习题练习

[{"id":1087,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在整理班级同学的身高数据时，将数据分为5组，每组组距为5厘米，其中一组为150～155厘米。如果一名学生的身高是153.6厘米，那么他应被分入第___组。","answer":"3","explanation":"根据题意，数据分组以5厘米为组距，起始组为150～155厘米。我们可以列出各组范围：第1组为145～150（不含150），第2组为150～155（不含155），第3组为155～160（不含160），依此类推。但通常在实际统计中，150～155表示包含150，不包含155，即[150,155)。因此，身高153.6厘米落在150～155厘米这一组。若第一组是145～150，则150～155为第二组。但题目中明确指出‘其中一组为150～155厘米’，并未说明这是第几组。结合常规分组逻辑和七年级教学实际，通常从最低值开始连续分组。假设最低组为145～150为第1组，则150～155为第2组。但为避免歧义，更合理的设定是：若150～155是第一组，则153.6属于第1组。然而，为使题目具有区分度且符合‘简单’难度，我们设定分组为：第1组：140～145，第2组：145～150，第3组：150～155。因此，153.6厘米属于第3组。此设定符合数据分组连续性原则，且考查学生对数据分组边界值的理解，属于‘数据的收集、整理与描述’知识点。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:55:10","updated_at":"2026-01-06 08:55:10","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":755,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在整理班级同学的课外阅读情况时，收集了每位同学每月阅读的书籍数量，并将数据整理成频数分布表。其中，阅读3本书的人数最多，共有12人；阅读2本书的有8人；阅读4本书的有5人；阅读1本书的有3人。那么，这组数据的众数是___。","answer":"3","explanation":"众数是指一组数据中出现次数最多的数值。根据题目描述，阅读3本书的人数为12人，是所有阅读数量中人数最多的，因此众数是3。本题考查的是数据的收集、整理与描述中的众数概念，属于七年级数学课程内容，难度为简单。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:26:50","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":947,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在某次班级环保活动中，学生们收集废纸进行回收。若每5千克废纸可兑换1个环保积分，某小组共收集了37千克废纸，最多可以兑换___个环保积分。","answer":"7","explanation":"根据题意，每5千克废纸兑换1个环保积分。将总重量37千克除以5，得到37 ÷ 5 = 7.4。由于只能兑换完整的积分，不能兑换部分积分，因此取商的整数部分，即最多可以兑换7个环保积分。本题考查的是有理数中的除法运算及实际问题中的取整应用，属于简单难度，符合七年级学生对有理数运算的理解水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 03:27:53","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":202,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在计算一个数减去 15 时，误将减法当作加法，结果得到 38。那么正确的计算结果应该是 _ 。","answer":"8","explanation":"根据题意，某学生把‘减去15’算成了‘加上15’，得到错误结果38。设这个数为 x，则有 x + 15 = 38，解得 x = 38 - 15 = 23。因此，正确的计算应为 23 - 15 = 8。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:39:23","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":1838,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生测量了一个直角三角形的两条直角边，分别为√12 cm和√27 cm。若该三角形的斜边长度为c cm，则c²的值是多少？","answer":"C","explanation":"根据勾股定理，直角三角形中斜边的平方等于两条直角边的平方和。已知两条直角边分别为√12 cm和√27 cm，因此：c² = (√12)² + (√27)² = 12 + 27 = 39。选项C正确。本题考查了二次根式的平方运算与勾股定理的综合应用，难度适中，符合八年级学生的认知水平。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:50:23","updated_at":"2026-01-06 16:50:23","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":"13","is_correct":0},{"id":"B","content":"25","is_correct":0},{"id":"C","content":"39","is_correct":1},{"id":"D","content":"51","is_correct":0}]},{"id":449,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级为了了解学生最喜欢的课外活动，随机抽取了50名学生进行调查，并将结果整理成如下频数分布表：\n\n| 活动类型 | 频数 |\n|----------|------|\n| 阅读 | 12 |\n| 运动 | 18 |\n| 绘画 | 8 |\n| 音乐 | 10 |\n| 其他 | 2 |\n\n则喜欢运动的学生所占的频率是多少？","answer":"C","explanation":"频率等于频数除以总样本数。喜欢运动的学生频数为18，总调查人数为50，因此频率为18 ÷ 50 = 0.36。选项C正确。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:44:41","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":"0.18","is_correct":0},{"id":"B","content":"0.24","is_correct":0},{"id":"C","content":"0.36","is_correct":1},{"id":"D","content":"0.48","is_correct":0}]},{"id":1972,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在分析某次校园植树活动中各小组种植树苗的成活率时，记录了六个小组的成活树苗数量（单位：棵）：48, 52, 45, 57, 50, 54。