某学生测量了班级10名同学的身高(单位:厘米),数据如下:152, 155, 148, 160, 158, 153, 157, 150, 156, 154。这组数据的众数是多少?
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[{"id":1971,"content":"某学生在研究某次学校科技节中各参赛小组完成项目所用时间时,记录了八个小组的数据(单位:分钟):28.5, 32.1, 26.8, 30.4, 29.7, 33.6, 27.9, 31.2。为了分析这组数据的集中趋势和离散程度,该学生先计算了平均数,再计算了各数据与平均数之差的绝对值,并求出这些绝对值的平均数(即平均绝对偏差,MAD)。请问这组数据的平均绝对偏差最接近以下哪个数值?","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"中等","answer":"B","explanation":"本题考查数据的收集、整理与描述中平均绝对偏差(MAD)的概念与计算。首先计算八个小组所用时间的平均数:(28.5 + 32.1 + 26.8 + 30.4 + 29.7 + 33.6 + 27.9 + 31.2) ÷ 8 = 240.2 ÷ 8 = 30.025。然后计算每个数据与平均数之差的绝对值:|28.5−30.025|=1.525,|32.1−30.025|=2.075,|26.8−30.025|=3.225,|30.4−30.025|=0.375,|29.7−30.025|=0.325,|33.6−30.025|=3.575,|27.9−30.025|=2.125,|31.2−30.025|=1.175。将这些绝对值相加:1.525 + 2.075 + 3.225 + 0.375 + 0.325 + 3.575 + 2.125 + 1.175 = 14.4。最后求平均绝对偏差:14.4 ÷ 8 = 1.8。1.8 最接近选项 B 的 1.7,因此答案为 B。","options":[{"id":"A","content":"1.5"},{"id":"B","content":"1.7"},{"id":"C","content":"1.9"},{"id":"D","content":"2.1"}]},{"id":2372,"content":"在一次校园绿化项目中,某学生负责测量一块三角形花坛的三边长度。他测得三边长分别为√12米、√27米和√75米。若他想用一根木条沿花坛边缘围一圈,则需要准备的木条最短长度为多少米?(结果保留最简二次根式)","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"C","explanation":"本题考查二次根式的化简与实数加法运算。首先将三个边长分别化简为最简二次根式:√12 = √(4×3) = 2√3;√27 = √(9×3) = 3√3;√75 = √(25×3) = 5√3。然后将三边相加求周长:2√3 + 3√3 + 5√3 = (2+3+5)√3 = 10√3。因此所需木条最短长度为10√3米,对应选项C。","options":[{"id":"A","content":"6√3"},{"id":"B","content":"8√3"},{"id":"C","content":"10√3"},{"id":"D","content":"12√3"}]},{"id":4,"content":"已知方程组{2x + 3y = 7, x - y = 1},则x = ____, y = ____。","type":"填空题","subject":"数学","grade":"初二","stage":"初中","difficulty":"中等","answer":"x = 2, y = 1","explanation":"由第二个方程得x = y + 1,代入第一个方程:2(y + 1) + 3y = 7,解得5y = 5,即y = 1,因此x = 2。","options":[]},{"id":2201,"content":"某学生在数轴上从原点出发,先向右移动5个单位长度,再向左移动8个单位长度。此时该学生所在位置所表示的数是___。","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"从原点出发向右移动5个单位,表示+5;再向左移动8个单位,表示-8。最终位置为5 + (-8) = -3,因此该学生所在位置表示的数是-3。","options":[{"id":"A","content":"3"},{"id":"B","content":"-3"},{"id":"C","content":"13"},{"id":"D","content":"-13"}]},{"id":761,"content":"在一次班级环保活动中,某学生收集了若干节废旧电池,其中一号电池比五号电池多8节,两种电池一共收集了20节。设五号电池有x节,则根据题意可列出一元一次方程:x + (x + 8) = 20。解这个方程,得到x = __。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"6","explanation":"根据题意,设五号电池有x节,则一号电池有(x + 8)节。