🧠:<think>Alright, so I have this problem here about an iOS developer optimizing their algorithm. It's split into two parts, and I need to figure out both. Let me take it step by step.Starting with the first part: The developer is working with 1000 unique integers and wants to minimize memory usage and typecasting. The memory usage formula given is M(n) = 4n + 2, where n is the number of integers stored. They want to know the total memory usage when the dataset is fully stored.Okay, so n is 1000 here. That should be straightforward. I just plug in 1000 into the formula. Let me write that out:M(1000) = 4 * 1000 + 2Calculating that, 4 times 1000 is 4000, plus 2 is 4002 bytes. Hmm, that seems pretty simple. I don't think I need to do anything else for the first part. Maybe double-check the formula? It says 4n + 2, so yeah, 4 per integer plus a constant 2. So 1000 integers would indeed be 4000 + 2 = 4002 bytes.Moving on to the second part: They want to model the execution time T(n) with the function T(n) = A n log(n) + B n. They give two data points: when n=100, T=500 ms, and when n=200, T=1200 ms. I need to find A and B, and then use those to estimate the execution time when n=1000.Alright, so this is a system of equations problem. Let me write down the equations based on the given data.For n=100:500 = A * 100 * log(100) + B * 100For n=200:1200 = A * 200 * log(200) + B * 200First, I need to figure out what log base they're using. In computer science, log is often base 2, but sometimes it's base 10 or natural log. The problem doesn't specify, so I might need to assume. Let me check if it's base 2 because that's common in algorithms.So, log base 2 of 100 is approximately... Well, 2^6 is 64, 2^7 is 128, so log2(100) is about 6.643856. Similarly, log2(200) is log2(100*2) = log2(100) + 1, so that's approximately 6.643856 + 1 = 7.643856.Alternatively, if it's natural log, ln(100) is about 4.60517 and ln(200) is about 5.29832. But in algorithm analysis, log is usually base 2. Let me proceed with log base 2.So, plugging in the values:First equation:500 = A * 100 * 6.643856 + B * 100Second equation:1200 = A * 200 * 7.643856 + B * 200Let me simplify these equations.First equation:500 = 664.3856 A + 100 BSecond equation:1200 = 1528.7712 A + 200 BNow, I can write this as a system:664.3856 A + 100 B = 500 ...(1)1528.7712 A + 200 B = 1200 ...(2)I can solve this system using substitution or elimination. Let's try elimination. Let's multiply equation (1) by 2 so that the coefficients of B become 200.Multiplying equation (1) by 2:1328.7712 A + 200 B = 1000 ...(1a)Now, subtract equation (1a) from equation (2):(1528.7712 A + 200 B) - (1328.7712 A + 200 B) = 1200 - 1000This simplifies to:200 A = 200So, A = 1.Now, plug A = 1 back into equation (1):664.3856 * 1 + 100 B = 500664.3856 + 100 B = 500Subtract 664.3856 from both sides:100 B = 500 - 664.3856100 B = -164.3856So, B = -164.3856 / 100 = -1.643856So, A is 1 and B is approximately -1.643856.Wait, that seems a bit odd because B is negative. Let me check my calculations again.First, when I multiplied equation (1) by 2, I think I might have made a mistake there.Wait, equation (1) is 664.3856 A + 100 B = 500Multiplying by 2 gives 1328.7712 A + 200 B = 1000Equation (2) is 1528.7712 A + 200 B = 1200Subtracting (1a) from (2):(1528.7712 - 1328.7712) A + (200 - 200) B = 1200 - 1000So, 200 A = 200 => A = 1. That's correct.Plugging A = 1 into equation (1):664.3856 * 1 + 100 B = 500So, 664.3856 + 100 B = 500100 B = 500 - 664.3856 = -164.3856So, B = -1.643856. Hmm, that's correct. So B is negative. That might be because the constants are such.Alternatively, maybe I should check if the log is base 10 or natural log. Let me try that.If log is base 10:log10(100) = 2, log10(200) ≈ 2.3010So, plugging into the equations:First equation:500 = A * 100 * 2 + B * 100 => 200 A + 100 B = 500Second equation:1200 = A * 200 * 2.3010 + B * 200 => 460.2 A + 200 B = 1200So, system:200 A + 100 B = 500 ...(1)460.2 A + 200 B = 1200 ...(2)Multiply equation (1) by 2:400 A + 200 B = 1000 ...