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11,875
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4
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预测中间值。
线性回归，最简单。 x=[0.4:0.05:0.6 0.8]; y=[28.313 26.307 23.807 21.211 18.569 8.111]; p=polyfit(x,y,1); polyval(p,[0.65:0.05:0.75]) ...
3 years ago | 0
| accepted