两种电池总数为20节,因此可列方程:x + (x + 8) = 20。化简得:2x + 8 = 20,两边同时减去8得:2x = 12,再两边同时除以2得:x = 6。所以五号电池有6节,符合题意且计算正确。","options":[]},{"id":2355,"content":"如图,在平面直角坐标系中,一次函数 y = kx + b 的图像经过点 A(2, 5) 和点 B(−1, −1)。若点 C(m, n) 也在此函数图像上,且满足 m² − 4m + 4 + |n − 5| = 0,则点 C 的坐标为( )。","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"B","explanation":"首先,利用点 A(2, 5) 和点 B(−1, −1) 求一次函数的解析式。由斜率公式得:k = (5 − (−1)) \/ (2 − (−1)) = 6 \/ 3 = 2。将 k = 2 和点 A(2, 5) 代入 y = kx + b,得 5 = 2×2 + b,解得 b = 1。因此函数解析式为 y = 2x + 1。接着分析条件 m² − 4m + 4 + |n − 5| = 0。注意到 m² − 4m + 4 = (m − 2)²,所以原式可化为 (m − 2)² + |n − 5| = 0。由于平方项和绝对值均为非负数,两者之和为 0 当且仅当每一项都为 0,故有 m − 2 = 0 且 n − 5 = 0,即 m = 2,n = 5。因此点 C 的坐标为 (2, 5),对应选项 B。验证该点是否在函数图像上:当 x = 2 时,y = 2×2 + 1 = 5,符合。故正确答案为 B。","options":[{"id":"A","content":"(0, 1)"},{"id":"B","content":"(2, 5)"},{"id":"C","content":"(4, 9)"},{"id":"D","content":"(1, 3)"}]},{"id":626,"content":"x + (x + 3) + 2x + x = 45","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":2521,"content":"某学生观察一个由三个全等的等边三角形拼接而成的轴对称图形(如图,未展示),若将该图形绕其对称中心旋转一定角度后能与原图形完全重合,则这个旋转角度最小为多少?","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"简单","answer":"C","explanation":"该图形由三个全等的等边三角形拼接而成,且具有轴对称性。由于等边三角形的每个内角为60°,三个三角形围绕中心拼接时,中心点周围的角度总和为360°,因此每个三角形占据120°的扇形区域。要使图形绕对称中心旋转后与自身重合,最小的旋转角度应等于其旋转对称的最小单位角度。因为图形具有三重旋转对称性(即每转120°重合一次),所以最小旋转角度为360° ÷ 3 = 120°。选项C正确。","options":[{"id":"A","content":"60°"},{"id":"B","content":"90°"},{"id":"C","content":"120°"},{"id":"D","content":"180°"}]},{"id":2768,"content":"考古学家在某遗址中发现了大量炭化稻谷、干栏式建筑遗迹和刻画符号的陶器,这些发现最有可能属于哪个新石器时代文化?","type":"选择题","subject":"历史","grade":"七年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"题干中提到的‘炭化稻谷’表明该地区以水稻种植为主,而水稻主要种植于长江流域;‘干栏式建筑’是适应潮湿环境的典型建筑形式,常见于南方地区;刻画符号的陶器也见于河姆渡遗址。河姆渡文化位于浙江余姚,属于长江流域的新石器时代文化,距今约7000年,符合上述特征。半坡文化位于黄河流域,以粟作农业和半地穴式房屋为特点;大汶口文化和红山文化也主要分布在北方,且不以水稻为主要作物。因此,正确答案为A。","options":[{"id":"A","content":"河姆渡文化"},{"id":"B","content":"半坡文化"},{"id":"C","content":"大汶口文化"},{"id":"D","content":"红山文化"}]},{"id":1772,"content":"某学生在平面直角坐标系中画出一个三角形ABC,其中点A的坐标为(2, 3),点B在x轴上,点C在y轴上,且三角形ABC的面积为6。若点B的横坐标为正,点C的纵坐标为正,则点B的坐标为____,点C的坐标为____。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"困难","answer":"(4, 0), (0, 3)","explanation":"设B(b, 0),C(0, c),利用三角形面积公式S = 1\/2 × |b| × |c| = 6,结合A(2,3)共面关系,解得b=4,c=3,故B(4,0),C(0,3)。","options":[]}]