(1a)Subtract (1a) from (2):(460.2 - 400) A + (200 - 200) B = 1200 - 100060.2 A = 200A = 200 / 60.2 ≈ 3.322Then plug A back into equation (1):200 * 3.322 + 100 B = 500664.4 + 100 B = 500100 B = 500 - 664.4 = -164.4B = -1.644So, same result, just different A and B. So regardless of log base, we get A=1 and B≈-1.644 if log is base 2, or A≈3.322 and B≈-1.644 if log is base 10.But the problem didn't specify the base. Hmm. In algorithm analysis, log is usually base 2, but in some contexts, it might be natural log. Wait, the problem says "log(n)", so in mathematics, log is often base e, but in computer science, it's base 2. Hmm.Wait, let me check the original problem statement. It says "execution time T(n) = A n log(n) + B n". It doesn't specify the base, so perhaps it's base 2? Or maybe it's natural log? Hmm.But in the first case, when I assumed log base 2, I got A=1 and B≈-1.644. Let me see if that makes sense.Wait, let's test the values.If A=1 and B≈-1.644, then for n=100:T(100) = 1 * 100 * log2(100) + (-1.644)*100log2(100)≈6.643856So, 100*6.643856 = 664.3856Minus 1.644*100 = 164.4So, 664.3856 - 164.4 ≈ 499.9856 ≈ 500 ms. That's correct.For n=200:T(200) = 1 * 200 * log2(200) + (-1.644)*200log2(200)≈7.643856200*7.643856≈1528.7712Minus 1.644*200=328.81528.7712 - 328.8≈1199.9712≈1200 ms. That's correct.So, if log is base 2, the values A=1 and B≈-1.644 work perfectly.If I had used log base 10, I would have gotten A≈3.322 and B≈-1.644, but let's test that.For n=100:T(100)=3.322*100*2 + (-1.644)*100=664.4 -164.4=500 ms.For n=200:T(200)=3.322*200*2.3010 + (-1.644)*200First, 3.322*200=664.4664.4*2.3010≈664.4*2.3≈1528.12Minus 1.644*200=328.81528.12 - 328.8≈1199.32≈1200 ms.So, both bases give the correct results. Hmm, so the problem is ambiguous because it doesn't specify the base. But in algorithm analysis, log is typically base 2, so I think the intended answer is A=1 and B≈-1.644.But the problem says "log(n)", which in mathematics is often natural log. Wait, no, in computer science, log is usually base 2. So, I think the intended base is 2.Therefore, A=1 and B≈-1.644.Now, to find the estimated execution time when n=1000.So, T(1000)=A*1000*log2(1000) + B*1000We know A=1, B≈-1.644First, compute log2(1000). Let's calculate that.We know that 2^10=1024, so log2(1000) is slightly less than 10. Specifically, log2(1000)=ln(1000)/ln(2)=6.907755/0.693147≈9.96578So, approximately 9.96578.So, T(1000)=1*1000*9.96578 + (-1.644)*1000Calculate each term:1000*9.96578=9965.78-1.644*1000=-1644So, total T(1000)=9965.78 -1644≈8321.78 msWhich is approximately 8321.78 milliseconds, or about 8.32 seconds.Wait, but let me check if I used the correct B. Earlier, I had B≈-1.644, but in the first calculation with log base 2, B was exactly -1.643856, which is approximately -1.644.So, yes, that's correct.Alternatively, if I use more precise numbers:log2(100)=6.64385619log2(200)=7.64385619So, solving the equations:Equation (1): 664.385619 A + 100 B = 500Equation (2): 1528.771238 A + 200 B = 1200We found A=1, B= (500 - 664.385619)/100= ( -164.385619)/100= -1.64385619So, B≈-1.643856Therefore, T(1000)=1*1000*log2(1000) + (-1.643856)*1000log2(1000)=ln(1000)/ln(2)=6.907755278/0.69314718056≈9.96578428466So, T(1000)=1000*9.96578428466 -1643.856≈9965.78428466 -1643.856≈8321.92828466 ms≈8321.93 msSo, approximately 8321.93 milliseconds, which is about 8.32 seconds.Alternatively, if I use more precise calculations, maybe 8321.93 ms.But let me see if I can write it more accurately.Alternatively, maybe I can keep more decimal places in B.B= -164.385619/100= -1.64385619So, T(1000)=1000*log2(1000) + (-1.64385619)*1000log2(1000)= approximately 9.965784284662087So, 1000*9.965784284662087=9965.784284662087Minus 1643.85619=9965.784284662087 -1643.85619=8321.928094662087 msSo, approximately 8321.93 ms.So, rounding to the nearest whole number, it's 8322 ms.Alternatively, if we keep it to two decimal places, 8321.93 ms.But the problem didn't specify the precision, so maybe we can write it as approximately 8322 ms.Alternatively, if we use the exact fractions, but I think 8322 ms is acceptable.Wait, but let me check if I made any mistake in the calculation.Wait, when I calculated log2(1000), I used natural log divided by natural log of 2, which is correct. So, that's accurate.So, yes, T(1000)=~8322 ms.Alternatively, if I use log base 10, let's see what happens.If log is base 10, then:log10(1000)=3So, T(1000)=A*1000*3 + B*1000From earlier, A≈3.322, B≈-1.644So, T(1000)=3.322*3000 + (-1.644)*1000=9966 -1644=8322 msSame result! Interesting. So regardless of the base, the result is the same because the constants A and B adjust accordingly.Wait, that's fascinating. So, whether log is base 2 or base 10, the execution time for n=1000 is the same because the constants A and B are scaled accordingly.So, in both cases, T(1000)=8322 ms.Therefore, the estimated execution time is approximately 8322 milliseconds.So, summarizing:1. Memory usage for 1000 integers is 4002 bytes.2. The constants are A=1 and B≈-1.644 (if log is base 2), or A≈3.322 and B≈-1.644 (if log is base 10). However, regardless of the base, the execution time for n=1000 is approximately 8322 ms.But since the problem didn't specify the base, and in algorithm analysis, log is usually base 2, I think the intended answer is A=1 and B≈-1.644, leading to T(1000)=8322 ms.Wait, but in the first part, the memory usage is straightforward. 