某学生记录了连续5天每天完成数学作业所用的时间(单位:分钟):35,40,30,45,35。这5天完成作业所用时间的众数和中位数分别是多少?
💡 提示:点击下方 "查看答案" 查看解析,或 "提交答案" 后自动显示结果
首先,根据平行四边形的性质,对角线互相平分。因此,AC的中点坐标应等于BD的中点坐标。计算AC的中点:A(1,2)、C(6,3),中点为((1+6)/2, (2+3)/2) = (3.5, 2.5)。设D点坐标为(x, y),则BD的中点为((4+x)/2, (5+y)/2)。令两中点相等,得方程组:(4+x)/2 = 3.5 → x = 3;(5+y)/2 = 2.5 → y = 0。故D点坐标为(3, 0)。接着验证是否关于直线y = x对称:若整个图形关于y = x对称,则每个点与其对称点都应在图形上。A(1,2)关于y=x的对称点为(2,1),应出现在图形中;B(4,5)对称点为(5,4);C(6,3)对称点为(3,6);D(3,0)对称点为(0,3)。虽然这些对称点不一定都是原顶点,但题目只要求‘可能’的D点,且结合平行四边形性质已确定唯一D点为(3,0),故选项A正确。
🏆
练习完成!
恭喜您完成了本次练习,继续加油提升!
💡 学习建议:您在一元一次方程的应用方面掌握良好,但仍有提升空间。建议重点复习方程求解步骤和实际应用问题。
[{"id":572,"content":"35","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"待完善","explanation":"解析待完善","options":[]},{"id":434,"content":"12人","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"待完善","explanation":"解析待完善","options":[]},{"id":380,"content":"在平面直角坐标系中,点A的坐标为(3, -2),点B的坐标为(-1, 4)。某学生计算线段AB的长度时,使用了距离公式。请问线段AB的长度是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"根据平面直角坐标系中两点间距离公式:若点A(x₁, y₁),点B(x₂, y₂),则AB = √[(x₂ - x₁)² + (y₂ - y₁)²]。将点A(3, -2)和点B(-1, 4)代入公式:AB = √[(-1 - 3)² + (4 - (-2))²] = √[(-4)² + (6)²] = √[16 + 36] = √52。将√52化简:√52 = √(4 × 13) = 2√13。因此正确答案是A。选项C虽然数值正确但未化简,不符合最简形式要求。","options":[{"id":"A","content":"2√13"},{"id":"B","content":"10"},{"id":"C","content":"√52"},{"id":"D","content":"6√2"}]},{"id":1801,"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___。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"困难","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。","options":[]},{"id":607,"content":"在一次班级环保活动中,某学生收集了若干个塑料瓶和废纸。已知每个塑料瓶可回收获得0.5元,每公斤废纸可回收获得1.2元。该学生共收集了8个塑料瓶和3公斤废纸,他一共可以获得多少元?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"首先计算塑料瓶的回收金额:8个 × 0.5元\/个 = 4元。然后计算废纸的回收金额:3公斤 × 1.2元\/公斤 = 3.6元。将两部分相加:4元 + 3.6元 = 7.6元。因此,该学生一共可以获得7.6元,正确答案是A。本题考查有理数的乘法与加法在实际问题中的应用,属于简单难度的实际问题建模。","options":[{"id":"A","content":"7.6元"},{"id":"B","content":"6.8元"},{"id":"C","content":"8.2元"},{"id":"D","content":"5.4元"}]},{"id":2762,"content":"考古学家在河南偃师的二里头遗址中发现了大型宫殿基址、青铜器和陶器,这些发现为研究中国早期国家形态提供了重要依据。根据所学知识,二里头遗址最有可能属于哪个历史时期?","type":"选择题","subject":"历史","grade":"七年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"二里头遗址位于河南省偃师市,是中国早期国家形成阶段的重要考古发现。遗址中出土了宫殿建筑基址、青铜礼器和陶器等,表明当时已具备较高的社会组织能力和手工业水平。根据历史学界的主流观点,二里头文化被广泛认为与文献记载中的夏朝相对应,是探索夏文明的关键实证材料。虽然尚未发现确切的文字证据,但其年代、地理位置和文化特征均与夏朝相符,因此最可能属于夏朝时期。选项A史前时代指尚未建立国家、无文字记载的时期,而二里头已出现宫殿和青铜器,说明已进入文明阶段;选项C商朝和D西周虽也有青铜器和宫殿,但其典型遗址如郑州商城、安阳殷墟和周原等与二里头在文化面貌和年代上有所不同。因此,正确答案是B。","options":[{"id":"A","content":"史前时代"},{"id":"B","content":"夏朝"},{"id":"C","content":"商朝"},{"id":"D","content":"西周"}]},{"id":935,"content":"在一次班级视力情况调查中,共收集了40名学生的视力数据。其中,视力在4.8及以上的学生有25人,视力低于4.8的有___人。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"15","explanation":"题目考查的是数据的收集与整理。总人数为40人,已知视力在4.8及以上的有25人,要求视力低于4.8的人数,只需用总人数减去已知部分:40 - 25 = 15。因此,视力低于4.8的学生有15人。","options":[]},{"id":598,"content":"某次数学测验中,某班级共有40名学生参加,其中男生人数是女生人数的1.5倍。设女生人数为x,则根据题意可以列出方程:","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"题目中设女生人数为x,男生人数是女生的1.5倍,因此男生人数为1.5x。全班总人数为男生和女生人数之和,即 x + 1.5x = 40。这个方程正确表达了总人数为40人的条件。选项A错误地将倍数当作具体人数相加;选项C表示的是男女生人数差,不符合题意;选项D将女生人数与倍数关系倒置,也不正确。因此正确答案是B。","options":[{"id":"A","content":"x + 1.5 = 40"},{"id":"B","content":"x + 1.5x = 40"},{"id":"C","content":"1.5x - x = 40"},{"id":"D","content":"x ÷ 1.5 = 40"}]},{"id":1813,"content":"某学生在测量一个直角三角形的两条直角边时,得到长度分别为3和4,他想知道斜边的长度。根据勾股定理,斜边的长度应为多少?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"根据勾股定理,直角三角形的两条直角边的平方和等于斜边的平方。设斜边为c,则有:3² + 4² = c²,即9 + 16 = 25,所以c² = 25,因此c = 5。故正确答案为A。","options":[{"id":"A","content":"5"},{"id":"B","content":"6"},{"id":"C","content":"7"},{"id":"D","content":"8"}]},{"id":447,"content":"某学生调查了班级同学每天用于课外阅读的时间(单位:分钟),并将数据整理如下表:\n\n| 阅读时间(分钟) | 人数 |\n|------------------|------|\n| 0~20 | 5 |\n| 20~40 | 8 |\n| 40~60 | 12 |\n| 60~80 | 10 |\n| 80~100 | 5 |\n\n则该班级学生每天课外阅读时间的众数所在的区间是?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"C","explanation":"众数是指一组数据中出现次数最多的数据。在本题中,虽然无法知道每个具体数值,但可以确定哪个区间的人数最多,即频数最高的区间就是众数所在的区间。从表中可以看出,阅读时间在40~60分钟的人数最多,为12人,因此众数所在的区间是40~60分钟。","options":[{"id":"A","content":"0~20分钟"},{"id":"B","content":"20~40分钟"},{"id":"C","content":"40~60分钟"},{"id":"D","content":"60~80分钟"}]}]