某学生在整理班级同学的身高数据时,发现最高身高为172厘米,最矮身高为148厘米,则这组数据的极差是___厘米。
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根据题意,设第二组收集的垃圾重量为x千克,则第一组为(x + 3.5)千克。两组共收集12.7千克,因此可列方程:x + (x + 3.5) = 12.7。化简得:2x + 3.5 = 12.7。两边同时减去3.5,得2x = 9.2。再两边同时除以2,得x = 4.6。所以第二组收集的垃圾重量为4.6千克。
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[{"id":9,"content":"电解水的化学方程式为______,反应类型为______反应。","type":"填空题","subject":"化学","grade":"初三","stage":"初中","difficulty":"中等","answer":"2H₂O → 2H₂↑ + O₂↑, 分解","explanation":"电解水生成氢气和氧气,是一种分解反应。","options":[]},{"id":131,"content":"一个长方形的长比宽多5厘米,若其周长为30厘米,则这个长方形的宽是______厘米。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"5","explanation":"设长方形的宽为x厘米,则长为(x + 5)厘米。根据长方形周长公式:周长 = 2 × (长 + 宽),代入得:2 × (x + x + 5) = 30,即2 × (2x + 5) = 30,化简得4x + 10 = 30,解得4x = 20,x = 5。因此,宽为5厘米。本题结合代数设未知数与一元一次方程求解,符合初一学生对方程和几何基础的学习要求。","options":[]},{"id":2156,"content":"某学生在数轴上标记了三个有理数:-1.5、0.8 和 -2\/3。若将这三个数按从小到大的顺序排列,正确的结果是?","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"D","explanation":"首先比较负数:-1.5 比 -2\/3(约等于 -0.67)更小,因为它在数轴上更靠左;0.8 是正数,最大。因此从小到大的顺序是 -1.5 < -2\/3 < 0.8。选项 D 正确。","options":[{"id":"A","content":"-1.5, -2\/3, 0.8"},{"id":"B","content":"-2\/3, -1.5, 0.8"},{"id":"C","content":"0.8, -2\/3, -1.5"},{"id":"D","content":"-1.5, -2\/3, 0.8"}]},{"id":259,"content":"一个多边形的内角和是1260°,则这个多边形从一个顶点出发可以画出___条对角线。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"中等","answer":"6","explanation":"首先根据多边形内角和公式:(n - 2) × 180° = 内角和。设边数为n,则 (n - 2) × 180 = 1260,解得 n - 2 = 7,n = 9。这是一个九边形。从一个顶点出发可以画出的对角线条数为 n - 3,即 9 - 3 = 6 条。因为不能连接自己和相邻的两个顶点,所以减去3。","options":[]},{"id":237,"content":"某学生在计算一个数减去 35 时,误将减法当作加法计算,结果得到 82。那么正确的计算结果应该是____。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"12","explanation":"该学生误将减法当作加法,即把原数加上 35 得到 82。设原数为 x,则有 x + 35 = 82,解得 x = 82 - 35 = 47。正确的计算应是 47 减去 35,即 47 - 35 = 12。因此正确答案是 12。","options":[]},{"id":2508,"content":"某学生在一张纸上画了一个半径为3 cm的圆,然后以该圆的圆心为中心,将整个图形绕点O逆时针旋转60°。旋转后,原圆上的一点P移动到点P'。若连接点P和点P',则线段PP'的长度最接近以下哪个值?(参考数据:sin30°=0.5,cos30°≈0.87)","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"本题考查旋转与圆的性质。由于圆以圆心O为中心旋转60°,点P在圆上,OP = OP' = 半径 = 3 cm,且∠POP' = 60°。因此,△POP'是等边三角形(两边相等且夹角为60°),所以PP' = OP = 3 cm。故正确答案为A。","options":[{"id":"A","content":"3 cm"},{"id":"B","content":"3√3 cm"},{"id":"C","content":"6 cm"},{"id":"D","content":"3√2 cm"}]},{"id":2165,"content":"某学生在数轴上标记了三个有理数点A、B、C,其中点A表示的数是-3\/4,点B位于点A右侧且与点A的距离为5\/6,点C位于点B左侧且与点B的距离为1\/3。