4n +2, n=1000, so 4000 +2=4002 bytes.Yes, that's correct.So, final answers:1. Total memory usage is 4002 bytes.2. A=1, B≈-1.644, and estimated execution time for n=1000 is approximately 8322 ms.But let me write the exact value of B as -1.643856, which is approximately -1.644.Alternatively, if I want to keep more decimal places, but probably -1.644 is sufficient.So, to present:1. M(1000)=4002 bytes.2. A=1, B≈-1.644, and T(1000)=8322 ms.Alternatively, if I want to write B as a fraction, since 164.3856/100=1.643856, which is approximately 1.644.But I think decimal form is fine.So, I think that's it.</think>

🧠:<think>Alright, so I have this problem about Dr. Smith analyzing gender representation in children's toys. It's divided into two parts. Let me tackle them one by one.Starting with part 1: I need to calculate the probabilities P(M), P(F), and P(N). These are the probabilities that a randomly selected toy portrays a traditionally masculine role, feminine role, or a neutral role, respectively.First, let me understand the data given. There are three categories of toys based on marketing: boys, girls, and neutral. Each category has a certain number of toys and how many portray masculine, feminine, or neutral roles.So, for the boys' toys: 200 toys in total, with 120 portraying masculine roles. That leaves 80 toys that are either feminine or neutral. But wait, the problem doesn't specify how many of the boys' toys are feminine or neutral. Hmm, maybe I need to assume that the rest are neutral? Or perhaps the problem is structured such that each toy is categorized into one of the three roles: masculine, feminine, or neutral. Let me check the problem statement again.Looking back, it says: "the number of toys marketed towards boys, girls, and neutrally. She also documents the frequency of specific gender-stereotyped roles these toys and games portray." So, for each type of toy (boys, girls, neutral), she documents how many portray masculine, feminine, or neutral roles.Wait, the numbers given are:- 200 toys marketed towards boys, with 120 portraying traditionally masculine roles. So, does that mean the remaining 80 are either feminine or neutral? But the problem doesn't specify. Similarly, for girls' toys: 180 total, 150 feminine roles, so 30 remaining. For neutral toys: 120 total, 40 masculine, 30 feminine, so 50 neutral.Wait, hold on. Let me parse this again.The dataset includes:- 200 toys marketed towards boys, with 120 portraying traditionally masculine roles.So, for boys' toys, 120 are masculine, and the rest (200 - 120 = 80) must be either feminine or neutral. But the problem doesn't specify how many. Hmm. Maybe I need to assume that the rest are neutral? Or perhaps the problem is structured such that each toy is categorized into one of the three roles, so for boys' toys, 120 masculine, and the rest are neutral? Because it doesn't mention any feminine roles in boys' toys.Wait, actually, looking again: "the frequency of specific gender-stereotyped roles these toys and games portray." So, perhaps each toy is assigned to one role: masculine, feminine, or neutral. So, for boys' toys, 120 are masculine, and the remaining 80 are neutral. Similarly, for girls' toys, 150 are feminine, and the remaining 30 are neutral. For neutral toys, 40 are masculine, 30 are feminine, and the rest (120 - 40 - 30 = 50) are neutral.Is that the correct interpretation? Let me think. If that's the case, then for boys' toys: 120 masculine, 80 neutral. For girls' toys: 150 feminine, 30 neutral. For neutral toys: 40 masculine, 30 feminine, 50 neutral.So, to find the total number of toys that portray each role, I can sum across the marketing categories.Total masculine roles: boys' toys (120) + neutral toys (40) = 160.Total feminine roles: girls' toys (150) + neutral toys (30) = 180.Total neutral roles: boys' toys (80) + girls' toys (30) + neutral toys (50) = 160.Wait, let me verify:Boys' toys: 200 total, 120 M, 80 N.Girls' toys: 180 total, 150 F, 30 N.Neutral toys: 120 total, 40 M, 30 F, 50 N.So, total M: 120 + 40 = 160.Total F: 150 + 30 = 180.Total N: 80 + 30 + 50 = 160.Total toys: 200 + 180 + 120 = 500.So, total M + F + N = 160 + 180 + 160 = 500, which matches. Good.Therefore, the probabilities are:P(M) = 160 / 500P(F) = 180 / 500P(N) = 160 / 500Let me compute these:160 / 500 = 0.32180 / 500 = 0.36160 / 500 = 0.32So, P(M) = 0.32, P(F) = 0.36, P(N) = 0.32.Wait, but let me double-check. Is the total number of toys 500? 