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数学
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[{"id":1037,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某班级在一次数学测验中,男生有15人,女生有20人。老师随机抽取了部分学生进行成绩分析,共抽取了10人。如果采用分层抽样的方法,且按男女生人数比例抽取,那么应抽取男生____人。","answer":"30\/7","explanation":"本题考查数据的收集、整理与描述中的分层抽样方法。分层抽样要求每一层抽取的样本数与该层在总体中的比例相同。男生占总人数的比例为 15 \/ (15 + 20) = 15 \/ 35 = 3\/7。总抽取人数为10人,因此应抽取男生人数为 10 × (3\/7) = 30\/7。虽然实际抽样中人数应为整数,但本题仅考查比例计算,因此答案为分数形式 30\/7。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 06:07:51","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":6,"subject":"物理","grade":"初二","stage":"初中","type":"选择题","content":"下列现象中,属于光的反射现象的是?","answer":"C","explanation":"平面镜成像是光的反射现象,水中倒影也是光的反射现象。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-08-29 16:33:04","updated_at":"2025-08-29 16:33:04","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":"日食和月食","is_correct":0},{"id":"B","content":"小孔成像","is_correct":0},{"id":"C","content":"平面镜成像","is_correct":1},{"id":"D","content":"海市蜃楼","is_correct":0}]},{"id":522,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生在整理班级同学的课外阅读时间时,收集了以下数据(单位:小时):3, 5, 4, 6, 3, 7, 5, 4, 3, 6。他将这些数据按从小到大的顺序排列后,发现中位数是4.5。如果再加入一个数据4,那么新的数据组的中位数是多少?","answer":"A","explanation":"原数据有10个数:3, 3, 3, 4, 4, 5, 5, 6, 6, 7。按从小到大排列后,第5个数是4,第6个数是5,中位数是(4+5)÷2=4.5。加入一个4后,新数据组有11个数:3, 3, 3, 4, 4, 4, 5, 5, 6, 6, 7。此时数据个数为奇数,中位数是第6个数,即4。因此新的中位数是4。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:25:10","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":"4","is_correct":1},{"id":"B","content":"4.25","is_correct":0},{"id":"C","content":"4.5","is_correct":0},{"id":"D","content":"5","is_correct":0}]},{"id":2036,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园计划修建一个等腰三角形花坛,设计要求其底边长为6米,且从顶点到底边的垂直距离(即高)为4米。施工过程中,工人需要验证花坛两侧是否对称,于是测量了从顶点到底边两个端点的距离。若花坛符合设计要求,则这两个距离应相等,并且满足勾股定理。现测得其中一侧的长度为5米,则该花坛是否符合设计要求?若符合,其周长为多少?","answer":"A","explanation":"根据题意,等腰三角形底边为6米,高为4米,从顶点向底边作高,将底边平分为两段,每段3米。利用勾股定理计算腰长:腰² = 高² + (底边\/2)² = 4² + 3² = 16 + 9 = 25,因此腰长为√25 = 5米。题目中测得一侧为5米,与设计一致,说明符合要求。周长 = 5 + 5 + 6 = 16米。因此正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 10:42:49","updated_at":"2026-01-09 10:42:49","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":"符合,周长为16米","is_correct":1},{"id":"B","content":"符合,周长为18米","is_correct":0},{"id":"C","content":"不符合,因为高应为3米","is_correct":0},{"id":"D","content":"不符合,因为腰长应为√13米","is_correct":0}]},{"id":1484,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究一个关于温度变化与时间关系的实际问题时,收集了一周内每天的最高气温和最低气温数据(单位:℃),并将这些数据整理如下表。已知这一周每天的平均气温是当天最高气温与最低气温的平均值,且整周的平均气温为 18℃。此外,该学生发现,若将每天的最低气温增加 2℃,则新的整周平均气温将变为 19℃。若最高气温的总和比最低气温的总和多 42℃,求这一周内最低气温的总和是多少?","answer":"设这一周内每天的最高气温分别为 H₁, H₂, ..., H₇,最低气温分别为 L₁, L₂, ..., L₇。\n\n根据题意,每天的平均气温为 (Hᵢ + Lᵢ)\/2,整周的平均气温为 18℃,因此:\n\n(1\/7) × Σ[(Hᵢ + Lᵢ)\/2] = 18\n\n两边同乘以 7 得:\nΣ[(Hᵢ + Lᵢ)\/2] = 126\n\n两边同乘以 2 得:\nΣ(Hᵢ + Lᵢ) = 252 → ΣHᵢ + ΣLᵢ = 252 (方程①)\n\n又已知:若每天最低气温增加 2℃,则新的最低气温总和为 Σ(Lᵢ + 2) = ΣLᵢ + 14\n\n此时新的每天平均气温为 (Hᵢ + Lᵢ + 2)\/2,整周平均气温为 19℃,故:\n\n(1\/7) × Σ[(Hᵢ + Lᵢ + 2)\/2] = 19\n\n两边同乘以 7 得:\nΣ[(Hᵢ + Lᵢ + 2)\/2] = 133\n\n两边同乘以 2 得:\nΣ(Hᵢ + Lᵢ + 2) = 266\n\n即:ΣHᵢ + ΣLᵢ + 14 = 266 (因为共7天,每天加2,总和加14)\n\n代入方程①:252 + 14 = 266,验证成立,说明信息一致。\n\n再根据题意:最高气温的总和比最低气温的总和多 42℃,即:\n\nΣHᵢ = ΣLᵢ + 42 (方程②)\n\n将方程②代入方程①:\n(ΣLᵢ + 42) + ΣLᵢ = 252\n2ΣLᵢ + 42 = 252\n2ΣLᵢ = 210\nΣLᵢ = 105\n\n答:这一周内最低气温的总和是 105℃。","explanation":"本题综合考查了数据的收集、整理与描述、有理数的运算、整式的加减以及一元一次方程的建立与求解。