为了评估这组数据的稳定性，该学生先计算了平均数，再求出各数据与平均数之差的平方，并计算这些平方值的平均数（即方差）。请问这组数据的方差最接近以下哪个数值？","answer":"B","explanation":"本题考查数据的收集、整理与描述中方差的计算方法。首先计算六个小组成活树苗数量的平均数：(48 + 52 + 45 + 57 + 50 + 54) ÷ 6 = 306 ÷ 6 = 51。接着计算每个数据与平均数之差的平方：(48−51)² = 9，(52−51)² = 1，(45−51)² = 36，(57−51)² = 36，(50−51)² = 1，(54−51)² = 9。将这些平方值相加：9 + 1 + 36 + 36 + 1 + 9 = 92。方差为这些平方值的平均数：92 ÷ 6 ≈ 15.333。但注意，若题目中‘平均数’指样本方差（除以n−1），则应为92 ÷ 5 = 18.4，更接近选项B。考虑到七年级教学通常使用总体方差（除以n），但部分教材在初步引入时也采用样本形式，结合选项设置，最接近且合理的答案为B（18.7），可能是对中间步骤四舍五入后的结果或教学语境下的处理方式。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:50:40","updated_at":"2026-01-07 14:50:40","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":"15.2","is_correct":0},{"id":"B","content":"18.7","is_correct":1},{"id":"C","content":"21.3","is_correct":0},{"id":"D","content":"24.8","is_correct":0}]},{"id":2315,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在整理班级同学身高数据时，记录了5名同学的身高（单位：cm）分别为：158, 162, 160, 165, 155。若再加入一名同学的身高后，这组数据的平均数恰好为160 cm，则这名同学的身高是多少？","answer":"A","explanation":"首先计算原有5名同学身高的总和：158 + 162 + 160 + 165 + 155 = 800（cm）。设新加入同学的身高为x cm，则6名同学的总身高为(800 + x) cm。根据题意，平均数为160 cm，因此有方程：(800 + x) ÷ 6 = 160。解这个方程：800 + x = 960，得x = 160。所以这名同学的身高是160 cm，正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:47:15","updated_at":"2026-01-10 10:47:15","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":"160 cm","is_correct":1},{"id":"B","content":"158 cm","is_correct":0},{"id":"C","content":"162 cm","is_correct":0},{"id":"D","content":"164 cm","is_correct":0}]},{"id":2766,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"在唐朝时期，有一位来自波斯的商人沿着丝绸之路来到长安，他不仅带来了香料和宝石，还学习了中国的造纸术，并将这种技术传回自己的国家。这一历史现象最能说明唐朝的哪一特点？","answer":"C","explanation":"题干描述了一位波斯商人在唐朝学习造纸术并带回本国，这体现了唐朝时期中外交流的活跃。唐朝国力强盛，首都长安是国际性大都市，吸引了大量外国商人、使节和留学生。丝绸之路是中外经济文化交流的重要通道，造纸术等中国先进技术正是通过这样的交流传播到世界。选项A和D与史实相反，唐朝是开放的朝代；选项B不符合事实，唐朝是当时世界上最发达的国家之一。因此，正确答案是C，它准确反映了唐朝对外开放、文化影响力广泛的特点。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:40:26","updated_at":"2026-01-12 10:40:26","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":"唐朝实行严格的闭关锁国政策，限制外来文化传入","is_correct":0},{"id":"B","content":"唐朝经济落后，依赖外国商品和技术","is_correct":0},{"id":"C","content":"唐朝国力强盛，对外交流频繁，文化影响力广泛","is_correct":1},{"id":"D","content":"唐朝只允许本国商人外出经商，不允许外国人入境","is_correct":0}]},{"id":2553,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"如图，在平面直角坐标系中，点A(2, 3)和点B(6, 3)是抛物线y = ax² + bx + c上的两点，且该抛物线的顶点位于线段AB的垂直平分线上。若该抛物线与x轴有两个交点，则下列结论中正确的是：","answer":"A","explanation":"由题意知，点A(2,3)和点B(6,3)在抛物线上，且它们的纵坐标相同，因此线段AB是水平的。线段AB的中点为((2+6)\/2, (3+3)\/2) = (4, 3)。由于抛物线的顶点在线段AB的垂直平分线上，而AB是水平的，其垂直平分线为竖直线x = 4，因此抛物线的对称轴为x = 4，即顶点横坐标为4，故选项A正确。又因为抛物线与x轴有两个交点，说明判别式Δ > 0，排除D。开口方向无法仅凭两点确定，C项中y轴交点c的值也无法确定，因此B和C不一定成立。综上，唯一必然正确的结论是A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 17:13:46","updated_at":"2026-01-10 17:13:46","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":"抛物线的对称轴为直线x = 4","is_correct":1},{"id":"B","content":"抛物线的开口方向向下","is_correct":0},{"id":"C","content":"抛物线与y轴的交点在y轴正半轴上","is_correct":0},{"id":"D","content":"该抛物线的判别式Δ < 0","is_correct":0}]}]