200 + 180 + 120 = 500. Yes. And the total M, F, N are 160, 180, 160, which add to 500. So, yes, the probabilities are correct.Moving on to part 2: Dr. Smith wants to use a chi-square test for independence. She needs a 3x3 contingency table of observed frequencies.So, the rows are the type of toy marketing: boys, girls, neutral.The columns are the gender roles: masculine, feminine, neutral.So, the table will look like this:``` | Masculine | Feminine | Neutral | Total-----------------------------------------------Boys' toys | 120 | ? | 80 | 200Girls' toys | ? | 150 | 30 | 180Neutral toys | 40 | 30 | 50 | 120-----------------------------------------------Total Masculine| 160 | | |Total Feminine | | 180 | |Total Neutral | | | 160 |Total | 500 | 500 | 500 | 500```Wait, but in the boys' toys row, we have 120 masculine and 80 neutral. So, the feminine role for boys' toys must be 0? Because 120 + 0 + 80 = 200. Similarly, for girls' toys, 150 feminine, 30 neutral, so masculine role is 0? Because 0 + 150 + 30 = 180. For neutral toys, 40 masculine, 30 feminine, 50 neutral.So, the observed frequencies table is:``` | Masculine | Feminine | Neutral-----------------------------------------------Boys' toys | 120 | 0 | 80Girls' toys | 0 | 150 | 30Neutral toys | 40 | 30 | 50```Yes, that makes sense. So, the observed frequencies are as above.Now, to perform the chi-square test for independence, we need to calculate the expected frequencies under the null hypothesis that marketing type and gender role are independent.The formula for expected frequency for each cell is:E_ij = (row total * column total) / grand totalSo, let's compute the expected frequencies for each cell.First, let's get the row totals and column totals.Row totals:Boys' toys: 200Girls' toys: 180Neutral toys: 120Column totals:Masculine: 160Feminine: 180Neutral: 160Grand total: 500Now, compute expected frequencies:For Boys' toys and Masculine:E = (200 * 160) / 500 = (32000) / 500 = 64For Boys' toys and Feminine:E = (200 * 180) / 500 = (36000) / 500 = 72For Boys' toys and Neutral:E = (200 * 160) / 500 = 64Wait, but wait, the row total for Boys' toys is 200, and the column totals are 160, 180, 160. So, yes, 200*160/500 = 64 for Masculine, 200*180/500=72 for Feminine, and 200*160/500=64 for Neutral. But wait, 64 + 72 + 64 = 200, which is correct.Similarly, for Girls' toys:Masculine: (180 * 160)/500 = (28800)/500 = 57.6Feminine: (180 * 180)/500 = (32400)/500 = 64.8Neutral: (180 * 160)/500 = 57.6Check: 57.6 + 64.8 + 57.6 = 180, correct.For Neutral toys:Masculine: (120 * 160)/500 = (19200)/500 = 38.4Feminine: (120 * 180)/500 = (21600)/500 = 43.2Neutral: (120 * 160)/500 = 38.4Check: 38.4 + 43.2 + 38.4 = 120, correct.So, the expected frequencies table is:``` | Masculine | Feminine | Neutral-----------------------------------------------Boys' toys | 64 | 72 | 64Girls' toys | 57.6 | 64.8 | 57.6Neutral toys | 38.4 | 43.2 | 38.4```Now, the chi-square statistic is calculated as the sum over all cells of (O - E)^2 / E.So, let's compute each cell's contribution.Starting with Boys' toys:Masculine: O=120, E=64(120 - 64)^2 / 64 = (56)^2 /64 = 3136 /64 = 49Feminine: O=0, E=72(0 - 72)^2 /72 = 5184 /72 = 72Neutral: O=80, E=64(80 - 64)^2 /64 = (16)^2 /64 = 256 /64 = 4Total for Boys' toys: 49 + 72 + 4 = 125Girls' toys:Masculine: O=0, E=57.6(0 - 57.6)^2 /57.6 = 3317.76 /57.6 = 57.6Feminine: O=150, E=64.8(150 - 64.8)^2 /64.8 = (85.2)^2 /64.8 = 7259.04 /64.8 ≈ 112Neutral: O=30, E=57.6(30 - 57.6)^2 /57.6 = (-27.6)^2 /57.6 = 761.76 /57.6 ≈ 13.222Total for Girls' toys: 57.6 + 112 + 13.222 ≈ 182.822Neutral toys:Masculine: O=40, E=38.4(40 - 38.4)^2 /38.4 = (1.6)^2 /38.4 = 2.56 /38.4 ≈ 0.0668Feminine: O=30, E=43.2(30 - 43.2)^2 /43.2 = (-13.2)^2 /43.2 = 174.24 /43.2 ≈ 4.03Neutral: O=50, E=38.4(50 - 38.4)^2 /38.4 = (11.6)^2 /38.4 = 134.56 /38.4 ≈ 3.504Total for Neutral toys: 0.0668 + 4.03 + 3.504 ≈ 7.6008Now, summing up all the contributions:Boys' toys: 125Girls' toys: ≈182.822Neutral toys: ≈7.6008Total chi-square statistic: 125 + 182.822 + 7.6008 ≈ 315.4228Wait, that seems really high. Let me double-check my calculations because a chi-square of over 300 seems extremely large, which might indicate a mistake.Looking back, let's recalculate each cell carefully.Starting with Boys' toys:Masculine: (120 - 64)^2 /64 = 56^2 /64 = 3136 /64 = 49. Correct.Feminine: (0 - 72)^2 /72 = 5184 /72 = 72. Correct.Neutral: (80 - 64)^2 /64 = 16^2 /64 = 256 /64 = 4. Correct.Total: 49 + 72 + 4 = 125.Girls' toys:Masculine: (0 - 57.6)^2 /57.6 = 57.6^2 /57.6 = 57.6. Correct.Feminine: (150 - 64.8)^2 /64.8 = (85.2)^2 /64.8.85.2 squared: 85.2 *85.2. Let's compute 85^2 =7225, 0.2^2=0.04, cross term 2*85*0.2=34. So, total is 7225 +34 +0.04=7259.04. Divided by 64.8: 7259.04 /64.8.Let me compute 64.8 * 112 = 7257.6. So, 7259.04 -7257.6=1.44. So, 112 + 1.44 /64.8 ≈112 +0.0222≈112.0222. So, approximately 112.0222.Neutral: (30 -57.6)^2 /57.6 = (-27.6)^2 /57.6 =761.76 /57.6.57.6 *13=748.8. 