某学生在分析某次校园植树活动中各小组种植树苗的成活率时,记录了六个小组的成活树苗数量(单位:棵):48, 52, 45, 57, 50, 54。为了评估这组数据的稳定性,该学生先计算了平均数,再求出各数据与平均数之差的平方,并计算这些平方值的平均数(即方差)。请问这组数据的方差最接近以下哪个数值?
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首先计算选择‘从不进行垃圾分类’的学生人数:总人数45减去‘经常’的27人和‘偶尔’的12人,即45 - 27 - 12 = 6人。然后用6除以总人数45,得到比例为6 ÷ 45 ≈ 0.1333,换算成百分比约为13.3%。因此正确答案是B。
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[{"id":6,"content":"下列现象中,属于光的反射现象的是?","type":"选择题","subject":"物理","grade":"初二","stage":"初中","difficulty":"简单","answer":"C","explanation":"平面镜成像是光的反射现象,水中倒影也是光的反射现象。","options":[{"id":"A","content":"日食和月食"},{"id":"B","content":"小孔成像"},{"id":"C","content":"平面镜成像"},{"id":"D","content":"海市蜃楼"}]},{"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":2019,"content":"在一次校园绿化设计中,工人师傅需要在一块矩形空地的对角线上铺设一条石板路。已知这块空地的长为12米,宽为5米。为了估算材料用量,一名学生想计算这条对角线的长度。请问该对角线的长度是多少?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"本题考查勾股定理的应用。矩形空地可看作一个长方形,其对角线将长方形分成两个直角三角形。根据勾股定理,对角线长度 c 满足 c² = a² + b²,其中 a = 12 米,b = 5 米。计算得:c² = 12² + 5² = 144 + 25 = 169,因此 c = √169 = 13 米。选项A正确。","options":[{"id":"A","content":"13米"},{"id":"B","content":"15米"},{"id":"C","content":"17米"},{"id":"D","content":"√119米"}]},{"id":2208,"content":"某学生在记录一周内每天气温变化时,发现某天的气温比前一天上升了3℃,记作+3℃;另一天比前一天下降了2℃,应记作多少?","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"B","explanation":"气温上升用正数表示,下降则用负数表示。题目中气温下降了2℃,应记作-2℃,因此正确答案是B。","options":[{"id":"A","content":"+2℃"},{"id":"B","content":"-2℃"},{"id":"C","content":"0℃"},{"id":"D","content":"2℃"}]},{"id":439,"content":"某学生在整理班级同学的课外阅读时间(单位:小时\/周)时,记录了以下数据:3, 5, 4, 6, 4, 7, 4, 5。如果将这组数据按从小到大的顺序排列,位于中间位置的两个数的平均数是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"首先将数据从小到大排序:3, 4, 4, 4, 5, 5, 6, 7。共有8个数据,是偶数个,因此中位数是中间两个数的平均数。中间两个数是第4个和第5个,即4和5。计算它们的平均数:(4 + 5) ÷ 2 = 9 ÷ 2 = 4.5。因此正确答案是B。","options":[{"id":"A","content":"4"},{"id":"B","content":"4.5"},{"id":"C","content":"5"},{"id":"D","content":"5.5"}]},{"id":2765,"content":"唐朝时期,一位外国使节来到长安,看到城内市场繁荣、街道整齐,还有来自不同国家的人穿着各异、使用不同语言交流。他惊叹于唐朝的开放与包容。这种局面最能体现唐朝哪一方面的特点?","type":"选择题","subject":"历史","grade":"七年级","stage":"初中","difficulty":"中等","answer":"C","explanation":"题目描述的是唐朝都城长安中外人士云集、市场繁荣、文化多元的场景,这直接反映了唐朝对外开放、积极与外国进行经济和文化交流的特点。唐朝实行开明的对外政策,长安作为国际大都市,吸引了大量外国商人、使节和留学生,体现了其文化包容性和中外交流的频繁。选项A、B、D虽然也是唐朝的特点,但与题干中‘外国使节’‘不同国家的人’等关键词不符,因此正确答案为C。","options":[{"id":"A","content":"选项A"},{"id":"B","content":"选项B"},{"id":"C","content":"选项C"},{"id":"D","content":"选项D"}]},{"id":591,"content":"6件","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":2416,"content":"如图,在平面直角坐标系中,点 A(1, 2)、B(4, 6)、C(7, 2) 构成三角形 ABC。若点 D 是点 A 关于直线 BC 的对称点,则点 D 的坐标最接近下列哪一项?(提示:可利用轴对称性质与一次函数求对称点)","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"C","explanation":"本题综合考查轴对称、一次函数、勾股定理与坐标几何知识。首先求直线 BC 的解析式:B(4,6)、C(7,2),斜率 k = (2−6)\/(7−4) = −4\/3,得直线 BC:y − 6 = −4\/3(x − 4),即 y = −(4\/3)x + 34\/3。点 A(1,2) 关于该直线的对称点 D 满足:AD 的中点在 BC 上,且 AD ⊥ BC。设 D(x,y),则中点 M((1+x)\/2, (2+y)\/2) 在 BC 上,代入直线方程得 (2+y)\/2 = −(4\/3)·((1+x)\/2) + 34\/3。又因 AD 斜率为 (y−2)\/(x−1),应与 BC 斜率 −4\/3 互为负倒数,即 (y−2)\/(x−1) = 3\/4。联立两个方程解得 x ≈ 11,y ≈ 4。因此点 D 坐标最接近 (11, 4)。","options":[{"id":"A","content":"(9, 6)"},{"id":"B","content":"(10, 5)"},{"id":"C","content":"(11, 4)"},{"id":"D","content":"(12, 3)"}]},{"id":1838,"content":"某学生测量了一个直角三角形的两条直角边,分别为√12 cm和√27 cm。若该三角形的斜边长度为c cm,则c²的值是多少?","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"C","explanation":"根据勾股定理,直角三角形中斜边的平方等于两条直角边的平方和。已知两条直角边分别为√12 cm和√27 cm,因此:c² = (√12)² + (√27)² = 12 + 27 = 39。选项C正确。本题考查了二次根式的平方运算与勾股定理的综合应用,难度适中,符合八年级学生的认知水平。","options":[{"id":"A","content":"13"},{"id":"B","content":"25"},{"id":"C","content":"39"},{"id":"D","content":"51"}]},{"id":436,"content":"3","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":[]}]