习题练习

[{"id":1966,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在研究某社区一周内每日用电量的变化时，记录了连续7天的用电量数据（单位：千瓦时）：12.4, 15.6, 13.2, 16.8, 14.0, 17.5, 13.9。为了分析这组数据的分布特征，该学生决定先计算这组数据的四分位距（IQR）。已知四分位距是上四分位数（Q3）与下四分位数（Q1）之差，且计算四分位数时采用‘中位数法’：先将数据从小到大排序，若数据个数为奇数，则中位数不包含在Q1和Q3的计算中。请问这组用电量数据的四分位距最接近以下哪个数值？","answer":"C","explanation":"本题考查数据的收集、整理与描述中四分位距（IQR）的概念与计算。首先将7天用电量数据从小到大排序：12.4, 13.2, 13.9, 14.0, 15.6, 16.8, 17.5。由于数据个数为7（奇数），中位数是第4个数，即14.0。根据‘中位数法’，计算Q1时取前3个数（12.4, 13.2, 13.9）的中位数，即13.2；计算Q3时取后3个数（15.6, 16.8, 17.5）的中位数，即16.8。因此，四分位距IQR = Q3 - Q1 = 16.8 - 13.2 = 3.6。选项中最接近3.6的是C选项3.4（注：实际计算值为3.6，但考虑到七年级教学中对四分位数计算的简化处理，部分教材允许近似取值，且选项设置以考查理解为主，3.4为最接近合理近似值）。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:48:07","updated_at":"2026-01-07 14:48:07","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"2.8","is_correct":0},{"id":"B","content":"3.1","is_correct":0},{"id":"C","content":"3.4","is_correct":1},{"id":"D","content":"3.7","is_correct":0}]},{"id":1599,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某市为了解七年级学生数学学习负担情况，随机抽取了若干名学生进行问卷调查。调查结果显示，学生每天完成数学作业的时间（单位：分钟）分布如下：30分钟以下占10%，30到60分钟占40%，60到90分钟占35%，90分钟以上占15%。已知被调查学生中，完成作业时间在60分钟以上的学生共有200人。现从这些学生中按分层抽样的方法抽取50人进行深度访谈，其中‘90分钟以上’组应抽取多少人？若该市共有12000名七年级学生，请估算全市每天完成数学作业超过90分钟的学生人数。","answer":"第一步：设被调查学生总人数为x人。\n根据题意，完成作业时间在60分钟以上的学生包括‘60到90分钟’和‘90分钟以上’两组，占比为35% + 15% = 50%。\n因此有：\n50% × x = 200\n即：\n0.5x = 200\n解得：x = 400\n所以被调查学生总人数为400人。\n\n第二步：计算‘90分钟以上’组的人数。\n该组占比15%，人数为：\n15% × 400 = 0.15 × 400 = 60（人）\n\n第三步：进行分层抽样，总样本为50人。\n分层抽样要求各组抽取人数比例与原群体一致。\n因此‘90分钟以上’组应抽取人数为：\n(60 \/ 400) × 50 = (3\/20) × 50 = 7.5\n由于人数必须为整数，且分层抽样通常四舍五入处理，但此处需保持总人数为50，应合理分配。\n更精确做法是按比例分配：\n各组人数分别为：\n- 30分钟以下：10% × 400 = 40人 → 抽取 (40\/400)×50 = 5人\n- 30到60分钟：40% × 400 = 160人 → 抽取 (160\/400)×50 = 20人\n- 60到90分钟：35% × 400 = 140人 → 抽取 (140\/400)×50 = 17.5人\n- 90分钟以上：60人 → 抽取 (60\/400)×50 = 7.5人\n将小数部分调整：17.5和7.5分别取18和7，或17和8。为使总和为50，可取：\n5 + 20 + 17 + 8 = 50\n因此‘90分钟以上’组应抽取8人。\n\n第四步：估算全市超过90分钟的学生人数。\n样本中‘90分钟以上’占比为15%，以此估计全市：\n12000 × 15% = 12000 × 0.15 = 1800（人）\n\n答：分层抽样中‘90分钟以上’组应抽取8人；全市估计有1800名学生每天完成数学作业超过90分钟。","explanation":"本题综合考查数据的收集、整理与描述中的百分比计算、分层抽样原理及用样本估计总体的统计思想。解题关键在于先通过已知部分人数反推总样本量，再根据各层比例进行分层抽样人数分配，注意实际抽样中人数必须为整数，需合理调整。最后利用样本比例推断总体数量，体现统计推断的基本方法。题目情境贴近学生实际，数据真实合理，考查学生综合运用统计知识解决实际问题的能力，难度较高。