761.76 -748.8=12.96. 12.96 /57.6=0.225. So, total is 13 +0.225=13.225.So, total for Girls' toys:57.6 +112.0222 +13.225≈182.8472.Neutral toys:Masculine: (40 -38.4)^2 /38.4=1.6^2 /38.4=2.56 /38.4≈0.0668.Feminine: (30 -43.2)^2 /43.2= (-13.2)^2 /43.2=174.24 /43.2=4.03.Neutral: (50 -38.4)^2 /38.4=11.6^2 /38.4=134.56 /38.4≈3.504.Total:0.0668 +4.03 +3.504≈7.6008.Adding all together:125 +182.8472 +7.6008≈315.448.Yes, that's correct. So, the chi-square statistic is approximately 315.45.But wait, that seems extremely high. A chi-square of over 300 is way beyond typical values. Let me think if I made a mistake in the expected frequencies.Wait, the expected frequencies are calculated correctly? Let me check one cell.For Boys' toys and Masculine: (200 *160)/500=64. Correct.Girls' toys and Feminine: (180 *180)/500=64.8. Correct.Neutral toys and Neutral: (120 *160)/500=38.4. Correct.So, the expected frequencies are correct.Then, the observed frequencies: Boys' toys have 120 M, 0 F, 80 N. Girls' toys:0 M,150 F,30 N. Neutral toys:40 M,30 F,50 N.So, the observed frequencies are correct.Calculations for each cell seem correct. So, the chi-square is indeed very high, indicating a strong association.But let's recall that the chi-square test is sensitive to sample size. Since the sample size is 500, even small deviations can lead to large chi-square values.But let me check if the formula was applied correctly.Yes, for each cell, (O - E)^2 / E.So, for Boys' toys and Masculine: (120 -64)^2 /64=49.Similarly, for Girls' toys and Feminine: (150 -64.8)^2 /64.8≈112.These are large contributions because the observed counts are much higher than expected in those cells, leading to large chi-square values.So, the chi-square statistic is approximately 315.45.Now, to determine if this is significant at the 0.05 level, we need to compare it to the critical value from the chi-square distribution table.The degrees of freedom for a 3x3 table is (3-1)*(3-1)=4.Looking up the critical value for chi-square with 4 degrees of freedom at alpha=0.05: the critical value is approximately 9.488.Since our calculated chi-square statistic is 315.45, which is much larger than 9.488, we can reject the null hypothesis. There is a significant association between the type of toy marketing and the portrayal of gender roles.But wait, the problem only asks to calculate the chi-square statistic, not to perform the hypothesis test. So, I think I just need to compute the chi-square statistic, which is approximately 315.45.However, let me check if I made a mistake in the calculation because 315 is unusually high. Let me recalculate one cell.Take Boys' toys, Feminine: O=0, E=72.(0 -72)^2 /72=5184 /72=72. Correct.Girls' toys, Masculine: O=0, E=57.6.(0 -57.6)^2 /57.6=3317.76 /57.6=57.6. Correct.Girls' toys, Feminine: O=150, E=64.8.(150 -64.8)=85.2. 85.2^2=7259.04. 7259.04 /64.8≈112.02. Correct.So, all calculations seem correct. Therefore, the chi-square statistic is indeed approximately 315.45.But wait, let me think about the structure of the data. The observed frequencies have zeros in some cells, which can affect the chi-square test. However, in this case, the expected frequencies are all above 5 (the rule of thumb for chi-square test applicability). Let's check:Expected frequencies:Boys' toys:64,72,64 – all above 5.Girls' toys:57.6,64.8,57.6 – all above 5.Neutral toys:38.4,43.2,38.4 – all above 5.So, the chi-square test is appropriate here.Therefore, the chi-square statistic is approximately 315.45.But let me compute it more precisely.Let me recalculate each cell with more precision.Boys' toys:Masculine: (120-64)^2 /64=56^2 /64=3136 /64=49.Feminine: (0-72)^2 /72=5184 /72=72.Neutral: (80-64)^2 /64=16^2 /64=256 /64=4.Total:49+72+4=125.Girls' toys:Masculine: (0-57.6)^2 /57.6=3317.76 /57.6=57.6.Feminine: (150-64.8)^2 /64.8=85.2^2=7259.04 /64.8=7259.04 ÷64.8.Let me compute 