1
/ 10
中等
第 1 题
已完成 0 题
练习设置
题目内容
数学
初中
选择题
请选择一道题目开始练习
我的笔记
答案解析
请选择一道题目查看解析
练习完成!
恭喜您完成了本次练习
0
总题数
0
正确
0
错误
答题卡
[{"id":2518,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生设计了一个圆形花坛,其边缘由一段抛物线形状的装饰带和一段圆弧拼接而成。已知抛物线的顶点在原点,且经过点 (2, -4),而圆弧所在的圆以原点为圆心,半径为 2。若装饰带与圆弧在点 (2, -4) 处平滑连接,则该抛物线的解析式为( )。","answer":"A","explanation":"题目中说明抛物线的顶点在原点,因此可设其解析式为 y = ax²。又已知该抛物线经过点 (2, -4),代入得:-4 = a × 2² → -4 = 4a → a = -1。因此抛物线的解析式为 y = -x²。虽然题目提到与圆弧连接,但问题仅要求求出抛物线解析式,且点 (2, -4) 确实在 y = -x² 上,而半径为 2 的圆上点 (2, -4) 并不在圆上(因为 2² + (-4)² = 20 ≠ 4),这说明‘平滑连接’在此题中仅为情境设定,不影响抛物线解析式的求解。关键信息是顶点在原点且过 (2, -4),由此唯一确定解析式为 y = -x²。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:49:55","updated_at":"2026-01-10 15:49:55","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":"y = -x²","is_correct":1},{"id":"B","content":"y = -2x²","is_correct":0},{"id":"C","content":"y = -x² + 4","is_correct":0},{"id":"D","content":"y = -2x² + 4","is_correct":0}]},{"id":775,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中,某学生收集了若干千克废纸。如果他将废纸重量的小数点向右移动一位,所得的新数比原数大27.9千克。那么他实际收集的废纸重量是___千克。","answer":"3.1","explanation":"设该学生收集的废纸重量为x千克。根据题意,将小数点向右移动一位相当于将原数乘以10,即得到10x。题目说明10x比x大27.9,因此可以列出方程:10x - x = 27.9,即9x = 27.9。解这个一元一次方程,得x = 27.9 ÷ 9 = 3.1。所以,他实际收集的废纸重量是3.1千克。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:52:14","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":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":556,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的身高数据时,制作了如下频数分布表:\n\n| 身高区间(cm) | 频数(人) |\n|----------------|------------|\n| 150~155 | 4 |\n| 155~160 | 6 |\n| 160~165 | 10 |\n| 165~170 | 8 |\n| 170~175 | 2 |\n\n若该学生想用这组数据绘制条形统计图,并要求每个条形的高度与对应区间的频数成正比,且已知160~165cm区间对应的条形高度为5厘米,那么155~160cm区间对应的条形高度应为多少厘米?","answer":"B","explanation":"题目考查的是数据的收集、整理与描述中的条形统计图绘制原理。条形的高度与频数成正比,因此可以通过比例关系求解。\n\n已知:160~165cm区间频数为10人,对应条形高度为5厘米。\n求:155~160cm区间频数为6人,对应条形高度为多少?\n\n设所求高度为x厘米,根据正比关系列比例式:\n10 : 5 = 6 : x\n即 10 \/ 5 = 6 \/ x\n2 = 6 \/ x\n解得 x = 6 \/ 2 = 3\n\n因此,155~160cm区间对应的条形高度应为3厘米。\n\n该题结合了频数分布表与统计图绘制,考查比例思想和实际应用能力,符合七年级‘数据的收集、整理与描述’知识点要求,难度适中,情境真实。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:17:45","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":"2厘米","is_correct":0},{"id":"B","content":"3厘米","is_correct":1},{"id":"C","content":"4厘米","is_correct":0},{"id":"D","content":"6厘米","is_correct":0}]},{"id":334,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"90°","answer":"答案待完善","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 15:39:49","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":640,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保活动,收集废纸和塑料瓶。已知每千克废纸可兑换0.8元,每千克塑料瓶可兑换1.2元。一名学生共收集了15千克废品,兑换后获得16元。若设该学生收集的废纸为x千克,则根据题意可列出一元一次方程为:","answer":"A","explanation":"设收集的废纸为x千克,则塑料瓶为(15 - x)千克。废纸每千克兑换0.8元,总价值为0.8x元;塑料瓶每千克兑换1.2元,总价值为1.2(15 - x)元。两者之和等于16元,因此方程为0.8x + 1.2(15 - x) = 16。选项A正确。选项B错误地将两种废品都设为x千克;选项C颠倒了废纸和塑料瓶的对应关系;选项D使用了减法,不符合实际兑换逻辑。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:07:00","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":"0.8x + 1.2(15 - x) = 16","is_correct":1},{"id":"B","content":"0.8x + 1.2x = 16","is_correct":0},{"id":"C","content":"0.8(15 - x) + 1.2x = 16","is_correct":0},{"id":"D","content":"0.8x - 1.2(15 - x) = 16","is_correct":0}]},{"id":492,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读时间数据时,记录了5名同学每周的阅读时间(单位:小时)分别为:3,5,4,6,7。