answer：Alright, so I have this problem here about a multiple choice test with 10 questions. Each correct answer gives 5 points, each unanswered question gives 1 point, and incorrect answers give nothing. I need to figure out how many integers between 30 and 50, inclusive, are not possible total scores. Hmm, okay, let me break this down step by step.First, let me understand the scoring system. For each question, you can either get 5 points if you answer correctly, 1 point if you leave it unanswered, or 0 if you answer incorrectly. So, for each question, the possible points are 0, 1, or 5. Since there are 10 questions, the maximum possible score is 10*5 = 50, and the minimum is 0. But we're only concerned with scores between 30 and 50, inclusive.I need to find all the integers in this range that cannot be achieved with any combination of correct, unanswered, and incorrect answers. So, essentially, I need to find all the numbers between 30 and 50 that can't be expressed as 5c + u, where c is the number of correct answers and u is the number of unanswered questions. Since each question is either correct, unanswered, or incorrect, the total number of questions is 10, so c + u + w = 10, where w is the number of incorrect answers. But since w doesn't contribute to the score, maybe I can ignore it for now and just focus on c and u.So, the score S can be written as S = 5c + u, with the constraints that c and u are non-negative integers and c + u ≤ 10. Because if c + u is less than 10, the remaining questions are incorrect, which don't affect the score.So, my task is to find all integers S between 30 and 50 such that there do not exist non-negative integers c and u with c + u ≤ 10 and S = 5c + u.Alternatively, for each S from 30 to 50, I can check if there exists c and u such that 5c + u = S and c + u ≤ 10. If not, then S is not achievable.Maybe I can approach this by fixing c and seeing what u can be. Since each correct answer gives 5 points, the maximum number of correct answers is 10, which would give 50 points. But since we're starting from 30, maybe I can find the minimum number of correct answers needed to reach 30.Let me see: 5c + u = S. If I want to get S=30, what's the minimum c? 30 divided by 5 is 6. So, if c=6, then u=0, and that's 6 correct answers, 0 unanswered, 4 incorrect. That works. Similarly, for S=31, c=6, u=1, so 6 correct, 1 unanswered, 3 incorrect. That's also possible. Similarly, S=32: c=6, u=2; S=33: c=6, u=3; S=34: c=6, u=4; S=35: c=7, u=0; S=36: c=7, u=1; and so on.Wait, but maybe I should check each S from 30 to 50 and see if it's achievable.Let me list all S from 30 to 50 and check each one.Starting with S=30:30: 5c + u = 30, c + u ≤10.Let me solve for c: c = (30 - u)/5. Since u must be an integer between 0 and 10 - c.Wait, maybe it's better to fix c and see if u is possible.c can be from 0 to 10. For each c, u = S - 5c must be between 0 and 10 - c.So, for S=30:c can be from 0 to 6 (since 5*6=30). Let's check c=6: u=30 - 5*6=0. c=6, u=0: total questions used=6, which is ≤10. So yes, achievable.Similarly, S=31:c can be from 0 to 6 (since 5*6=30 <31). Let's try c=6: u=31-30=1. c=6, u=1: total=7 ≤10. So yes.S=32:c=6: u=2. c=6, u=2: total=8 ≤10. Yes.S=33:c=6: u=3. c=6, u=3: total=9 ≤10. Yes.S=34:c=6: u=4. c=6, u=4: total=10 ≤10. Yes.S=35:c=7: u=0. c=7, u=0: total=7 ≤10. Yes.S=36:c=7: u=1. c=7, u=1: total=8 ≤10. Yes.S=37:c=7: u=2. c=7, u=2: total=9 ≤10. Yes.S=38:c=7: u=3. c=7, u=3: total=10 ≤10. Yes.S=39:c=7: u=4. But c=7, u=4: total=11 >10. Not