1227 Maths Gaby/Kevin

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Hi, Kevin. How are you? Are you all right? Yeah, Yeah, tired having a good holiday. Yeah, Yeah. Okay. All right. Well, and let me what this. Let me just plug in. I'm trying to find the right cable. Some reason. I don't I've got a new, well, different computer. I'm not sure which one is the right. Okay. And. Okay, there it is. All right. We are plugged in. Okay, so do you want to try and solve this? I know you said you find the pen annoying on the board, so if you want to, you can always do it on paper and pencil with paper and pencil, and then just tell me afterwards, do you want to try and solve however you like? You can use the board, obviously, but it's up to you whichever you prefer. I'll just tell you the answer. Just let me know if you are feeling unsure about how to solve it. Are you able to solve it to use a method? Yeah, okay. I'm just going to put nice. Yeah. Huge. Have you set up an equation, a pair of equations? Okay, well, I think a pair of equations is what's going to help you here. Okay, simultaneous equations. We've got horses plus clowns equals how many items? 45, right? Yeah, okay. So the number of horses and the number of clowns make 45. A horse se's got four feet. Yes, so four times. So four times the horse's feet. And the clown has got two feet. So two times the number of clowns would be the total number of feet Yeah Yeah which would be 124. Now, do you know how to solve? 哎呀。17 horses. All right, so if there's 17 horses, then there would have to be 28 clowns. Yeah. Does that fit? There's only there's only one correct answer. Oh, 56 plus 68. Seems to fit right. 56Yeah plus 68 fits. Okay. So in this kind of linear equation, there's only one correct answer. Yeah so you can use substitution. Yeah. You use substitution to check that your answer is correct. Yeah. Okay, right. So let's try a bit of substitution. How much are these equal to? You can work them out on paper if you want. So s is three, b is one, t is four, n is eight. What is the value of the expressions you see there? Do you know what to do? Yeah, okay, all right, let's make a start. Sorry. I'm just going to check out my calculator because I'm too lazy. Yeah, good job. Okay, so you just needed to add the one there. Yeah, there we go. Okay, good job. Right. Which one is the odd one out? Okay, Hey, this one. In each one, A, B, C, D or e, there is one odd one out. Can you tell one is a rearrangement of the other? As in they are all rearrangements. Four of them are rearrangements of the same equation and one of them is incorrect. Yeah. Yeah, okay. Adden. Feeling stuck. Yeah. So I would try and I would try and rearrange. Say a to see if it makes b. Can those two things be equal? Can a and b be equal in the first example? No. You sure I mean, if you expanded that. Let me just make the font smaller holon. If you expanded that. You would get. Oops. Wouldn't you? So couldn't you divide so if you divided everything there by five. It would come up as a over five is equal to b plus c, wouldn't it? Yeah, so I think I think these two are the same. Real. Okay. So can you make this one into this one by rearranging. I think so, right? You can just move, right? So have you done rearranging it all where you can make this? If this plus c, if we want it to go to the other side of the equation, what happens? Right? So that's just this backwards written backwards, right? So those two are the same as well, right? And this one is the same too. Can you tell me why? This one is just have if we have b and we send b over to the other side of the equation, you would get a over five minus b equals c right? So that's just that backwards. And this one is the odd one out. But I think by looking here. It feels normal. It should feel quite normal that these two do not really match. Can you see why? Okay, all right, try the second one. Try try going. So have you done rearranging equations? Okay, we'll try going from a to b and see if they match. All right, so here e is wrong. Yeah. So they do fit. Yeah okay. So then does does b fit with c or is c fit with a? No no, because c and b are both positive there, aren't they? And c and c is positive here and b is a negative there. Yeah. Hold on though, these these two are positive on same side. Hold on. Do these two really match? Three c. Equals 15 minus b. Do a and b really match? What? I'm not sure that's right. Look, because. Look, b and c are positive on both sides of the equation there. Would you agree? And here. And here. And here. It's only here. Where b is negative on the opposite side of the equation to c. That has to make it the odd one out, doesn't it? Yeah because look, if b and c are both positive when they're on opposite sides, then when they're on the same side, when they're on the same side as one another, one of them has to be negative in a rearrangement. Yeah so you