求解题高手高中数学pdf解题

来源：蜘蛛抓取(WebSpider) 时间：2015-10-16 13:53 标签： 求高手猜成语
求Gmat数学翻译+解题技巧高手!A certain characteristic in a large population has a distribution that is symmetric about the mean M,If 68 percent of the distribution lies within one standard deviation D of the mean, what percent of the distri_百度作业帮
求Gmat数学翻译+解题技巧高手!A certain characteristic in a large population has a distribution that is symmetric about the mean M,If 68 percent of the distribution lies within one standard deviation D of the mean, what percent of the distri
求Gmat数学翻译+解题技巧高手!A certain characteristic in a large population has a distribution that is symmetric about the mean M,If 68 percent of the distribution lies within one standard deviation D of the mean, what percent of the distribution is less than M+ D? 多谢啦!
某一特性在大人口基数的条件下具有关于平均值M对称的分布,如果68%分布于平均值的一个标准偏差D内,那么分布于小于M+D的概率是多少? The distribution is symmetric,
(因为分布是对称的)so
(所以)50% of the distribution is less than M
(有50%的分布是小于平均值M的)and
(并且)the distribution between M and M+D is half in one D,
(在M与M+D之间的分布是一个标准偏差D内的一半)68%/2 = 34% of the distribution lies between M and M+D.
(即68%/2 = 34%分布于M与M+D之间)so(所以)50% + 34%
= 84% of the distribution
is less than M+D
(50% + 34% = 84%分布于小于M+D)求不定积分∫(1＋lnx)/(xlnx)^2dx求高手解题要步骤谢谢_百度作业帮
求不定积分∫(1＋lnx)/(xlnx)^2dx求高手解题要步骤谢谢
求不定积分∫(1＋lnx)/(xlnx)^2dx求高手解题要步骤谢谢
∫(1＋lnx)/(xlnx)^2dx= ∫(1)/(xlnx)^2d(xlnx)= -1/(xlnx) + c ----------思路：(xlnx)' = lnx + 1
∫(1＋lnx)/(xlnx)^2dx=∫1/(xlnx)²d(xlnx)=-1/(xlnx)+cy=(Lnx)^x+x^(1/x)的导数 用对数求求导法做 求高手解题_百度作业帮
y=(Lnx)^x+x^(1/x)的导数 用对数求求导法做 求高手解题
y=(Lnx)^x+x^(1/x)的导数 用对数求求导法做 求高手解题
设y1=(lnx)^x则lny1=xlnlnxy1'/y1=x*(1/(xlnx)) + lnlnx = 1/(lnx)
lnlnx所以y1'=(lnx)^x * [1/(lnx)
lnlnx]设y2=x^(1/x)则lny2=(1/x)lnxy2'/y2=(x/x-lnx)/x^2=(1-lnx)/x^2所以y2'=x^(1/x)(1-lnx)/x^2=x^((1/x)-2)(1-lnx)所以y'=y1'+y2'=(lnx)^x * [1/(lnx)
x^((1/x)-2)(1-lnx)

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