64.8 *112=7257.6. So, 7259.04 -7257.6=1.44. So, 1.44 /64.8=0.022222. So, total is 112 +0.022222=112.022222.Neutral: (30-57.6)^2 /57.6= (-27.6)^2=761.76 /57.6=13.222222.Total:57.6 +112.022222 +13.222222=57.6 +125.244444≈182.844444.Neutral toys:Masculine: (40-38.4)^2 /38.4=1.6^2=2.56 /38.4≈0.0666667.Feminine: (30-43.2)^2 /43.2=(-13.2)^2=174.24 /43.2=4.03.Neutral: (50-38.4)^2 /38.4=11.6^2=134.56 /38.4≈3.504.Total:0.0666667 +4.03 +3.504≈7.6006667.Total chi-square=125 +182.844444 +7.6006667≈315.445111.So, approximately 315.45.Yes, that's correct.Therefore, the chi-square statistic is approximately 315.45.But let me think again: is this possible? Because in the observed table, the Boys' toys have a very high number of masculine roles, and Girls' toys have a very high number of feminine roles, which are both much higher than expected under independence. Similarly, the Neutral toys have a mix, but the observed counts are not extremely different from expected.So, the large chi-square is due to the strong association between marketing type and gender role.Therefore, the calculations are correct.So, summarizing:1. P(M)=0.32, P(F)=0.36, P(N)=0.32.2. The chi-square statistic is approximately 315.45.But wait, let me check if I should present it as 315.45 or round it to two decimal places, which would be 315.45.Alternatively, sometimes people round to three decimal places, but in this case, 315.45 is sufficient.Alternatively, maybe I should present it as 315.45, but perhaps the exact value is 315.445, which is approximately 315.45.Yes, that's correct.So, final answers:1. P(M)=0.32, P(F)=0.36, P(N)=0.32.2. Chi-square statistic≈315.45.But wait, let me check if I should use more precise decimal places.Alternatively, perhaps I should present it as 315.45.Yes, that's fine.So, to answer the questions:1. Calculate P(M), P(F), and P(N).2. Construct the contingency table and calculate the chi-square statistic.So, the contingency table is as above, and the chi-square statistic is approximately 315.45.</think>

🧠:<think>Alright, so I have this problem where a mother is setting up a monthly allowance system for her two kids, Alex and Jamie. They both participate in a math competition, and their allowance is based on their scores. The first part is about maximizing their total allowance, and the second part is about calculating a bonus based on their contributions to chores. Let me try to tackle each part step by step.Starting with the first problem: They have a total score of 150, meaning A + J = 150. The allowance L is given by L = 50 + (A² + J²)/20. I need to find the values of A and J that maximize L. Hmm, okay, so since L is a function of A and J, and we have a constraint that A + J = 150, this sounds like an optimization problem with a constraint. I remember that in calculus, we can use the method of Lagrange multipliers for such problems, but maybe there's a simpler way since it's just two variables.Alternatively, since A + J is fixed, maybe I can express one variable in terms of the other and then maximize the function. Let me try that. If A + J = 150, then J = 150 - A. So, substituting that into the allowance formula, we get:L = 50 + (A² + (150 - A)²)/20.Let me expand that. First, compute (150 - A)²:(150 - A)² = 150² - 2*150*A + A² = 22500 - 300A + A².So, plugging that back into L:L = 50 + (A² + 22500 - 300A + A²)/20= 50 + (2A² - 300A + 22500)/20.Let me simplify that:First, divide each term by 20:= 50 + (2A²)/20 - (300A)/20 + 22500/20= 50 + (A²)/10 - 15A + 1125.Now, combine the constants:50 + 1125 = 1175.So, L = 1175 + (A²)/10 - 15A.So, L is a quadratic function in terms of A. Since the coefficient of A² is positive (1/10), the parabola opens upwards, meaning the vertex is the minimum point. But we want to maximize L, so wait, that can't be right. If it's a parabola opening upwards, the minimum is at the vertex, but we need the maximum. Hmm, but since A is bounded between 0 and 150 (since A and J are both non-negative and sum to 150), the maximum must occur at one of the endpoints.Wait, hold on. Let me think again. If the function is quadratic in A, and it's opening upwards, then the minimum is at the vertex, but the maximum would be at the endpoints of the interval. So, to maximize L, we need to check the values of L when A is 0 and when A is 150.Let me compute L when A = 0:L = 1175 + (0)/10 - 15*0 = 1175.When A = 150:L = 1175 + (150²)/10 - 15*150= 1175 + (22500)/10 - 2250= 1175 + 2250 - 2250= 1175.Wait, so at both endpoints, L is 1175. So, does that mean that L is constant? That can't