如果他想用这组数据估计全班同学的平均阅读时间,并发现这组数据的平均数恰好等于中位数,那么他应该再添加一个数据,使得新的6个数据仍满足平均数等于中位数。这个添加的数据可能是多少?","answer":"C","explanation":"首先计算原始5个数据:3,5,4,6,7。按从小到大排列为:3,4,5,6,7。中位数为中间的数,即5。平均数为(3+4+5+6+7)÷5 = 25÷5 = 5,此时平均数等于中位数。现在要添加一个数据x,使新的6个数据的平均数仍等于中位数。6个数据的中位数是中间两个数的平均数。若添加x后,数据仍有序,且中位数仍为5,则中间两个数应为4和6,或5和5。若添加x=5,新数据为:3,4,5,5,6,7,中位数为(5+5)÷2=5,平均数为(3+4+5+5+6+7)÷6=30÷6=5,满足条件。其他选项如x=4,数据为3,4,4,5,6,7,中位数为(4+5)÷2=4.5,平均数为29÷6≈4.83,不等;x=6时,中位数为(5+6)÷2=5.5,平均数为31÷6≈5.17,也不等;x=3时,中位数为(4+5)÷2=4.5,平均数为28÷6≈4.67,不等。因此只有x=5满足条件。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:04:38","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":"3","is_correct":0},{"id":"B","content":"4","is_correct":0},{"id":"C","content":"5","is_correct":1},{"id":"D","content":"6","is_correct":0}]},{"id":2251,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在数轴上,点P表示的数是-3,点Q与点P之间的距离是7个单位长度,且点Q在原点的右侧。那么点Q表示的数是___。","answer":"B","explanation":"点P表示-3,点Q与点P相距7个单位长度。由于点Q在原点右侧,说明点Q表示的数是正数。从-3向右移动7个单位,计算为:-3 + 7 = 4。因此点Q表示的数是4。选项B正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 16:03:06","updated_at":"2026-01-09 16:03:06","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":"A","content":"-10","is_correct":0},{"id":"B","content":"4","is_correct":1},{"id":"C","content":"10","is_correct":0},{"id":"D","content":"-4","is_correct":0}]},{"id":1927,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级组织一次环保活动,收集可回收垃圾。第一周收集了x千克废纸,第二周收集的比第一周的2倍少3千克。已知两周共收集了17千克废纸,则第一周收集了多少千克?","answer":"C","explanation":"设第一周收集废纸x千克,则第二周收集了(2x - 3)千克。根据题意,两周共收集17千克,可列方程:x + (2x - 3) = 17。化简得3x - 3 = 17,移项得3x = 20,解得x = 7。因此第一周收集了7千克废纸。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:18:07","updated_at":"2026-01-07 13:18:07","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":"5","is_correct":0},{"id":"B","content":"6","is_correct":0},{"id":"C","content":"7","is_correct":1},{"id":"D","content":"8","is_correct":0}]},{"id":1784,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在平面直角坐标系中绘制了一个由四个点组成的四边形,其顶点坐标分别为 A(1, 2)、B(4, 6)、C(8, 3)、D(5, -1)。该学生通过测量和计算发现,这个四边形的对边长度分别相等,且对角线互相垂直。根据这些特征,该四边形最可能是以下哪种图形?","answer":"B","explanation":"首先,根据坐标计算四边形的边长:AB = √[(4-1)² + (6-2)²] = √(9+16) = 5;BC = √[(8-4)² + (3-6)²] = √(16+9) = 5;CD = √[(5-8)² + (-1-3)²] = √(9+16) = 5;DA = √[(1-5)² + (2+1)²] = √(16+9) = 5。四条边长度均为5,说明是菱形或正方形。再计算对角线AC和BD的斜率:AC斜率为(3-2)\/(8-1)=1\/7,BD斜率为(-1-6)\/(5-4)=-7。两斜率乘积为(1\/7)×(-7) = -1,说明对角线互相垂直。由于四条边相等且对角线垂直,符合菱形的判定条件。进一步验证是否为正方形:若为正方形,对角线应相等。计算AC = √[(8-1)²+(3-2)²]=√(49+1)=√50,BD = √[(5-4)²+(-1-6)²]=√(1+49)=√50,对角线相等。但还需验证角是否为直角。取向量AB=(3,4),向量AD=(-4,-3),点积为3×(-4)+4×(-3)=-12-12=-24≠0,说明角A不是直角,因此不是正方形。综上,该四边形是菱形。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 15:56:11","updated_at":"2026-01-06 15:56:11","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":"矩形","is_correct":0},{"id":"B","content":"菱形","is_correct":1},{"id":"C","content":"正方形","is_correct":0},{"id":"D","content":"等腰梯形","is_correct":0}]}]