answer：First, let's analyze each student's statement:1. **Lin's Statement**: "My score is not the highest and not the lowest." This means Lin's score is in the middle range.2. **Mika's Statement**: "I scored higher than both Neal and Otis." This indicates that Mika's score is higher than both Neal and Otis.3. **Neal's Statement**: "At least two of us scored higher than me." This suggests that there are at least two people who scored higher than Neal.4. **Otis's Statement**: "I did not score lower than Lin." This means Otis's score is equal to or higher than Lin's score.Now, let's try to rank the scores based on these statements.- From Mika's statement, we know that Mika > Neal and Mika > Otis.- From Lin's statement, Lin is neither the highest nor the lowest, so Lin must be in the middle.- From Neal's statement, at least two people scored higher than him. Since Mika is already higher than Neal, we need at least one more person to be higher than Neal. This could be Lin or Otis.- From Otis's statement, Otis did not score lower than Lin, so Otis ≥ Lin.Considering these points, we can deduce the following order:1. **Neal** is the lowest because both Mika and at least one other person (either Lin or Otis) scored higher than him.2. **Lin** is next because Lin is neither the highest nor the lowest.3. **Otis** is higher than Lin but not necessarily the highest.4. **Mika** is the highest because Mika scored higher than both Neal and Otis.Thus, the ranking from lowest to highest is: **Neal, Lin, Otis, Mika**.

answer：Okay, so I have this problem here about sampling methods, and I need to figure out which method is best for each scenario. Let me take it step by step.First, there are two problems:1. In a community of 4000 people, the ratio of young people, middle-aged people, and elderly people is 1:2:4. We need to draw a sample of 200 people to understand their physical condition.2. From a class of 45 students, 5 students need to be selected to participate in the student council.And the sampling methods given are:I. Simple random sampling II. Systematic sampling III. Stratified samplingI need to pair each problem with the correct sampling method.Starting with Problem 1: The community has 4000 people divided into three groups with a ratio of 1:2:4. That means if I add up the ratios, it's 1 + 2 + 4 = 7 parts. So, young people are 1/7 of the population, middle-aged are 2/7, and elderly are 4/7. Since the problem mentions understanding their physical condition, and physical condition can vary significantly between different age groups, it's important that the sample reflects the proportion of each age group in the community. If I use simple random sampling, there's a chance that the sample might not accurately represent each subgroup, especially if one group is much larger than the others. Stratified sampling, on the other hand, allows us to ensure that each subgroup is adequately represented in the sample. So, for Problem 1, stratified sampling (III) seems appropriate.Now, moving to Problem 2: We have a class of 45 students, and we need to select 5 students for the student council. This is a smaller population, and the sample size is relatively small compared to the total population. In such cases, simple random sampling is often sufficient because it's straightforward and ensures that every student has an equal chance of being selected. There's no mention of different subgroups or strata within the class that need special attention, so there's no need for stratified sampling here. Systematic sampling could also be an option, where you select every nth person from a list. But since the class size is small, simple random sampling is just as easy and avoids any potential biases that might come from the ordering of the list in systematic sampling.So, for Problem 2, simple random sampling (I) is the best choice.Putting it all together:- Problem 1: Stratified sampling (III)- Problem 2: Simple random sampling (I)Looking at the options provided:A: 1-III, 2-I B: 1-I, 2-II C: 1-II, 2-III D: 1-III, 2-IIThe correct pairing is option A.