解题关键在于将文字信息转化为代数表达式:首先利用平均气温的定义建立总和关系;其次通过‘最低气温增加2℃’这一变化条件,推导出新的总和表达式,并验证一致性;最后结合‘最高气温总和比最低气温总和多42℃’这一条件,设立方程求解。整个过程需要学生具备较强的信息转化能力和代数建模能力,属于困难难度的综合应用题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:57:35","updated_at":"2026-01-06 11:57:35","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":626,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"x + (x + 3) + 2x + x = 45","answer":"待完善","explanation":"解析待完善","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:52:29","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":195,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明买了3支铅笔和2本笔记本,共花费18元。已知每本笔记本比每支铅笔贵3元,设每支铅笔的价格为x元,则下列方程正确的是( )。","answer":"A","explanation":"设每支铅笔的价格为x元,根据题意,每本笔记本比每支铅笔贵3元,因此每本笔记本的价格为(x + 3)元。小明买了3支铅笔,总价为3x元;买了2本笔记本,总价为2(x + 3)元。两者相加等于总花费18元,因此方程为:3x + 2(x + 3) = 18。选项A正确。其他选项中,B错误地将笔记本价格设为比铅笔便宜,C和D则颠倒了铅笔和笔记本的数量与单价对应关系,均不符合题意。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:04:01","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":"3x + 2(x + 3) = 18","is_correct":1},{"id":"B","content":"3x + 2(x - 3) = 18","is_correct":0},{"id":"C","content":"3(x + 3) + 2x = 18","is_correct":0},{"id":"D","content":"3(x - 3) + 2x = 18","is_correct":0}]},{"id":1957,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生参加学校组织的‘健康生活’主题调查,记录了连续7天每天步行的步数(单位:千步),数据如下:6.2, 5.8, 7.1, 6.5, 6.9, 5.5, 7.3。若该学生希望估算自己一个月(按30天计算)的总步行步数,并假设每日步数服从这组数据的平均水平,则估算结果最接近以下哪个数值?","answer":"B","explanation":"本题考查数据的收集、整理与描述中利用样本平均数估计总体的应用。首先计算7天步行步数的平均数:(6.2 + 5.8 + 7.1 + 6.5 + 6.9 + 5.5 + 7.3) ÷ 7 = 45.3 ÷ 7 ≈ 6.471(千步\/天)。然后估算30天的总步数:6.471 × 30 ≈ 194.13(千步),最接近195千步。因此选项B正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:47:02","updated_at":"2026-01-07 14:47:02","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":"180千步","is_correct":0},{"id":"B","content":"195千步","is_correct":1},{"id":"C","content":"200千步","is_correct":0},{"id":"D","content":"210千步","is_correct":0}]},{"id":590,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级进行了一次数学测验,成绩分布如下表所示。已知成绩在70分到89分之间的学生人数占总人数的40%,而成绩在90分及以上的学生有12人,占总人数的20%。那么,成绩低于70分的学生有多少人?","answer":"B","explanation":"首先根据题意,90分及以上的学生占20%,共12人,因此总人数为 12 ÷ 20% = 12 ÷ 0.2 = 60人。成绩在70到89分之间的学生占40%,即 60 × 40% = 24人。那么低于70分的学生所占比例为 100% - 20% - 40% = 40%,对应人数为 60 × 40% = 24人。因此,成绩低于70分的学生有24人。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 20:28:30","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":"18人","is_correct":0},{"id":"B","content":"24人","is_correct":1},{"id":"C","content":"30人","is_correct":0},{"id":"D","content":"36人","is_correct":0}]},{"id":494,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某班级进行了一次数学测验,成绩分布如下表所示。根据表格信息,成绩在80分及以上的人数占总人数的百分比最接近以下哪个选项?\n\n| 分数段(分) | 人数 |\n|--------------|------|\n| 60以下 | 5 |\n| 60—69 | 8 |\n| 70—79 | 12 |\n| 80—89 | 15 |\n| 90—100 | 10 |","answer":"C","explanation":"首先计算总人数:5 + 8 + 12 + 15 + 10 = 50(人)。\n成绩在80分及以上的人数包括80—89和90—100两个分数段,共15 + 10 = 25(人)。\n所求百分比为:25 ÷ 50 × 100% = 50%。\n因此,正确答案是C选项。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:06:22","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":"25%","is_correct":0},{"id":"B","content":"40%","is_correct":0},{"id":"C","content":"50%","is_correct":1},{"id":"D","content":"60%","is_correct":0}]}]