can view it like that as well. Okay. Which values are positive on which side? All right, so let's try and do the next one and see if we've got a few more clues as to which one might be incorrect. These look. Okay. So they're not so the b's and the c's are not incorrect. They're all they're all the b's and c's are all identical there. They're on different sides of the equation to each other and they are all positive. These two are identical, aren't they? Almost it just you're right. Okay, so so look so a and c can not be the odd one ha I've already got that. Which one's the odd one out then? I mean, I've. Really got that into the same, okay? Yeah, it's got to be that one, hasn't it? We can see that these two are just rearrangements of each other. Yeah. Yeah. Okay. Last one. Which one's the odd one out? Cool. Yeah, nice. It's got a ba, right? Yeah. Okay, let's take okay, well, I'll try and give you an option. Do you want to go? Okay, well, let's do this because we just finish off pie charts with this. 嗯。19岁,10 60 60 90 10。Mama, you think I should have this? Why? I'm stuokay. So you need four numbers that turn right? So you can do it just kind of problem solving, or you can do it algebra. You can do it using algebra. You've got four numbers that add up to 60. Let's just take an algebraic method. Because I think I think you might actually just prefer the algebraic if you think about it as a number plus two, oops, plus two times the number, plus four times the same number, plus. That's it, right? Hold on, see. D could be x right and then c Oh. Yeah D D if d is x Yeah then c is two X B is four x. And a is atex because right? C is double D, B is double c and a is double b. So that's how it looks in algea. So that's how it looks in algebra. Are you okay with that? All right, so all of those add up to. Or have I done it wrong? That's three x. So 15x, Yeah, I just couldn't. I just doubted myself counting there, 15x equals 60. Right? So d would be so x is four. Yes, that was d. Let's just check, does four plus eight, that's twelve plus 16, that's 28 plus 32 make 60 right? Yeah. Okay. So then so then how much are these in angles, in degrees in terms of pie chart? So Yeah, 60 degrees, 60 people for 360 degrees. So it's 60 degrees per person. Yeah six degrees per person. So we've got 32 times six Yeah. Yeah. Well. Okay, except sorry. Yeah it's obviously it's sectsy so it does just be eight times six Yeah Yeah all right. That's okay once we know how to do it, isn't it? Yeah, right. Let's just revise let's just revise a little bit on histograms. Okay. So I think what we want to find out first is obviously we leart right that frequency density is always the y axis in histograms Yeah and frequency density represents the height of each bar. So the height the height of a bar times the width of a bar, gives us the bar area and the air and the area was the frequency, right? Okay, so this bar is ten wide. It seems to be going from 2.5 to 7.5. Oh no, it's not ten wide. Five wide sorry, I beg upon. So it's five wide going from 2.5. Okay, so so five times 1050 so 50 cats kittens. Yeah, Yeah. So 5050 kittens took between. So we can describe that as 50 kittens taking between 2.5 and 7.5 minutes. Okay, so can you tell me the frequency of the other bars? Yeah. Right? Okay, so that's all the frequencies. Very nicely done. And basically we now need to calculate an estimate for the meantime. So look, the reason we put estimate is not because we're not calculating exactly, right? You're calculating. Your calculations are done exactly. You're not rounding up or rounding down. The reason it's called estimate is because Yeah, okay, so you've got loads of data in here. Yeah. So it could be all those things could be anywhere. Yeah. The data points could be the data points could actually all be down here like around the eleven minute mark. We don't know. Yeah. So so do you have any thoughts on calculating an estimate for the mean? You would do it this set. I don't know if you use groufrequency at all groufrequency tables, but basically you're saying 50 cats took between two and a half and seven and a half minutes? Yeah. But the total time in minutes of all of those 50 cats, the sensible way to calculate it is to take the 50 cats and say that they, and basically approximate, say that they estimate that they all took five minutes each. Do you see why? We're we're going to basically say, look, we're going to take a guess and say that all of those 50 cats took five minutes each. Why would I say five? Oh, don't know. Is it is it a better guess to say they took five minutes than they all let's just guess, they all took two and a half minutes? Or let's let's just guess they all took seven and a half minutes. If if I say, okay, I'm going to assume they all took seven and a half minutes. That's saying they all took the maximum amount of time in that bar. Yeah. And if I say they all took two