allowed. So, try c=8: 5*8=40, which is more than 39, so u would be negative. Not allowed. So, is there another c?Wait, c=7: u=4 is too much, but c=6: 5*6=30, so u=9. c=6, u=9: total=15 >10. Not allowed. So, is there a c between 6 and 7? No, c must be integer. So, seems like S=39 is not achievable? Wait, hold on.Wait, maybe I made a mistake. Let me check again.For S=39:We can try different c values.c=7: u=39 - 35=4. c=7, u=4: total=11. That's over 10. Not allowed.c=8: 5*8=40 >39, so u would be negative.c=6: 5*6=30, u=9. c=6, u=9: total=15 >10. Not allowed.c=5: 5*5=25, u=14. That's way over.Wait, so is 39 achievable? Maybe not. Hmm.Wait, maybe I can try c=5: 25 points, then u=14, but that's too many unanswered. Similarly, c=4: 20, u=19: too much. So, seems like 39 is not achievable.Wait, but let me think differently. Maybe using more correct answers and fewer unanswered.Wait, 39 is 5*7 + 4, but that's 11 questions. Alternatively, 5*8 + (-1), which is invalid. So, no.So, 39 is not achievable. Hmm.Wait, let me check S=40:c=8: u=0. c=8, u=0: total=8 ≤10. Yes.S=41:c=8: u=1. c=8, u=1: total=9 ≤10. Yes.S=42:c=8: u=2. c=8, u=2: total=10 ≤10. Yes.S=43:c=8: u=3. c=8, u=3: total=11 >10. Not allowed.c=9: 5*9=45, which is more than 43. So, u would be negative.c=7: 35, u=8. c=7, u=8: total=15 >10. Not allowed.c=6: 30, u=13: too much.So, S=43: not achievable.Wait, is there another way? Maybe c=5: 25, u=18: nope. So, seems like 43 is not achievable.S=44:c=8: u=4. c=8, u=4: total=12 >10. Not allowed.c=9: u=44 -45= -1: invalid.c=7: u=44 -35=9. c=7, u=9: total=16 >10. Not allowed.c=6: u=44 -30=14: too much.So, 44 is not achievable.S=45:c=9: u=0. c=9, u=0: total=9 ≤10. Yes.S=46:c=9: u=1. c=9, u=1: total=10 ≤10. Yes.S=47:c=9: u=2. c=9, u=2: total=11 >10. Not allowed.c=10: 50, which is more than 47. So, u would be negative.c=8: 40, u=7. c=8, u=7: total=15 >10. Not allowed.c=7: 35, u=12: too much.So, 47 is not achievable.S=48:c=9: u=3. c=9, u=3: total=12 >10. Not allowed.c=10: 50, u=-2: invalid.c=8: u=48 -40=8. c=8, u=8: total=16 >10. Not allowed.c=7: u=48 -35=13: too much.So, 48 is not achievable.S=49:c=9: u=4. c=9, u=4: total=13 >10. Not allowed.c=10: 50, u=-1: invalid.c=8: u=49 -40=9. c=8, u=9: total=17 >10. Not allowed.c=7: u=49 -35=14: too much.So, 49 is not achievable.S=50:c=10: u=0. c=10, u=0: total=10 ≤10. Yes.So, compiling the results:From 30 to 50, inclusive, the scores that are not achievable are 39, 43, 44, 47, 48, 49.Wait, let me double-check each of these to make sure I didn't miss any possible combinations.Starting with S=39:Is there a way to get 39? Let's see:If I have 7 correct answers: 35 points. Then I need 4 more points, which would require 4 unanswered questions. But 7+4=11 questions, which is more than 10. So, that's not possible.If I have 8 correct answers: 40 points, which is more than 39. So, can't do that.If I have 6 correct answers: 30 points. Then I need 9 more points, which would require 9 unanswered questions. 6+9=15, which is way over 10. So, no.Similarly, 5 correct answers: 25, needing 14 unanswered: too much.So, 39 is indeed not achievable.S=43:Trying 8 correct: 40, needing 3 more: 8+3=11: too much.7 correct: 35, needing 8 more: 7+8=15: too much.9 correct: 45, which is over 43.6 correct: 30, needing 13: too much.So, 43 is not achievable.S=44:8 correct: 40, needing 4 more: 12 total: too much.7 correct: 35, needing 9: 16 total: too much.9 correct: 45, over 44.6 correct: 30, needing 14: too much.So, 44 is not achievable.S=47:9 correct: 45, needing 