题干

某商场“五一”期间为进行有奖销售活动,设立了一个可以自由转动的转盘.商场规定:顾客购物100元以上就能获得一次转动转盘的机会,当转盘停止时,指针落在哪一区域就可以获得相应的奖品.下表是此次活动中的一组统计数据:
转动转盘的次数n
100
200
400
500
800
1000
落在“可乐”区域的次数m
60
122
240
298

604
落在“可乐”区域的频率
m
n
0.6
0.61
0.6

0.59
0.604
(1)完成上述表格;(结果全部精确到0.1)
(2)请估计当n很大时,频率将会接近 ,假如你去转动该转盘一次,你获得“可乐”的概率约是  ;(结果全部精确到0.1)
(3)转盘中,表示“洗衣粉”区域的扇形的圆心角约是多少度?
上一题 下一题 0.0难度 选择题 更新时间:2017-03-03 10:47:49

答案(点此获取答案解析)

解:(1)298÷500≈0.6;0.59×800=472;

(2)估计当n很大时,频率将会接近0.6,假如你去转动该转盘一次,你获得“可乐”的概率约是0.6;

(3)(1﹣0.6)×360°=144°,

所以表示“洗衣粉”区域的扇形的圆心角约是144°.

故答案为0.6,0.6.

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