and a half minutes, I'm saying they were all the quickest time inside of that bar. Yeah. So five is dead in the middle, right? And this is the same this is the same idea as when you're doing grouped frequency. Have you if you heard that before group frequency? Yeah, we're basically saying, look, this is different times all squished together. Yeah this is not a bar chart where you're literally like there's one thing that's some how many? Yeah, it's a group. So so we're going for the middle because that's the best we can do in terms of saying, okay, some were probably less, some were probably more. Let's go for the middle. Yeah. So so there. So there you're saying the 50 cats took 250 minutes in total. Yeah. So you're going to guess that the you're going to estimate that the 150 cats in the next group took how much time. 750, I think you should go for ten no, ten minutes, right? What you're saying is in the next group, we've got 150 cats. And where's the middle time ten minutes? Yeah. You're making a guess of how long you're making an estimate of how long in total it took all of those cats in minutes. Yeah, okay. So what you sound like you're unsure. We're going for right we're trying to what would happen right? If you what happens when you calculate the mean on anything? Oh, I know. Yeah, I get Yeah, okay. So if you calculate the mean on anything, right, this is cats going over an obstacle and we timed it in our imaginations. Yeah so ordinarily you would have to add up all the times and divide by how many cats there were, right? If you only had five cats and it was like two minutes, three minutes, four minutes, five minutes and six minutes, you would add up all of those minutes and say, we divide by five. That's how many minutes on average. Mean, Yeah, it took them. But here we've got loads more data, 475 cats, and we've got loads more times. Yeah. So we're not going to do that. We're going to estimate the mean, yes. So you've taken 50 cats estimated at taking five minutes each, 150 cats estimated at taking ten minutes each. What's the next one? What's the next Yeah, what's the next estimate? Morning times. So the next group is 150 cats, took between twelve and a half minutes and 20 minutes. So what's the middle? So what's halfway between twelve and a half, right? We've got twelve and a half here point five, okay, 16.5 or 16.25, is it. Twelve and a half. 16.5. 16.5, do you say 0.2, 5.5, but that's not halfway, is it? Between what's halfway between twelve and a half halfway? Between 2:12.5. And 20, it's got to be two either two point a 0.25 or a.75. Just let's look on our calculator. I haven't got one. Okay all right, so I'm gonna to add them up and divide by 2:20 plus 12.5 divide by 2:16.25Yeah. Okay, right. And then it's times times that by 150. 2437.5Yeah. Okay. So what calculation do I need to do for the last one? I'll do it on my calculator. Yeah 125 times 22 point lovely. Okay so that gives me but just write down 2812.5Yeah so you would then add all of you would add all of these up. And divide by what then? Or all groups, but how many cats? 475, right? Because we're going for the average time for per cat. Yeah Yeah right. So I'm just doing not my calculator if you haven't got one. So I've added up 250, 15 hundred tooth up and those two other numbers. The total time estimate for all cats is seven zero minutes. Yeah divide by 475 cats. 14.7 minutes is the mean estimate of time taken for cap 14.7, right? 14.7 is about here. Does that feel like we did an okay estimate for our mean. I think it's okay, right? Look at the frequencies on this side added up in area and the frequencies on that side. It looks like the same area, it looks like a similar amount of area on each side of that mean line, doesn't it? Yeah. So that's a way of checking that you've got a good estimate of the mean. It's pretty much in the middle right of our Yeah. If you look at the blocks, the blocks on this side look as simyeah. Okay, so we did, we haven't done a Cray, obviously. If your mean line ended up here, yoube like, Oh, I've crazily done something. Yes, I've done something that's made my mean go really, really skewed to the wrong side. Yeah, okay. Well, look, we did some difficult substitution. We did some difficult rearrangement. We needed to do some. We did some problem solving there. We needed to do some well, that was simultaneous equations. Yeah in kind of the context of a question. Yeah when you're problem solving, I think you can always think to yourself, can I put this into algebra, right? Because algebra is your is your best friend here. Like algebra is a really good tool for problem solving using mystery numbers. Yeah. And then we didn't quite get onto the last thing, which would have been this, which would have been other simultaneous equations. But never mind have a great rest of your day and I will see you next time, Kevin. Alright, right. Cheers. Bye.