be. Let me check my calculations.Wait, when I substituted J = 150 - A into L, I had:L = 50 + (A² + (150 - A)²)/20.Which expanded to:50 + (2A² - 300A + 22500)/20.Then, dividing each term:50 + (A²)/10 - 15A + 1125.So, 50 + 1125 is 1175, so L = 1175 + (A²)/10 - 15A.Wait, but when I plug in A = 0, I get L = 1175 + 0 - 0 = 1175.When A = 150, L = 1175 + (22500)/10 - 15*150= 1175 + 2250 - 2250= 1175.So, indeed, at both ends, L is 1175. Hmm, so does that mean that L is constant regardless of A and J as long as A + J = 150? That seems counterintuitive because the formula for L is (A² + J²)/20, which would vary depending on how A and J are distributed.Wait, maybe I made a mistake in simplifying. Let me double-check.Starting again:L = 50 + (A² + J²)/20.Given that A + J = 150, so J = 150 - A.Thus,L = 50 + (A² + (150 - A)²)/20.Compute (150 - A)²:150² = 22500, (150 - A)² = 22500 - 300A + A².So,L = 50 + (A² + 22500 - 300A + A²)/20= 50 + (2A² - 300A + 22500)/20.Divide each term by 20:= 50 + (2A²)/20 - (300A)/20 + 22500/20= 50 + (A²)/10 - 15A + 1125.Combine constants:50 + 1125 = 1175.So, L = 1175 + (A²)/10 - 15A.Wait, so that's correct. So, L is a quadratic function in A, opening upwards, with vertex at A = -b/(2a) where a = 1/10 and b = -15.So, vertex at A = -(-15)/(2*(1/10)) = 15 / (1/5) = 75.So, the vertex is at A = 75, which is the minimum point. So, the minimum L is at A = 75, and since the parabola opens upwards, the maximum L occurs at the endpoints, which are A = 0 and A = 150, both giving L = 1175.Wait, so that suggests that regardless of how A and J are split, as long as their sum is 150, the allowance L is always 1175? That seems odd because if A and J are both 75, then L would be 50 + (75² + 75²)/20 = 50 + (5625 + 5625)/20 = 50 + 11250/20 = 50 + 562.5 = 612.5. But wait, that contradicts the earlier calculation where L was 1175. Hmm, something is wrong here.Wait, hold on. When I substituted A = 75, I get:L = 1175 + (75²)/10 - 15*75= 1175 + 5625/10 - 1125= 1175 + 562.5 - 1125= 1175 - 562.5= 612.5.But earlier, when I plugged in A = 0 or A = 150, I got L = 1175. So, that suggests that L is 1175 at the endpoints and 612.5 at the vertex. That's a huge difference. So, actually, the maximum L is 1175, and the minimum is 612.5.Wait, so that means that to maximize L, we need to set A or J to 0, making the other 150. So, either A = 150, J = 0 or A = 0, J = 150. That would give the maximum allowance of 1175.But that seems counterintuitive because if both score 75, their allowance is 612.5, which is less than 1175. So, the mother's formula actually incentivizes one child to score as high as possible while the other scores zero? That seems odd, but mathematically, that's what the function is doing.Wait, let me check the original formula again. L = 50 + (A² + J²)/20. So, if A and J are both 75, then (75² + 75²) = 2*(5625) = 11250. Divided by 20 is 562.5. Plus 50 is 612.5. But if A is 150 and J is 0, then (150² + 0) = 22500. Divided by 20 is 1125. Plus 50 is 1175. So, yeah, that's correct.So, the allowance is maximized when one child scores 150 and the other scores 0. That's interesting. So, the function is such that the allowance is maximized when one child's score is as high as possible, and the other's is as low as possible. So, in this case, the optimal scores are A = 150, J = 0 or A = 0, J = 150.But wait, the problem says "their scores" that maximize the allowance. So, does that mean both A and J? Or can one be 150 and the other 0? The problem doesn't specify any constraints on individual scores, just that A + J = 150. So, technically, the maximum occurs when one is 150 and the other is 0.But let me think again. Maybe I made a mistake in interpreting the problem. The problem says "their scores A and J that maximize their total monthly allowance L." So, it's possible that both A and J are variables, but under the constraint A + J = 150. So, perhaps the maximum occurs when one is 150 and the other is 0.Alternatively, maybe I should consider that both A and J are positive, but the problem doesn't specify that. So, if we allow A or J to be zero, then yes, the maximum is 1175. But if both A and J have to be positive, then the maximum would still be approached as one approaches 150 and the other approaches 0.But the problem doesn't specify that they have to have positive scores, so I think the answer is that either A = 150 and J = 0 or A = 0 and J = 150.Wait, but let me think about this again. If both A and J are 75, the allowance is 612.5, which is much less than 1175. So, the function is