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:50:16","updated_at":"2026-01-06 12:50:16","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1869,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为优化公交线路，对一条主干道的车流量进行了连续7天的观测，记录每天上午8:00至9:00的车辆通行数量（单位：辆），数据如下：312，298，305，310，307，299，304。交通部门计划根据这组数据预测未来某周的车流量，并设定一个合理的通行能力标准。已知该道路的设计通行能力为每天平均车流量的1.2倍，且要求实际车流量不超过设计通行能力的90%才算安全运行。若未来某周的车流量服从本次观测的平均水平，请通过计算判断该道路在未来是否满足安全运行要求。若不能满足，则至少需要将设计通行能力提升到当前观测平均车流量的多少倍（精确到0.01）才能满足安全要求？","answer":"解：\n\n第一步：计算7天观测数据的平均车流量。\n\n平均车流量 = (312 + 298 + 305 + 310 + 307 + 299 + 304) ÷ 7\n= (2135) ÷ 7\n= 305（辆）\n\n第二步：计算当前设计通行能力。\n\n设计通行能力 = 平均车流量 × 1.2 = 305 × 1.2 = 366（辆）\n\n第三步：计算安全运行上限（即设计通行能力的90%）。\n\n安全上限 = 366 × 90% = 366 × 0.9 = 329.4（辆）\n\n第四步：比较实际平均车流量与安全上限。\n\n实际平均车流量为305辆，小于329.4辆，因此当前道路满足安全运行要求。\n\n但题目要求判断“若不能满足”的情况下的处理方式，因此需进一步分析假设情形。\n\n然而根据计算，305 < 329.4，满足安全要求，故当前无需提升。\n\n但为完整解答问题，假设未来车流量上升至等于安全上限临界值，我们反向求解所需的设计通行能力倍数。\n\n设所需设计通行能力为平均车流量的k倍，则：\n\n安全上限 = k × 305 × 0.9 ≥ 305（因实际车流量为305）\n\n即：k × 305 × 0.9 ≥ 3...","explanation":"本题综合考查数据的收集与整理（计算平均数）、有理数运算、一元一次不等式的应用。解题关键在于理解‘安全运行’的定义：实际车流量 ≤ 设计通行能力 × 90%。先通过平均数反映典型车流量，再建立不等式模型求解最小安全倍数。难点在于将实际问题转化为数学不等式，并理解倍数关系的逻辑链条。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:41:09","updated_at":"2026-01-07 09:41:09","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":2181,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在一次数学测验中，一名学生记录了连续五天的气温变化情况（单位：摄氏度），以0℃为标准，高于0℃记为正，低于0℃记为负。这五天的气温分别为：+3，-2，+1，-4，+2。若将这五个有理数按从小到大的顺序排列，则排在第三位的数是（ ）。","answer":"B","explanation":"首先将五个有理数按从小到大的顺序排列：-4，-2，+1，+2，+3。其中-4最小，其次是-2，第三位是+1。因此，排在第三位的数是+1。本题考查有理数的大小比较及排序能力，符合七年级学生对有理数顺序的理解要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 14:21:04","updated_at":"2026-01-09 14:21:04","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"-2","is_correct":0},{"id":"B","content":"+1","is_correct":1},{"id":"C","content":"-4","is_correct":0},{"id":"D","content":"+2","is_correct":0}]},{"id":642,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次校园植物观察活动中，某学生记录了5种植物的高度（单位：厘米），分别为12、15、18、15、20。这组数据的中位数是____。","answer":"15","explanation":"首先将这组数据按从小到大的顺序排列：12、15、15、18、20。由于数据个数为5（奇数个），中位数就是位于中间位置的数，即第3个数。第3个数是15，因此这组数据的中位数是15。本题考查的是数据的收集、整理与描述中的中位数概念，属于七年级数学课程内容。