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{
    "header_icon": "fas fa-crown",
    "course_title_en": "1227 Maths Gaby\/Kevin Lesson Log",
    "course_title_cn": "1227 数学 Gaby\/Kevin 课程记录",
    "course_subtitle_en": "Mathematics Lesson - Algebra and Statistics Review",
    "course_subtitle_cn": "数学课程 - 代数与统计复习",
    "course_name_en": "Maths",
    "course_name_cn": "数学",
    "course_topic_en": "Simultaneous Equations, Algebraic Manipulation, and Histogram Mean Estimation",
    "course_topic_cn": "联立方程、代数重排和直方图均值估计",
    "course_date_en": "N\/A (Date not explicitly stated in dialogue)",
    "course_date_cn": "日期未在对话中明确说明",
    "student_name": "Kevin",
    "teaching_focus_en": "Reviewing and applying methods for solving simultaneous equations, practicing algebraic rearrangement, and estimating the mean from grouped frequency data (histograms).",
    "teaching_focus_cn": "复习和应用解联立方程的方法,练习代数重排,以及从分组频率数据(直方图)中估计均值。",
    "teaching_objectives": [
        {
            "en": "Successfully apply simultaneous equations to a word problem (horses and clowns).",
            "cn": "成功将联立方程应用于文字题(马和小丑)。"
        },
        {
            "en": "Identify the odd one out among several rearranged algebraic equations.",
            "cn": "在多个重排的代数方程中识别出不一致的那个。"
        },
        {
            "en": "Estimate the mean from grouped frequency data presented in a histogram.",
            "cn": "从直方图中展示的分组频率数据中估计均值。"
        }
    ],
    "timeline_activities": [
        {
            "time": "Beginning",
            "title_en": "Setup and Context Setting",
            "title_cn": "设置与情境建立",
            "description_en": "Teacher dealt with technical setup (plugging in) and offered student choice on note-taking method (board vs. paper).",
            "description_cn": "教师处理了技术设置(插拔设备),并为学生提供了记录方式的选择(板书或纸笔)。"
        },
        {
            "time": "Activity 1",
            "title_en": "Simultaneous Equations Word Problem",
            "title_cn": "联立方程文字题",
            "description_en": "Solving a problem involving horses (4 feet) and clowns (2 feet) summing to 45 animals and 124 feet, using simultaneous equations.",
            "description_cn": "使用联立方程解决一个涉及马(4只脚)和小丑(2只脚)的总数为45只动物,总脚数为124只的问题。"
        },
        {
            "time": "Activity 2",
            "title_en": "Algebraic Substitution Check",
            "title_cn": "代数代入检验",
            "description_en": "Checking the correctness of an answer using substitution with given variable values (s=3, b=1, t=4, n=8).",
            "description_cn": "使用给定的变量值 (s=3, b=1, t=4, n=8) 进行代入检验答案的正确性。"
        },
        {
            "time": "Activity 3",
            "title_en": "Odd One Out: Algebraic Rearrangement",
            "title_cn": "找出不同项:代数重排",
            "description_en": "Identifying which equation is not a rearrangement of the others by analyzing the positions and signs of variables (b and c).",
            "description_cn": "通过分析变量(b和c)的位置和符号,找出哪个方程不是其他方程的重排。"
        },
        {
            "time": "Activity 4",
            "title_en": "Problem Solving: Ratios and Sums (leading to Algebra)",
            "title_cn": "问题解决:比例与求和(导向代数)",
            "description_en": "Solving a problem requiring four numbers that sum to 60, based on multiplicative relationships (doubling), solved using algebra (x, 2x, 4x, 8x).",
            "description_cn": "解决一个要求四个数之和为60的问题,基于倍数关系(加倍),使用代数方法(x, 2x, 4x, 8x)求解。"
        },
        {
            "time": "Activity 5",
            "title_en": "Pie Chart Angle Calculation",
            "title_cn": "饼图角度计算",
            "description_en": "Converting the derived numbers from Activity 4 into degrees for a pie chart context (6 degrees per person\/unit).",
            "description_cn": "将活动4中得出的数字转换为饼图所需的角度(每人\/单位6度)。"
        },
        {
            "time": "Activity 6",
            "title_en": "Histogram Mean Estimation",
            "title_cn": "直方图均值估计",
