15x42x2=42x(15x2)用语言描述

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（1）解不等式：x+13?2≥2x?42；（2）解方程：43[34(15x?2)?6]＝1_百度知道
（1）解不等式：x+13?2≥2x?42；（2）解方程：43[34(15x?2)?6]＝1
（1）解不等式：x+13?2≥2x?42；（2）解方程：43[34(15x?2)?6]＝1．
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1）去分母得，2（x+1）-12≥3（2x-4），去括号得:normal">12;wordSpacing，x-2）-8=1:1px">15x-2-8=1，移项得，x=11，系数化为1得，2x-6x≥-12-2+12，合并同类项得:wordWwordWrap:normal">15x=1+2+8;wordWrap:normal"><table cellpadding="-1" cellspacing="-1" style="margin-right:wordWrap，-4x≥-2，化系数为1得，2x+2-12≥6x-12
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我们会通过消息、邮箱等方式尽快将举报结果通知您。Higher than Second-Order Approximations via Two-Stage Sampling on JSTOR
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Higher than Second-O...
Journal Article
Nitis Mukhopadhyay
Sankhyā: The Indian Journal of Statistics, Series A ()
Vol. 61, No. 2 (Jun., 1999), pp. 254-269
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Stable URL: http://www.jstor.org/stable/
Page Count: 16
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We consider the classical fixed-width (= 2d) confidence interval estimation problem for the mean μ of a normal population whose variance $\sigma ^{2}$ is unknown, but it is assumed that σ > $\sigma _{L}$ where $\sigma _{L}$ (> 0) is known. Under these circumstances, the seminal two-stage procedure of Stein () has been recently modified by Mukhopadhyay and Duggan (1997), and that modified methodology was shown to enjoy asymptotic second-order characteristics, similar to those found in Woodroofe (1977) and Ghosh and Mukhopadhyay (1981) in the case of the purely sequential estimation strategies, that is, expanding E(N) and the coverage probability respectively up to the orders o(1) and $o(d^{2})$ as d → 0. In Theorem 1.1, we first obtain expansions of both lower and upper bounds of E(N) up to the order $O(d^{6})$. In Theorem 1.2, we then provide expansions of the lower and upper bounds for the coverage probability associated with the two-stage procedure of Mukhopadhyay and Duggan (1997) up to the order $o(d^{4})$, whereas Theorem 1.3 further sharpens this order of approximation to $O(d^{6})$. These results amount to what may be referred as the third-order approximations and beyond via double sampling. Such results are not available in the case of any existing purely sequential and other multistage estimation strategies.
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设x∈(0,1】,求函数y=(3x^6+15x^2+2)÷(2x^6+15x^4+3)的值域
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代入x∈（0,1]后(3x^6+15x^2+2)∈(2,20](2x^6+15x^4+3)∈(3,20]∴ y∈(2/3,1]
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扫描下载二维码下列对方程的变形中.正确的是( ) A.由x-12-2x+33=1去分母.得3=1B.将1.5x0.4-15-x2=0.5化为15x4-15-x2=5C.由3x-2(x-3)=2去括号.得3x-2x-3=2D.由3x-2=3+2x移项.得3x-2x=3+2 题目和参考答案——精英家教网——
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下列对方程的变形中，正确的是（　　）
A、由去分母，得3（x-1）-2（2x+3）=1B、将化为C、由3x-2（x-3）=2去括号，得3x-2x-3=2D、由3x-2=3+2x移项，得3x-2x=3+2
考点：解一元一次方程
专题：计算题
分析：各项中方程变形得到结果，即可做出判断．
解答：解：A、由去分母，得3（x-1）-2（2x+3）=6，错误；B、将-=0.5化为-=0.5，错误；C、由3x-2（x-3）=2去括号，得3x-2x+6=2，错误；D、由3x-2=3+2x移项，得3x-2x=3+2，正确，故选D
点评：此题考查了解一元一次方程，其步骤为：去分母，去括号，移项合并，把未知数系数化为1，求出解．
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