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:
| 成绩区间(分) | 频数(人) |
|----------------|------------|
| 60 ≤ x < 70 | 5 |
| 70 ≤ x < 80 | 12 |
| 80 ≤ x < 90 | 18 |
| 90 ≤ x ≤ 100 | 10 |
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?
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首先计算小明购买5本笔记本的总花费:每本8元,5本就是 8 × 5 = 40 元。他付了50元,所以应找回的钱是 50 - 40 = 10 元。因此正确答案是A。
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[{"id":1990,"content":"某学生在纸上画了一个边长为6 cm的正方形ABCD,以顶点A为原点建立平面直角坐标系,AB边在x轴正方向,AD边在y轴正方向。若在正方形内部随机取一点P,则点P到x轴的距离小于3 cm的概率是多少?","type":"选择题","subject":"数学","grade":"九年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"本题考查概率初步与几何图形的综合应用。正方形边长为6 cm,面积为6×6=36 cm²。点P到x轴的距离即为其纵坐标y的值。要求y < 3,即在正方形下半部分(从y=0到y=3)的区域中取点。该区域是一个长为6 cm、宽为3 cm的矩形,面积为6×3=18 cm²。因此,所求概率为18\/36=1\/2。","options":[{"id":"A","content":"1\/2"},{"id":"B","content":"1\/3"},{"id":"C","content":"2\/3"},{"id":"D","content":"3\/4"}]},{"id":206,"content":"一个三角形的三个内角分别是50度、60度和_空白处_度。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"70","explanation":"三角形的内角和恒等于180度。已知两个角分别是50度和60度,将这两个角相加得到50 + 60 = 110度。用180度减去110度,得到第三个角的度数为180 - 110 = 70度。因此,空白处应填写70。","options":[]},{"id":911,"content":"在一次环保活动中,某学生收集了不同种类的垃圾,其中可回收垃圾占总量的3\/8,厨余垃圾占总量的1\/4,有害垃圾占0.125,其余为其他垃圾。如果其他垃圾的重量是2.5千克,那么这次收集垃圾的总重量是___千克。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"10","explanation":"首先将各部分垃圾所占比例统一为分数形式:可回收垃圾占3\/8,厨余垃圾占1\/4 = 2\/8,有害垃圾占0.125 = 1\/8。将这些比例相加:3\/8 + 2\/8 + 1\/8 = 6\/8 = 3\/4。因此,其他垃圾占总量的1 - 3\/4 = 1\/4。已知其他垃圾为2.5千克,设总重量为x千克,则有(1\/4)x = 2.5,解得x = 2.5 × 4 = 10。所以总重量是10千克。","options":[]},{"id":834,"content":"在一次环保活动中,某班级学生共收集了120千克废纸。已知男生每人收集2.5千克,女生每人收集3千克,全班共有45人参与。设男生有x人,则女生有___人,根据题意可列出一元一次方程为___。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"45 - x, 2.5x + 3(45 - x) = 120","explanation":"全班共45人,男生有x人,则女生人数为总人数减去男生人数,即45 - x。男生每人收集2.5千克,共收集2.5x千克;女生每人收集3千克,共收集3(45 - x)千克。总收集量为120千克,因此可列方程:2.5x + 3(45 - x) = 120。本题考查了一元一次方程的实际应用,涉及有理数运算和方程建模,符合七年级数学课程要求。","options":[]},{"id":2458,"content":"在一次数学实践活动中,某学生测量了一块等腰三角形花坛的两条腰长均为5米,底边上的高为4米。若要在花坛中铺设一条从顶点到底边中点的装饰带,则这条装饰带的长度为____米。","type":"填空题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"中等","answer":"4","explanation":"等腰三角形底边上的高、中线、顶角平分线三线合一,因此从顶点到底边中点的线段就是高,长度为4米。","options":[]},{"id":1031,"content":"在一次班级环保活动中,某学生收集了若干节废旧电池。他将这些电池分成3组,第一组比第二组多2节,第三组比第二组少1节,三组共收集了14节电池。设第二组有x节电池,则可列出一元一次方程为:___。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"x + (x + 2) + (x - 1) = 14","explanation":"设第二组有x节电池,则第一组有(x + 2)节,第三组有(x - 1)节。根据题意,三组总数为14节,因此方程为x + (x + 2) + (x - 1) = 14。该题考查学生根据实际问题建立一元一次方程的能力,属于一元一次方程知识点的简单应用。","options":[]},{"id":250,"content":"一个长方形的长比宽多5厘米,若其周长为38厘米,则这个长方形的宽是___厘米。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"中等","answer":"7","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),代入已知条件得:2 × (x + x + 5) = 38。化简得:2 × (2x + 5) = 38,即4x + 10 = 38。解得4x = 28,x = 7。因此,长方形的宽是7厘米。","options":[]},{"id":537,"content":"某学生在整理班级同学最喜欢的课外活动调查数据时,将收集到的信息绘制成扇形统计图。已知喜欢阅读的同学所占的圆心角为72度,那么喜欢阅读的同学占全班人数的百分比是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"C","explanation":"扇形统计图中,整个圆的圆心角为360度,代表全班100%的人数。喜欢阅读的同学对应的圆心角是72度,因此所占百分比为:72 ÷ 360 × 100% = 0.2 × 100% = 20%。所以正确答案是C。","options":[{"id":"A","content":"10%"},{"id":"B","content":"15%"},{"id":"C","content":"20%"},{"id":"D","content":"25%"}]},{"id":2011,"content":"在一次班级组织的户外测量活动中,某学生使用测角仪和卷尺测量了一块三角形空地ABC。他测得∠A = 60°,AB = 8米,AC = 6米。为了验证测量准确性,他根据这些数据计算出BC的长度。若该三角形满足余弦定理,则BC的长度最接近以下哪个值?(结果保留一位小数)","type":"选择题","subject":"数学","grade":"八年级","stage":"初中","difficulty":"简单","answer":"A","explanation":"本题考查余弦定理在三角形中的应用,属于勾股定理的拓展内容,符合八年级数学知识范围。已知两边及其夹角,可直接使用余弦定理:BC² = AB² + AC² - 2·AB·AC·cos∠A。代入数据:BC² = 8² + 6² - 2×8×6×cos60° = 64 + 36 - 96×0.5 = 100 - 48 = 52。因此,BC = √52 ≈ 7.211,保留一位小数约为7.2米。故正确答案为A。","options":[{"id":"A","content":"7.2米"},{"id":"B","content":"7.6米"},{"id":"C","content":"8.0米"},{"id":"D","content":"8.4米"}]},{"id":470,"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:15, 20, 25, 30, 35, 40, 45。如果再加入一个数据后,这组数据的中位数变为30,那么加入的这个数据可能是多少?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"原数据有7个数,按从小到大排列为:15, 20, 25, 30, 35, 40, 45。中位数是第4个数,即30。加入一个数据后,总共有8个数,中位数是第4个和第5个数的平均数。要使中位数为30,则第4个和第5个数的平均数必须为30。若加入30,则新数据为:15, 20, 25, 30, 30, 35, 40, 45,此时第4个数是30,第5个数也是30,中位数为(30+30)÷2=30,符合条件。其他选项加入后,中位数均不等于30。因此正确答案是B。","options":[{"id":"A","content":"25"},{"id":"B","content":"30"},{"id":"C","content":"35"},{"id":"D","content":"40"}]}]