            "description_en": "Calculating frequency density, frequency from bar area, and estimating the overall mean time using midpoints for grouped frequency data.",
            "description_cn": "计算频率密度,从柱体面积计算频率,并使用分组频率数据的中点来估计总体平均时间。"
        }
    ],
    "vocabulary_en": "Simultaneous equations, substitution, rearrange, expression, frequency density, midpoint, grouped frequency, estimate, mean.",
    "vocabulary_cn": "联立方程, 代入法, 重排, 表达式, 频率密度, 中点, 分组频率, 估计, 均值。",
    "concepts_en": "Solving linear systems, algebraic manipulation rules, principles of histogram area relating to frequency, and estimating means from grouped data using midpoints.",
    "concepts_cn": "解线性方程组,代数重排规则,直方图面积与频率的关系原理,以及使用中点估计分组数据的均值。",
    "skills_practiced_en": "Problem-solving in context, algebraic manipulation (rearranging and substituting), critical analysis of equations, and statistical estimation.",
    "skills_practiced_cn": "情境中的问题解决,代数操作(重排和代入),对方程的批判性分析,以及统计估计。",
    "teaching_resources": [
        {
            "en": "Visual aid for the horse and clown problem (implied context).",
            "cn": "马和小丑问题的视觉辅助材料(暗示)。"
        },
        {
            "en": "Set of multiple-choice questions requiring algebraic rearrangement comparison.",
            "cn": "需要进行代数重排比较的选择题组。"
        },
        {
            "en": "Histogram graph depicting time taken by kittens, used for mean estimation.",
            "cn": "展示小猫花费时间的直方图,用于均值估计。"
        }
    ],
    "participation_assessment": [
        {
            "en": "Kevin was generally engaged, showing willingness to attempt problems, although sometimes hesitant when methods were complex (e.g., rearrangement logic).",
            "cn": "Kevin总体上参与度较高,愿意尝试解决问题,尽管在方法复杂时(例如重排逻辑)有时会犹豫。"
        }
    ],
    "comprehension_assessment": [
        {
            "en": "Strong understanding of setting up simultaneous equations for the initial word problem. Understanding of using midpoints for histogram mean estimation was good after teacher clarification.",
            "cn": "对初始文字题的联立方程的建立理解深刻。经过教师澄清后,对直方图均值估计中使用中点的理解良好。"
        }
    ],
    "oral_assessment": [
        {
            "en": "Kevin responded clearly to direct questions, though occasional pauses indicated deep processing of multi-step mathematical logic.",
            "cn": "Kevin对直接提问的回答清晰,但偶尔的停顿表明他对多步数学逻辑正在深入思考。"
        }
    ],
    "written_assessment_en": "N\/A (Focus was on procedural demonstration and verbal justification)",
    "written_assessment_cn": "不适用(重点在于程序演示和口头论证)",
    "student_strengths": [
        {
            "en": "Proficient in setting up basic linear equations from context (e.g., horse\/clown problem).",
            "cn": "擅长根据情境(如马\/小丑问题)建立基础线性方程。"
        },
        {
            "en": "Good application of algebraic thinking when establishing the relationship between the four numbers (x, 2x, 4x, 8x).",
            "cn": "在建立四个数(x, 2x, 4x, 8x)之间的关系时,代数思维应用良好。"
        },
        {
            "en": "Quickly grasped the concept of frequency density = area in histograms.",
            "cn": "很快理解了直方图中频率密度=面积的概念。"
        }
    ],
    "improvement_areas": [
        {
            "en": "Rigor in algebraic rearrangement, particularly when dealing with negative signs across the equals sign (odd one out activity).",
            "cn": "代数重排的严谨性,特别是在处理等号两边的负号时(找出不同项活动)。"
        },
        {
            "en": "Confidence in justifying the choice of the midpoint for mean estimation in grouped data without immediate teacher prompting.",
            "cn": "在没有即时教师提示的情况下,自信地为分组数据的均值估计选择中点提供理由。"
        }
    ],
    "teaching_effectiveness": [
        {
            "en": "The teacher effectively scaffolded the complex histogram mean estimation by breaking it down into defining the midpoint and understanding the 'estimate' terminology.",
            "cn": "教师通过分解定义中点和理解‘估计’术语,有效地搭起了复杂直方图均值估计的脚手架。"
        }
    ],
    "pace_management": [