2 more: 11 total: too much.8 correct: 40, needing 7: 15 total: too much.10 correct: 50, over 47.7 correct: 35, needing 12: too much.So, 47 is not achievable.S=48:9 correct: 45, needing 3: 12 total: too much.8 correct: 40, needing 8: 16 total: too much.10 correct: 50, over 48.7 correct: 35, needing 13: too much.So, 48 is not achievable.S=49:9 correct: 45, needing 4: 13 total: too much.8 correct: 40, needing 9: 17 total: too much.10 correct: 50, over 49.7 correct: 35, needing 14: too much.So, 49 is not achievable.Therefore, the impossible scores between 30 and 50 are 39, 43, 44, 47, 48, 49. That's 6 numbers.Wait, let me check if I missed any other numbers. Let me go through each score from 30 to 50 and see if I marked them correctly.30: achievable (6c, 0u)31: achievable (6c,1u)32: achievable (6c,2u)33: achievable (6c,3u)34: achievable (6c,4u)35: achievable (7c,0u)36: achievable (7c,1u)37: achievable (7c,2u)38: achievable (7c,3u)39: not achievable40: achievable (8c,0u)41: achievable (8c,1u)42: achievable (8c,2u)43: not achievable44: not achievable45: achievable (9c,0u)46: achievable (9c,1u)47: not achievable48: not achievable49: not achievable50: achievable (10c,0u)So, the impossible scores are 39,43,44,47,48,49: that's 6 numbers.Wait, but let me check S=44 again. Is there a way to get 44?If I have 8 correct answers: 40 points. Then I need 4 more points, which would require 4 unanswered. So, 8+4=12 questions, which is over 10. Not allowed.If I have 7 correct answers: 35 points. Then I need 9 more points, which would require 9 unanswered. 7+9=16: way over.If I have 9 correct answers: 45 points. That's over 44.If I have 6 correct answers: 30 points. Then I need 14 unanswered: too much.So, yes, 44 is not achievable.Similarly, S=47:8 correct: 40, needing 7: 15 total: too much.9 correct: 45, needing 2: 11 total: too much.10 correct: 50: over.So, 47 is not achievable.Same with 48 and 49.Therefore, the total number of impossible scores between 30 and 50 is 6.Wait, but let me check S=39 again. Maybe I can have some incorrect answers mixed in.Wait, the total number of questions is 10. So, if I have c correct, u unanswered, and w incorrect, with c + u + w =10.But the score is only dependent on c and u: S=5c + u.So, even if I have incorrect answers, it doesn't affect the score, but it does affect the total number of questions used.So, for S=39, if I have c=7, u=4, then w=10 -7 -4= -1: which is invalid.Wait, no, w can't be negative. So, that's not allowed.If I have c=6, u=9: w=10 -6 -9= -5: invalid.c=5, u=14: w=10 -5 -14= -9: invalid.So, no way to get 39 without exceeding the total number of questions.Therefore, 39 is indeed not achievable.Similarly, for the others.So, my conclusion is that there are 6 integers between 30 and 50 that are not possible total scores.**Final Answer**The number of integers between 30 and 50 that are not possible total scores is boxed{6}.

answer：To determine when Kolya, Seryozha, and Vanya will all meet at the cinema again, I need to find the least common multiple (LCM) of their visiting intervals. Kolya visits every 3 days, Seryozha every 7 days, and Vanya every 5 days. First, I'll list the prime factors of each interval:- 3 is already a prime number.- 5 is also a prime number.- 7 is a prime number as well.Since all three numbers are prime and have no common factors other than 1, the LCM is simply the product of these three numbers.Calculating the product:3 × 5 × 7 = 105Therefore, all three boys will meet at the cinema again in 105 days.

