct.Yes, everything checks out.</think>

🧠:<think>Alright, so I have this problem about a coin collector who has 50 coins from four different periods: Ancient Greece, Ancient Rome, Medieval Europe, and the Renaissance. I need to figure out how many coins he has from each period. Let me try to break this down step by step.First, the problem gives me some relationships between the numbers of coins from each period. It says:1. The number of Ancient Roman coins is twice the number of Ancient Greek coins.2. The number of Medieval European coins is three times the number of Ancient Greek coins.3. The number of Renaissance coins is four less than the number of Medieval European coins.And the total number of coins is 50. So, I can set up some equations based on this information.Let me assign variables to each period to make it easier. Let's say:- Let G be the number of Ancient Greek coins.- Then, the number of Ancient Roman coins would be 2G (since it's twice the Greek coins).- The number of Medieval European coins would be 3G (three times the Greek coins).- The number of Renaissance coins would be 3G - 4 (four less than Medieval European coins).So, now I have expressions for each period in terms of G. The total number of coins is the sum of all these, which is 50. So, I can write the equation:G + 2G + 3G + (3G - 4) = 50Let me simplify this equation step by step.First, combine like terms:G + 2G is 3G.3G + 3G is 6G.6G - 4 is the total so far.So, 6G - 4 = 50Now, I need to solve for G.Add 4 to both sides:6G = 54Divide both sides by 6:G = 9So, the number of Ancient Greek coins is 9.Now, let's find the number of coins for each period:- Ancient Roman coins: 2G = 2*9 = 18- Medieval European coins: 3G = 3*9 = 27- Renaissance coins: 3G - 4 = 27 - 4 = 23Wait, let me check if these add up to 50.9 (Greek) + 18 (Roman) + 27 (Medieval) + 23 (Renaissance) = 9 + 18 is 27, 27 + 27 is 54, 54 + 23 is 77. Hmm, that's way more than 50. That can't be right. I must have made a mistake somewhere.Let me go back and check my equations.I set G as Ancient Greek, then Roman is 2G, Medieval is 3G, and Renaissance is 3G - 4.Total coins: G + 2G + 3G + (3G - 4) = 50Wait, adding those up: G + 2G is 3G, plus 3G is 6G, plus (3G -4) is 9G -4. So, 9G -4 = 50.Oh! I see where I messed up earlier. I thought 3G + 3G was 6G, but actually, it's 3G (from Roman) plus 3G (from Medieval) plus another 3G (from Renaissance, but wait, Renaissance is 3G -4, so it's not exactly 3G). Let me recast the equation correctly.Wait, no, the total is G (Greek) + 2G (Roman) + 3G (Medieval) + (3G -4) (Renaissance). So that's G + 2G + 3G + 3G -4 = 9G -4.So, 9G -4 = 50.Then, adding 4 to both sides: 9G = 54.Divide by 9: G = 6.Ah, okay, so G is 6, not 9. I see, I must have miscalculated earlier when I thought G was 9. Let me recalculate with G=6.So, Ancient Greek coins: 6Ancient Roman coins: 2G = 12Medieval European coins: 3G = 18Renaissance coins: 3G -4 = 18 -4 = 14Now, let's add them up: 6 + 12 + 18 +14.6 +12 is 18, 18 +18 is 36, 36 +14 is 50. Perfect, that's correct.So, the number of coins from each period is:- Ancient Greek: 6- Ancient Roman: 12- Medieval Europe: 18- Renaissance: 14Alright, that seems to check out.Now, moving on to the second part of the problem. The collector wants to display his coins in a rectangular glass case. The case's length is three times its width. Each coin occupies a square space of 4 square centimeters. The total area of the case is exactly enough to accommodate all 50 coins without any gaps or overlaps. I need to determine the dimensions of the glass case.First, let's figure out the total area required for the coins. Since each coin takes up 4 cm², 50 coins would take up 50 * 4 = 200 cm².So, the area of the glass case is 200 cm².The case is rectangular, with length three times its width. Let me denote the width as W. Then, the length would be 3W.The area of a rectangle is length multiplied by width, so:Area = Length * Width = 3W * W = 3W²We know the area is 200 cm², so:3W² = 200To find W, divide both sides by 3:W² = 200 / 3 ≈ 66.666...Then, take the square root of both sides:W = √(200 / 3)Let me compute that. √(200) is approximately 14.142, and √3 is approximately 1.732. So, 14.142 / 1.732 ≈ 8.164 cm.But let me do it more accurately.200 divided by 3 is approximately 66.6667.The square root of 66.6667 is approximately 8.164965809.So, W ≈ 8.165 cm.Then, the length is 3W ≈ 3 * 8.165 ≈ 24.495 cm.But since we're dealing with physical dimensions, it's better to express this in exact terms rather than approximate decimals.So, W² = 200 / 3, so W = √(200/3). Let me rationalize that.√(200/3) can be written as √(200)/√3 = (10√2)/√3. To rationalize the denominator, multiply numerator and denominator by √3:(10√2 * √3) / (√3 * √3) = (10√6)/3.So, W = (10√6)/3 cm.Similarly, the length is 3W = 3*(10√6)/3 = 10√6 cm.So, the exact dimensions are:Width: (10√6)/3 cmLength: 10√6 cmBut let me check if these make sense.Compute the area: Length * Width = 10√6 * (10√6)/3 = (100 * 6)/3 = 600 / 3 = 200 cm², which matches the required area. So, that's correct.Alternatively, if we need decimal approximations, we can compute √6 ≈ 2.449.So, Width ≈ (10 * 2.449)/3 ≈ 24.49 / 3 ≈ 8.163 cmLength ≈ 10 * 2.449 ≈ 24.49 cmSo, approximately 8.16 cm by 24.49 cm.But since the problem doesn't specify whether to provide exact values or approximate, and given that √6 is irrational, it's better to present the exact values.Therefore, the dimensions are (10√6)/3 cm in width and 10√6 cm in length.Let me recap:1. Number of coins: - Ancient Greek: 6 - Ancient Roman: 12 - Medieval Europe: 18 - Renaissance: 142. Dimensions of the glass case: - Width: (10√6)/3 cm - Length: 10√6 cmI think that's all. Let me just double-check my calculations to make sure I didn't make any errors.For the coins:G = 6Roman: 2*6=12Medieval: 3*6=18Renaissance: 18-4=14Total: 6+12=18, 18+18=36, 36+14=50. Correct.For the case:Area needed: 50*4=200 cm²Let width = W, length = 3WArea: 3W²=200 => W²=200/3 => W=√(200/3)=10√6/3Length=3W=10√6Yes, that's correct.So, I think I've got it right.Final Answer1. The number of coins from each period is: - Ancient Greece: boxed{6} - Ancient Rome: boxed{12} - Medieval Europe: boxed{18} - Renaissance: boxed{14}2. The dimensions of the glass case are: - Width: boxed{dfrac{10sqrt{6}}{3} text{cm}} - Length: boxed{10sqrt{6} text{cm}}</think>