        {
            "en": "The pace was generally appropriate, allowing for depth in statistical concepts, though the transition between the rearrangement section and the pie chart section felt slightly abrupt.",
            "cn": "节奏总体适中,允许在统计概念上深入探讨,但从代数重排部分到饼图部分的过渡感觉略显突然。"
        }
    ],
    "classroom_atmosphere_en": "Supportive and interactive. The teacher frequently checked for understanding and encouraged student input, even when the student was unsure.",
    "classroom_atmosphere_cn": "支持性和互动性强。教师经常检查理解情况并鼓励学生发表意见,即使学生感到不确定。",
    "objective_achievement": [
        {
            "en": "All primary objectives were addressed, with the student demonstrating successful application, particularly in setting up and executing the algebraic methods.",
            "cn": "所有主要目标都得到了解决,学生展示了成功的应用,特别是在建立和执行代数方法方面。"
        }
    ],
    "teaching_strengths": {
        "identified_strengths": [
            {
                "en": "Excellent scaffolding when introducing abstract statistical estimation (histogram mean).",
                "cn": "在引入抽象的统计估计(直方图均值)时,提供了出色的脚手架支持。"
            },
            {
                "en": "Effective use of contextual problems (horses\/clowns) to anchor algebraic concepts.",
                "cn": "有效地利用情境问题(马\/小丑)来锚定代数概念。"
            }
        ],
        "effective_methods": [
            {
                "en": "Guiding the student to justify rearrangement logic by comparing variable positions across the equals sign.",
                "cn": "通过比较等号两边的变量位置,引导学生为重排逻辑提供理由。"
            },
            {
                "en": "Connecting estimation back to the definition of the mean (total time \/ total count).",
                "cn": "将估计与均值的定义(总时间\/总计数)联系起来。"
            }
        ],
        "positive_feedback": [
            {
                "en": "Positive reinforcement was given immediately upon correct calculation or logical breakthrough ('Good job,' 'That's it, right?').",
                "cn": "在计算正确或逻辑突破后立即给予积极的强化反馈(‘干得好’,‘对吧?’)。"
            }
        ]
    },
    "specific_suggestions": [
        {
            "icon": "fas fa-equals",
            "category_en": "Algebraic Manipulation & Logic",
            "category_cn": "代数重排与逻辑",
            "suggestions": [
                {
                    "en": "Practice rearranging complex equations, specifically focusing on how moving terms affects their signs. Review the rule: 'What happens to a term when it crosses the equals sign?'",
                    "cn": "练习重排复杂方程,特别关注移动项如何影响其符号。复习规则:“当一项跨过等号时会发生什么?”"
                }
            ]
        },
        {
            "icon": "fas fa-chart-bar",
            "category_en": "Statistics: Grouped Data",
            "category_cn": "统计学:分组数据",
            "suggestions": [
                {
                    "en": "Create flashcards for the steps to calculate the estimated mean from a histogram to build procedural fluency.",
                    "cn": "制作从直方图中计算估计均值的步骤卡片,以建立程序熟练度。"
                }
            ]
        }
    ],
    "next_focus": [
        {
            "en": "If time allows, review the final planned topic (implied second set of simultaneous equations) or solidify the logic behind identifying the 'odd one out' in algebraic forms.",
            "cn": "如果时间允许,复习最后计划的主题(暗示的第二组联立方程)或巩固代数形式中识别‘不同项’背后的逻辑。"
        }
    ],
    "homework_resources": [
        {
            "en": "Worksheet focusing on rearranging equations involving multiple variables (b, c) to isolate one variable.",
            "cn": "以重排涉及多个变量(b, c)的方程来分离一个变量的练习题。"
        },
        {
            "en": "Practice problems converting grouped frequency data tables into histograms and calculating the mean estimate.",
            "cn": "练习将分组频率数据表转换为直方图并计算均值估计的习题。"
        }
    ]
}
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生成时间: 2025-12-30 03:12:10

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