answer：To determine the equation of the line that passes through the point (0, 4) and is perpendicular to the line x - 3y - 7 = 0, I'll follow these steps:First, I'll find the slope of the given line. The equation x - 3y - 7 = 0 can be rewritten in slope-intercept form (y = mx + b) to easily identify the slope. Solving for y, I get y = (1/3)x - 7/3, so the slope (m₁) of the given line is 1/3.Since the desired line is perpendicular to this, its slope (m₂) must be the negative reciprocal of m₁. Therefore, m₂ = -3.Now that I have the slope of the desired line and a point it passes through (0, 4), I can use the point-slope form of a line to write its equation. Plugging in the values, I get y - 4 = -3(x - 0), which simplifies to y = -3x + 4.Finally, I'll rearrange this equation to match the standard form. Bringing all terms to one side, the equation becomes 3x + y - 4 = 0.Looking at the answer choices, option A matches this equation.

answer：First, I need to calculate the total time Gunther spends on each cleaning task.Vacuuming the carpets takes 45 minutes.Dusting the furniture takes 60 minutes.Mopping the kitchen floors takes 30 minutes.Brushing each cat takes 5 minutes, and since there are three cats, this amounts to 15 minutes.Adding all these tasks together: 45 + 60 + 30 + 15 equals 150 minutes.Gunther has 3 hours of free time, which is 180 minutes.Subtracting the total cleaning time from his available time: 180 - 150 equals 30 minutes.Therefore, Gunther will have 30 minutes of free time left after cleaning.

answer：Alright, so I have this problem here about machine parts and their dimensions. It says that under normal conditions, the dimensions follow a normal distribution, which is N(μ, σ²). Then, in a normal experiment, when 1000 parts are taken, I need to find the possible number of parts that do not fall within the size range of (μ - 3σ, μ + 3σ). Hmm, okay.First, I remember that in a normal distribution, most of the data lies within three standard deviations from the mean. Specifically, I think about the empirical rule, which states that approximately 99.7% of the data falls within μ ± 3σ. So, that means only about 0.3% of the data lies outside that range. Let me write that down. The probability that a single part does not fall within (μ - 3σ, μ + 3σ) is 1 - 0.997 = 0.003, or 0.3%. So, for each part, there's a 0.3% chance it's outside that range.Now, if we have 1000 parts, we can model the number of parts outside this range using the binomial distribution. The binomial distribution gives the probability of having exactly k successes (in this case, parts outside the range) in n trials (1000 parts), with the probability of success on a single trial being p (0.003).The expected number of parts outside the range would be n * p, which is 1000 * 0.003 = 3. So, on average, we'd expect 3 parts to be outside the range. But the question is asking for the possible number, not just the expected number. Hmm, so it's not just the mean; we need to consider the distribution around that mean. Since the number of parts outside the range is a binomial random variable, we can describe the possible number by looking at the variance and standard deviation.The variance of a binomial distribution is n * p * (1 - p). Plugging in the numbers, that's 1000 * 0.003 * 0.997 ≈ 1000 * 0.002991 ≈ 2.991. So, the variance is approximately 2.991, which means the standard deviation is the square root of that, which is about √2.991 ≈ 1.73.So, the number of parts outside the range can be approximated as a normal distribution with mean 3 and standard deviation approximately 1.73. But wait, the binomial distribution can also be approximated by a Poisson distribution when n is large and p is small, which is the case here. The Poisson distribution has the parameter λ = n * p = 3, so the possible number of parts outside the range can be modeled by a Poisson distribution with λ = 3.In a Poisson distribution, the probability of k occurrences is given by P(k) = (e^{-λ} * λ^k) / k!. So, the possible number of parts outside the range can be 0, 1, 2, 3, etc., with probabilities decreasing as k increases.But the question is asking for the possible number, not the probabilities. So, in practical terms, how many parts could we expect to be outside the range? Well, the mean is 3, but due to the variance, the actual number could be more or less. To get a sense of the possible range, we can consider the standard deviation. Since the standard deviation is about 1.73, we can say that the number of parts outside the range is likely within 3 ± 2*1.73, which is approximately 3 ± 3.46. So, that