answer：Okay, so I have this problem about an arithmetic progression. Let me try to figure it out step by step. First, the problem says that we have an arithmetic progression {a_n} starting from n=1, with a_1 being positive. That means the first term is greater than zero. Then, it gives a condition: 5 times the 13th term is equal to 6 times the 19th term. I need to find the smallest integer n such that the nth term is negative. Alright, let's recall what an arithmetic progression is. It's a sequence where each term after the first is obtained by adding a constant difference. So, if the first term is a_1 and the common difference is d, then the nth term can be written as:a_n = a_1 + (n - 1)dGot that. So, for example, the 13th term would be a_1 + 12d, and the 19th term would be a_1 + 18d.The problem gives me that 5 times the 13th term equals 6 times the 19th term. Let me write that equation out:5 * a_13 = 6 * a_19Substituting the expressions for a_13 and a_19, I get:5*(a_1 + 12d) = 6*(a_1 + 18d)Okay, let's expand both sides:5a_1 + 60d = 6a_1 + 108dHmm, now I can subtract 5a_1 from both sides:60d = a_1 + 108dWait, that doesn't seem right. Let me check my expansion again. Wait, no, actually, when I expand 5*(a_1 + 12d), it's 5a_1 + 60d, and 6*(a_1 + 18d) is 6a_1 + 108d. So, subtracting 5a_1 from both sides:60d = a_1 + 108dWait, that would mean:60d - 108d = a_1Which simplifies to:-48d = a_1So, a_1 = -48dBut the problem says that a_1 is greater than zero. So, a_1 > 0, which means that -48d > 0. Therefore, d must be negative because multiplying a negative number by -48 gives a positive result.So, d is negative. That makes sense because if the common difference is negative, the terms of the arithmetic progression will eventually become negative as n increases.Now, I need to find the smallest integer n such that a_n < 0. Let's write the expression for a_n again:a_n = a_1 + (n - 1)dWe know that a_1 = -48d, so substituting that in:a_n = -48d + (n - 1)dLet me factor out the d:a_n = d*(-48 + n - 1) = d*(n - 49)So, a_n = d*(n - 49)We need a_n < 0. Since d is negative, let's see when d*(n - 49) is less than zero.If I have a negative number multiplied by (n - 49), the result will be negative if (n - 49) is positive because a negative times a positive is negative. Conversely, if (n - 49) is negative, then d*(n - 49) would be positive because a negative times a negative is positive.So, to have a_n < 0, we need (n - 49) > 0. That is:n - 49 > 0Which simplifies to:n > 49Since n has to be an integer, the smallest integer greater than 49 is 50. Therefore, the smallest integer n such that a_n < 0 is 50.Let me just double-check my steps to make sure I didn't make a mistake.1. Wrote the general term of the arithmetic progression: a_n = a_1 + (n - 1)d. Correct.2. Expressed a_13 and a_19 in terms of a_1 and d: a_13 = a_1 + 12d and a_19 = a_1 + 18d. Correct.3. Plugged into the given equation: 5*(a_1 + 12d) = 6*(a_1 + 18d). Correct.4. Expanded and simplified: 5a_1 + 60d = 6a_1 + 108d. Then subtracted 5a_1: 60d = a_1 + 108d. Then subtracted 108d: -48d = a_1. Correct.5. Since a_1 > 0, then d must be negative. Correct.6. Expressed a_n in terms of d: a_n = d*(n - 49). Correct.7. Determined that for a_n < 0, since d is negative, (n - 49) must be positive, so n > 49. Therefore, n = 50. Correct.I think all the steps are solid. So, yeah, the answer should be 50.**Final Answer**The smallest integer ( n ) such that ( a_n < 0 ) is boxed{50}.