designed such that the allowance is maximized when one child's score is as high as possible, and the other's is as low as possible. That seems like a very competitive system, but mathematically, that's correct.So, to answer the first part, the scores that maximize L are A = 150 and J = 0, or A = 0 and J = 150.Now, moving on to the second part. The bonus B is given by B = (h/2)(√A + √J), and h = 20. So, B = (20/2)(√A + √J) = 10(√A + √J). We need to calculate this bonus based on the previously calculated optimal scores A and J.From the first part, the optimal scores are either (150, 0) or (0, 150). So, let's compute B for both cases.Case 1: A = 150, J = 0.Then, B = 10(√150 + √0) = 10(√150 + 0) = 10√150.√150 can be simplified as √(25*6) = 5√6 ≈ 5*2.449 ≈ 12.245.So, B ≈ 10*12.245 ≈ 122.45 dollars.Case 2: A = 0, J = 150.Similarly, B = 10(√0 + √150) = 10(0 + √150) = 10√150 ≈ 122.45 dollars.So, in both cases, the bonus is approximately 122.45 dollars.But let me compute it exactly. √150 = √(25*6) = 5√6. So, B = 10*5√6 = 50√6.Since √6 is approximately 2.449, 50*2.449 ≈ 122.45.So, the exact value is 50√6 dollars, which is approximately 122.45 dollars.But the problem says to determine the additional bonus they receive based on the previously calculated optimal scores. So, since the optimal scores are either (150,0) or (0,150), the bonus is 50√6 dollars.Wait, but let me think again. If the optimal scores are (150,0) or (0,150), then the bonus is 50√6 in both cases. So, that's the answer.But just to make sure, let me check if there's another way to interpret the problem. Maybe the mother wants to maximize both L and B? But the problem says in the first part to find the scores that maximize L, and in the second part, based on those scores, calculate B. So, it's two separate parts.So, in the first part, we found that to maximize L, one child should score 150 and the other 0. Then, in the second part, using those scores, we calculate B, which is 50√6.Alternatively, if we had chosen different scores, say both 75, then B would be 10(√75 + √75) = 10*2√75 = 20√75 = 20*5√3 = 100√3 ≈ 173.2 dollars, which is higher than 50√6 ≈ 122.45. So, interestingly, if we had chosen equal scores, the bonus would be higher, but the allowance L would be lower.But since the first part is only about maximizing L, regardless of B, the optimal scores for L are (150,0) or (0,150), leading to a lower B. So, the bonus is 50√6.Wait, but let me think again. Is there a way to maximize both L and B? Or is it that the first part is independent of the second? The problem says in the first part, find the scores that maximize L, and in the second part, based on those scores, calculate B. So, it's two separate steps.Therefore, the answer for the first part is A = 150, J = 0 or vice versa, and the bonus is 50√6.But let me just make sure I didn't make any mistakes in the calculations.For the first part:Given A + J = 150, and L = 50 + (A² + J²)/20.Expressed in terms of A, L = 50 + (A² + (150 - A)²)/20.Expanding that, we get L = 50 + (2A² - 300A + 22500)/20 = 50 + (A²)/10 - 15A + 1125 = 1175 + (A²)/10 - 15A.This is a quadratic in A, opening upwards, so the minimum is at A = 75, and the maximum at the endpoints A = 0 or A = 150, giving L = 1175.So, that's correct.For the second part, using A = 150 and J = 0, B = 10(√150 + √0) = 10√150 = 10*5√6 = 50√6.Yes, that's correct.So, the final answers are:1. A = 150, J = 0 or A = 0, J = 150.2. Bonus B = 50√6 dollars.But let me write it in the required format.</think>

🧠:<think>First, I need to determine the initial number of church members when John joined 55 years ago. The church has been growing at a 2% annual rate, and it currently has 300 members.I'll use the formula for exponential growth:[ A = P times (1 + r)^t ]Where:- ( A ) is the current number of members (300),- ( P ) is the initial number of members (what we're solving for),- ( r ) is the annual growth rate (2% or 0.02),- ( t ) is the time in years (55).Rearranging the formula to solve for ( P ):[ P = frac{A}{(1 + r)^t} ]Next, I'll calculate the initial investment in the charitable fund 20 years ago. The fund has grown to 100,000 with a continuous annual return of 5%.For continuous growth, the formula is:[ A = P times e^{rt} ]Where:- ( A ) is the current amount (100,000),- ( P ) is the initial investment (what we're solving for),- ( r ) is the continuous growth rate (5% or 0.05),- ( t ) is the time in years (20),- ( e ) is the base of the natural logarithm.Rearranging the formula to solve for ( P ):[ P = frac{A}{e^{rt}} ]</think>