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:08:56","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":2235,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在数轴上从原点出发，先向右移动5个单位长度，再向左移动8个单位长度，接着向右移动3个单位长度，最后向左移动4个单位长度。此时该学生所在位置的数与其相反数之和为___。","answer":"0","explanation":"首先计算该学生在数轴上的最终位置：从原点0开始，向右移动5个单位到达+5，再向左移动8个单位到达-3，接着向右移动3个单位到达0，最后向左移动4个单位到达-4。因此，最终位置表示的数是-4。一个数与其相反数之和恒为0，即-4 + 4 = 0。本题综合考查了数轴上的正负数移动、有理数加减运算以及相反数的性质，符合七年级正负数章节的拓展要求，难度较高。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:39:22","updated_at":"2026-01-09 14:39:22","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1080,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在一次环保活动中，某学生收集了可回收垃圾和不可回收垃圾共12千克，其中可回收垃圾比不可回收垃圾多4千克。设不可回收垃圾为x千克，则可列出一元一次方程为：______。","answer":"x + (x + 4) = 12","explanation":"设不可回收垃圾为x千克，根据题意，可回收垃圾比不可回收垃圾多4千克，因此可回收垃圾为(x + 4)千克。两者总重量为12千克，所以方程为x + (x + 4) = 12。该题考查一元一次方程的实际建模能力，属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:54:06","updated_at":"2026-01-06 08:54:06","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":321,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛，共收集了50份有效问卷。根据统计结果，喜欢‘垃圾分类’主题的有28人，喜欢‘节约用水’主题的有25人，同时喜欢两个主题的有12人。那么，只喜欢其中一个主题的学生共有多少人？","answer":"A","explanation":"本题考查数据的收集、整理与描述中的集合思想。设喜欢‘垃圾分类’的人数为A = 28，喜欢‘节约用水’的人数为B = 25，两者都喜欢的人数为A ∩ B = 12。只喜欢‘垃圾分类’的人数为28 - 12 = 16人，只喜欢‘节约用水’的人数为25 - 12 = 13人。因此，只喜欢其中一个主题的学生总数为16 + 13 = 29人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:37:48","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"29","is_correct":1},{"id":"B","content":"30","is_correct":0},{"id":"C","content":"31","is_correct":0},{"id":"D","content":"32","is_correct":0}]},{"id":1801,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在平面直角坐标系中，点A(2, 3)、B(6, 7)，线段AB的中点为M。若点P(x, y)满足PM = 5且x + y = 10，则点P的横坐标x的可能值为___。","answer":"4或8","explanation":"先求中点M(4,5)，设P(x,10−x)，利用距离公式列方程(x−4)²+(5−x)²=25，化简得x²−12x+32=0，解得x=4或8。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 16:15:51","updated_at":"2026-01-06 16:15:51","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":713,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了教室里5盏灯的功率，分别为40瓦、60瓦、40瓦、100瓦和40瓦。这组数据的中位数是____瓦。","answer":"40","explanation":"首先将这组数据按从小到大的顺序排列：40、40、40、60、100。共有5个数据，是奇数个，因此中位数是正中间的那个数，即第3个数，为40瓦。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:49:55","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]}]