answer：Alright, so I have this problem about a sequence where the sum of the first n terms, S_n, is given by S_n = 2a_n - n. I need to find the general formula for the nth term, a_n, and then use that to find the smallest positive integer m such that the sum T_n of another sequence b_n is less than m/20 for all natural numbers n.Starting with part 1: Finding a_n.I know that S_n is the sum of the first n terms, so S_n = a_1 + a_2 + ... + a_n. The problem gives me S_n = 2a_n - n. Hmm, that's interesting because it relates the sum to the nth term itself.Maybe I can find a relationship between S_n and S_{n-1}. I remember that S_n = S_{n-1} + a_n. So, substituting the given formula, S_n = 2a_n - n and S_{n-1} = 2a_{n-1} - (n-1). Therefore, S_n = S_{n-1} + a_n implies:2a_n - n = 2a_{n-1} - (n - 1) + a_nLet me write that out:2a_n - n = 2a_{n-1} - n + 1 + a_nSimplify the right side:2a_n - n = a_n + 2a_{n-1} - n + 1Subtract a_n from both sides:a_n - n = 2a_{n-1} - n + 1Add n to both sides:a_n = 2a_{n-1} + 1Okay, so I have a recursive formula: a_n = 2a_{n-1} + 1. That looks like a linear recurrence relation. Maybe I can solve it.First, I need the initial condition. When n=1, S_1 = a_1. According to the formula, S_1 = 2a_1 - 1. So:a_1 = 2a_1 - 1Subtract a_1 from both sides:0 = a_1 - 1So, a_1 = 1.Now, with the recurrence a_n = 2a_{n-1} + 1 and a_1 = 1, I can try to find a closed-form solution.This is a nonhomogeneous linear recurrence. The homogeneous part is a_n = 2a_{n-1}, which has the solution a_n = C*2^n. For the particular solution, since the nonhomogeneous term is constant, I can assume a constant particular solution, say a_n = K.Plugging into the recurrence:K = 2K + 1Subtract 2K:-K = 1 => K = -1So the general solution is a_n = C*2^n - 1. Now apply the initial condition a_1 = 1:1 = C*2^1 - 1 => 1 = 2C - 1 => 2C = 2 => C = 1Therefore, the general formula is a_n = 2^n - 1.Wait, let me verify this. For n=1, a_1 = 2^1 - 1 = 1, which matches. For n=2, S_2 = a_1 + a_2 = 1 + (2^2 -1) = 1 + 3 = 4. According to the formula, S_2 = 2a_2 - 2 = 2*3 - 2 = 6 - 2 = 4. That works. For n=3, a_3 = 2^3 -1 =7, so S_3 = 1 + 3 +7=11. Using the formula, S_3=2*7 -3=14 -3=11. Perfect, seems correct.So part 1 is done: a_n = 2^n -1.Moving on to part 2: Let b_n = 2^n / (a_n a_{n+1}), and T_n is the sum of the first n terms of b_n. We need to find the smallest positive integer m such that T_n < m/20 for all n in natural numbers.First, let's write down b_n:b_n = 2^n / (a_n a_{n+1}) = 2^n / [(2^n -1)(2^{n+1} -1)]Hmm, that looks a bit complicated. Maybe I can simplify it using partial fractions or something.Let me try to express b_n as a difference of two fractions. Suppose:b_n = A / (2^n -1) + B / (2^{n+1} -1)But wait, since the denominator is (2^n -1)(2^{n+1} -1), maybe it's better to write it as:b_n = C / (2^n -1) - D / (2^{n+1} -1)Let me try that.So, let's assume:2^n / [(2^n -1)(2^{n+1} -1)] = C / (2^n -1) - D / (2^{n+1} -1)Multiply both sides by (2^n -1)(2^{n+1} -1):2^n = C(2^{n+1} -1) - D(2^n -1)Let me expand the right side:2^n = C*2^{n+1} - C - D*2^n + DGroup like terms:2^n = (2C)2^n - D*2^n - C + DFactor 2^n:2^n = (2C - D)2^n + (-C + D)Now, equate coefficients on both sides:For 2^n: 1 = 2C - DFor constants: 0 = -C + DSo, we have a system:1) 2C - D = 12) -C + D = 0From equation 2: D = CSubstitute into equation 1: 2C - C = 1 => C = 1Thus, D = C =1Therefore, b_n = 1/(2^n -1) - 1/(2^{n+1} -1)Wow, that's a telescoping series! So, when we sum b_n from 1 to n, most terms will cancel out.Let me write out T_n:T_n = sum_{k=1}^n b_k = sum_{k=1}^n [1/(2^k -1) - 1/(2^{k+1} -1)]This telescopes to:T_n = [1/(2^1 -1) - 1/(2^{2} -1)] + [1/(2^2 -1) - 1/(2^{3} -1)] + ... + [1/(2^n -1) - 1/(2^{n+1} -1)]All the intermediate terms cancel, so we're left with:T_n = 1/(2^1 -1) - 1/(2^{n+1} -1) = 1/1 - 1/(2^{n+1} -1) = 1 - 1/(2^{n+1} -1)So, T_n = 1 - 1/(2^{n+1} -1)We need to find the smallest positive integer m such that T_n < m/20 for all n in natural numbers.So, 1 - 1/(2^{n+1} -1) < m/20We need this inequality to hold for all n >=1.First, let's analyze T_n. As n increases, 2^{n+1} -1 increases, so 1/(2^{n+1} -1) decreases, approaching 0. Therefore, T_n approaches 1 as n approaches infinity.So, T_n is an increasing sequence approaching 1. Therefore, the maximum value of T_n is approaching 1, but never exceeding 1.Therefore, to have T_n < m/20 for all n, we need m/20 to be greater than the supremum of T_n, which is 1. So, m/20 >=1 => m >=20.But m must be an integer, so the smallest positive integer m is 20.Wait, let me check for n=1: T_1 = 1 - 1/(2^2 -1) = 1 -1/3 = 2/3 ≈0.666...20/20=1, which is greater than 2/3.For n=2: T_2=1 -1/(2^3 -1)=1 -1/7≈0.857...Still less than 1.As n increases, T_n approaches 1, so m/20 must be at least 1, so m must be at least 20.Therefore, the smallest positive integer m is 20.I think that's the answer.**Final Answer**The smallest positive integer ( m ) is boxed{20}.

answer：First, I need to determine how many photographs Debby took at each location during her vacation. She took 60% of the total photographs at the zoo, 25% at the museum, and the remaining 15% at the art gallery.Next, I'll calculate the number of photographs she decided to keep from each location. At the zoo, she kept 70% of the photographs she took. At the museum, she deleted 50% of the photographs, which means she kept 50% of them. At the art gallery, she kept all the photographs she took.After calculating the number of photographs kept from each location, I'll add them up to find the total number of photographs Debby has after the deletions.

answer：First, I need to determine A's initial investment in the business. I'll denote A's initial investment as ( X ) rupees.A invested the capital for the entire year, which is 12 months. B joined after 5 months, so B's investment period is ( 12 - 5 = 7 ) months.The profit is divided in the ratio 2:3, meaning A's share of the profit is 2 parts and B's share is 3 parts. This ratio should be equivalent to the ratio of their investments multiplied by the time for which the capital was invested.I can set up the equation as follows:[frac{X times 12}{9000 times 7} = frac{2}{3}]Next, I'll solve for ( X ):[X times 12 = frac{9000 times 7 times 2}{3}][X times 12 = frac{126000}{3}][X times 12 = 42000][X = frac{42000}{12}][X = 3500]Therefore, A's initial investment in the business was Rs. 3500.