would give us a range from about -0.46 to 6.46. Since the number of parts can't be negative, the lower bound is 0, and the upper bound is roughly 6 or 7.But wait, in reality, the Poisson distribution is skewed, so the upper tail is longer. So, while it's unlikely, there's a small probability of having more than 6 or 7 parts outside the range. However, for practical purposes, especially in quality control, we often consider the range within a few standard deviations from the mean.Alternatively, if we use the binomial distribution, we can calculate the probabilities for different numbers of parts. For example, the probability that exactly 0 parts are outside is P(0) = C(1000,0)*(0.003)^0*(0.997)^1000 ≈ e^{-3} ≈ 0.0498, or about 4.98%. Similarly, P(1) ≈ e^{-3} * 3 ≈ 0.1494, or 14.94%, and so on.But again, the question is about the possible number, not the probability. So, in terms of possible numbers, it's any non-negative integer, but practically, it's most likely around 3, with some variation.Wait, maybe the question is expecting a specific number, like the expected value? But it says "possible number," which is a bit vague. Maybe it's asking for the range or the typical number. Alternatively, perhaps it's referring to the number of parts that are expected to be outside, which is 3. But since it's a probability, it's not a fixed number, so the possible number could be 0, 1, 2, 3, 4, etc., but the most probable is around 3.But maybe the question is more straightforward. It says "the possible number of parts that do not fall within the size range." So, in a normal distribution, about 0.3% are outside, so for 1000 parts, that's 3 parts. So, the possible number is 3.But wait, the question says "the possible number," which might imply a range rather than a single number. So, perhaps it's asking for the number of parts that could be outside, considering some confidence interval.If we consider the 95% confidence interval, which is approximately mean ± 2 standard deviations, so 3 ± 2*1.73 ≈ 3 ± 3.46. So, the number of parts outside the range is likely between 0 and 6 or 7.But since the question is a bit ambiguous, it's hard to say. Maybe it's just expecting the expected value, which is 3. Or perhaps it's expecting the range based on the 3σ rule, but that's about the data itself, not the count.Wait, no, the 3σ rule is about the data points, not the count of outliers. So, in this case, the count of outliers follows a binomial distribution, which we approximated as Poisson with λ=3.So, the possible number of parts outside is a random variable with mean 3 and standard deviation ~1.73. So, the possible number is any integer from 0 upwards, but the majority of the probability mass is around 0 to 6 or 7.But the question is in Chinese, and the translation is "the possible number of parts that do not fall within the size range of (μ - 3σ, μ + 3σ) is." So, maybe it's expecting a specific number, like 3, or a range.Alternatively, maybe it's expecting the number based on the 3σ rule, which is 0.3% of 1000, which is 3. So, the possible number is 3.But considering that it's a probability, it's not fixed. So, maybe the answer is that the number is approximately 3, but it could vary.Wait, but in the context of an exam question, it's likely expecting the expected value, which is 3. So, the possible number is 3.Alternatively, maybe it's expecting the number of standard deviations, but no, the question is about the count.So, to sum up, the expected number is 3, but the possible number could be 0,1,2,3,4,5,6, etc., but the most probable is around 3.But since the question says "the possible number," maybe it's expecting the expected value, which is 3. So, I think the answer is 3.But wait, let me double-check. The probability that a single part is outside is 0.003, so for 1000 parts, the expected number is 3. So, yes, 3 is the expected number. But the question is about the possible number, which could be more or less.But in the context of an exam, unless it specifies a confidence interval or something, it's likely expecting the expected value, which is 3.So, I think the answer is 3.**Final Answer**The possible number of parts that do not fall within the size range is boxed{3}.

answer：To find the greatest common divisor of 1248 and 585, I'll use the Euclidean algorithm.First, I divide 1248 by 585. The quotient is 2 and the remainder is 78.Next, I divide 585 by 78. The quotient is 7 and the remainder is 39.Then, I divide 78 by 39. The quotient is 2 and the remainder is 0.Since the remainder is now 0, the last non-zero remainder is 39, which is the greatest common divisor.