某学生在计算两个有理数的乘积时,先确定了结果的符号,再计算绝对值的乘积。已知这两个数分别为 -3/4 和 2/5,该学生正确地完成了符号判断和绝对值计算,但最终写出的结果却比正确答案多了一个负号。请问该学生可能犯的错误是什么?
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要计算这组数据的平均分,需要将所有分数相加,然后除以人数。计算过程如下:78 + 85 + 90 + 82 + 88 + 87 = 510。总人数为6人,因此平均分为510 ÷ 6 = 85(分)。所以正确答案是B选项。本题考查的是数据的收集、整理与描述中的平均数计算,属于七年级数学课程内容,难度简单,符合学生认知水平。
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[{"id":2004,"content":"某学生测量了一块直角三角形铁片的边长,其中两条直角边分别为5 cm和12 cm,他需要计算斜边的长度以确定是否适合放入一个边长为13 cm的正方形槽中。请问这块铁片的斜边长度是多少?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"根据勾股定理,在直角三角形中,斜边的平方等于两条直角边的平方和。设斜边为c,则有:c² = 5² + 12² = 25 + 144 = 169。因此,c = √169 = 13(cm)。所以斜边长为13 cm,正好可以放入边长为13 cm的正方形槽中。","options":[{"id":"A","content":"10 cm"},{"id":"B","content":"13 cm"},{"id":"C","content":"15 cm"},{"id":"D","content":"17 cm"}]},{"id":272,"content":"在一次班级调查中,某学生记录了10名同学每天用于课外阅读的时间(单位:分钟),数据如下:25,30,35,40,40,45,50,55,60,65。这组数据的中位数和众数分别是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"首先将数据按从小到大顺序排列(已排好):25,30,35,40,40,45,50,55,60,65。共有10个数据,为偶数个,因此中位数是第5个和第6个数据的平均数,即(40 + 45) ÷ 2 = 85 ÷ 2 = 42.5。众数是出现次数最多的数,其中40出现了两次,其余数均只出现一次,因此众数是40。所以正确答案是A。","options":[{"id":"A","content":"中位数是42.5,众数是40"},{"id":"B","content":"中位数是40,众数是42.5"},{"id":"C","content":"中位数是45,众数是40"},{"id":"D","content":"中位数是40,众数是45"}]},{"id":1984,"content":"某学生在纸上画了一个边长为10 cm的正方形ABCD,并以顶点A为圆心、AB为半径画了一个四分之一圆。若将该四分之一圆绕点A顺时针旋转90°,则旋转过程中该四分之一圆所扫过的区域面积是多少?(π取3.14)","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"简单","answer":"C","explanation":"本题考查旋转与圆的综合应用,重点在于理解扇形旋转过程中扫过区域的构成。初始四分之一圆的半径为10 cm,圆心角为90°。当它绕圆心A顺时针旋转90°时,其轨迹形成一个半径为10 cm、圆心角为180°的扇形(即半圆)。这是因为旋转过程中,原四分之一圆的每条半径都扫过一个90°的角,整体叠加后形成一个半圆形区域。该半圆的面积为(1\/2) × π × r² = (1\/2) × 3.14 × 10² = 157 cm²。因此,扫过的区域面积为157 cm²。","options":[{"id":"A","content":"78.5 cm²"},{"id":"B","content":"100 cm²"},{"id":"C","content":"157 cm²"},{"id":"D","content":"235.5 cm²"}]},{"id":429,"content":"某学生记录了连续5天的气温(单位:℃),分别为:-2,3,0,-1,4。这5天气温的平均值是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"A","explanation":"求一组数据的平均值,需要将这组数据相加,然后除以数据的个数。本题中,气温数据为:-2,3,0,-1,4。首先计算总和:-2 + 3 + 0 + (-1) + 4 = 4。共有5个数据,因此平均值为 4 ÷ 5 = 0.8。所以正确答案是A。本题考查的是数据的收集、整理与描述中的平均数计算,属于七年级数学中的基础内容,难度为简单。","options":[{"id":"A","content":"0.8"},{"id":"B","content":"1.0"},{"id":"C","content":"1.2"},{"id":"D","content":"1.4"}]},{"id":2257,"content":"在数轴上,点A表示的数是-3,点B表示的数是5。若点C是线段AB的中点,则点C表示的数是___。","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"点C是线段AB的中点,其表示的数为点A和点B所表示数的平均数。计算过程为:(-3 + 5) ÷ 2 = 2 ÷ 2 = 1。因此,点C表示的数是1,对应选项A。","options":[{"id":"A","content":"1"},{"id":"B","content":"2"},{"id":"C","content":"-1"},{"id":"D","content":"4"}]},{"id":2441,"content":"某学生测量了一块直角三角形草地的两条直角边,分别为√12米和√27米。他计划在斜边上每隔1米种一棵树,包括两个端点。若每棵树占地忽略不计,则最多可以种多少棵树?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"B","explanation":"首先,利用勾股定理计算斜边长度。已知两条直角边分别为√12米和√27米。将根式化简:√12 = 2√3,√27 = 3√3。根据勾股定理,斜边c满足:c² = (2√3)² + (3√3)² = 4×3 + 9×3 = 12 + 27 = 39,因此c = √39米。接下来,计算在长度为√39米的线段上,每隔1米种一棵树(包括两个端点)最多可种多少棵。由于√36 = 6,√49 = 7,所以6 < √39 < 7,即斜边长度约为6.24米。从起点开始,每隔1米种一棵树,位置为0米、1米、2米、…、6米,共7个点(因为6 ≤ √39 < 7,第7棵树在6米处仍在线段上)。因此最多可种7棵树。","options":[{"id":"A","content":"6棵"},{"id":"B","content":"7棵"},{"id":"C","content":"8棵"},{"id":"D","content":"9棵"}]},{"id":346,"content":"某学生在整理班级同学的课外阅读时间数据时,记录了10名同学每周阅读的小时数分别为:3,5,4,6,3,7,5,4,5,6。这组数据的众数是多少?","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。将数据从小到大排列为:3,3,4,4,5,5,5,6,6,7。其中,3出现2次,4出现2次,5出现3次,6出现2次,7出现1次。因此,出现次数最多的数是5,所以这组数据的众数是5。","options":[{"id":"A","content":"3"},{"id":"B","content":"4"},{"id":"C","content":"5"},{"id":"D","content":"6"}]},{"id":611,"content":"在一次班级数学测验中,某学生记录了5名同学的数学成绩(单位:分)如下:82,76,90,88,74。如果老师要求将这组数据按从小到大的顺序排列,并找出中位数,那么中位数是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"B","explanation":"首先将5个成绩按从小到大的顺序排列:74,76,82,88,90。由于数据个数为5(奇数个),中位数就是位于正中间的那个数,即第3个数。因此,中位数是82。本题考查的是数据的整理与描述中的中位数概念,属于简单难度,符合七年级数学课程标准要求。","options":[{"id":"A","content":"76"},{"id":"B","content":"82"},{"id":"C","content":"88"},{"id":"D","content":"90"}]},{"id":2451,"content":"某学生在研究一个轴对称图形时,发现其对称轴是直线x=3,且图形上一点A的坐标为(5, 4)。若点A关于该对称轴的对称点为B,则点B的横坐标为___。","type":"填空题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"1","explanation":"对称轴为x=3,点A(5,4)到对称轴的距离为5−3=2,对称点B在对称轴另一侧相同距离处,故横坐标为3−2=1。","options":[]},{"id":1080,"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:______。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","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。","options":[]}]