若点C表示的数为x,则x的值可能是多少?","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"困难","answer":"D","explanation":"首先,点A表示-3\/4,点B在点A右侧5\/6单位,因此点B表示的数为:-3\/4 + 5\/6 = (-9\/12 + 10\/12) = 1\/12。点C在点B左侧1\/3单位,因此点C表示的数为:1\/12 - 1\/3 = 1\/12 - 4\/12 = -3\/12 = -1\/4。因此正确答案是D。本题综合考查了有理数在数轴上的表示、加减运算及通分能力,符合七年级有理数章节的难点要求。","options":[{"id":"A","content":"-1\/12"},{"id":"B","content":"1\/4"},{"id":"C","content":"-5\/12"},{"id":"D","content":"-1\/4"}]},{"id":1905,"content":"某次环保活动中,某学生收集了若干废旧电池。第一天他收集了总数的1\/3,第二天收集了剩下的1\/2,此时还剩下24节电池未收集。请问他一共需要收集多少节废旧电池?","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"C","explanation":"设总共需要收集的废旧电池数量为x节。第一天收集了总数的1\/3,即(1\/3)x,剩下(2\/3)x。第二天收集了剩下部分的1\/2,即(1\/2)×(2\/3)x = (1\/3)x。此时总共已收集(1\/3)x + (1\/3)x = (2\/3)x,剩余部分为x - (2\/3)x = (1\/3)x。根据题意,剩余24节,因此(1\/3)x = 24,解得x = 72。故正确答案为C。","options":[{"id":"A","content":"48"},{"id":"B","content":"60"},{"id":"C","content":"72"},{"id":"D","content":"96"}]},{"id":1800,"content":"某班级组织一次数学知识竞赛,参赛学生的成绩被整理成频数分布表如下:\n\n| 成绩区间(分) | 频数(人) |\n|----------------|------------|\n| 60 ≤ x < 70 | 5 |\n| 70 ≤ x < 80 | 12 |\n| 80 ≤ x < 90 | 18 |\n| 90 ≤ x ≤ 100 | 10 |\n\n已知该班参赛学生总人数为45人,且所有成绩均为整数。若将成绩按从高到低排列,则第23名学生的成绩最可能落在哪个区间?","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"中等","answer":"C","explanation":"本题考查数据的整理与描述中的频数分布及中位数思想的应用。总人数为45人,将成绩从高到低排列,第23名是正中间的位置,即中位数所在位置。\n\n首先计算累计频数(从高分段开始累加):\n- 90 ≤ x ≤ 100:10人(第1~10名)\n- 80 ≤ x < 90:18人 → 累计10 + 18 = 28人(第11~28名)\n\n因此,第23名落在第11到第28名之间,即属于“80 ≤ x < 90”这一组。\n\n虽然不能确定具体分数,但根据分组数据的中位数估计方法,第23名最可能落在80到90分区间内。\n\n故正确答案为C。","options":[{"id":"A","content":"60 ≤ x < 70"},{"id":"B","content":"70 ≤ x < 80"},{"id":"C","content":"80 ≤ x < 90"},{"id":"D","content":"90 ≤ x ≤ 100"}]},{"id":2446,"content":"某校八年级开展‘数学建模’活动,研究校园内一座直角三角形花坛的围栏长度。已知花坛的两条直角边分别为√12米和√27米,现需在斜边上安装装饰灯带。若每米灯带成本为8元,则安装整条斜边灯带的总费用最接近以下哪个数值?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"B","explanation":"首先化简两条直角边:√12 = 2√3,√27 = 3√3。根据勾股定理,斜边c = √[(2√3)² + (3√3)²] = √[12 + 27] = √39 ≈ 6.245米。每米灯带8元,总费用为6.245 × 8 ≈ 49.96元,最接近48元。因此选B。本题综合考查二次根式化简与勾股定理的实际应用,难度适中。","options":[{"id":"A","content":"40元"},{"id":"B","content":"48元"},{"id":"C","content":"56元"},{"id":"D","content":"64元"}]}]