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题面有误
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DreamLand_zcb楼主2023/5/6 20:16

s=1n∑i=1n(bi−1n∑i=1nbi)s = \sqrt{\frac{1}{n}\sum_{i=1}^{n}(b_i -\frac{1}{n}\sum_{i=1}^{n}b_i)}

⇒\Rightarrow

s=1n∑i=1n(bi−1n∑j=1nbj)2s = \sqrt{\frac{1}{n}\sum_{i=1}^{n}(b_i -\frac{1}{n}\sum_{j=1}^{n}b_j)^2}

2023/5/6 20:16
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