均值置信限等效法,The equivalent method of the mean-value and confidence limits
1)The equivalent method of the mean-value and confidence limits均值置信限等效法
2)mean equivalent均值等效
1.Sometimes the comparison between the drug effects of the two kinds is needed with themean equivalent of two sided t-test;then the dose of a drug can be gradually extended.用均值等效二次单侧检验方法,并对其逐步推广,从而对处理组与对照组进行检验,确定是否等效。
3)advanced mean value method高等均值法
4)average equivalence method平均等效法
5)extreme equivalent method极限等效法
6)equivalent numerical method等效数值法
1.Introducing the concept and character of entropy and entropy weight,theequivalent numerical method based on entropy weights was established,and applied the model for evaluating the eutrophication of Shanzi Reservoir comprehensively.介绍了熵和熵权的概念与性质,结合等效数值评价方法,对福州山仔水库富营养化进行综合评价,评价结果与物元分析模型、属性识别理论方法、模糊隶属函数法进行对比,结果表明,利用熵权等效数值法进行富营养化综合评价是可行的。
英文短句/例句
1.Evaluation on Stability of Surrounding Rock Mass Based on Equivalent Numerical Method等效数值法在围岩稳定性评价中的应用
2.Evaluation of underground ore-rock stability based on equivalent value method基于等效数值法的地下开采矿岩稳定性评价
3.Application of Equivalent Numerical Method Based on Entropy Weights in Water Environmental Quality Assessment基于熵权的等效数值法在水环境质量综合评价中的应用
4.Naturally Fractured Reservoir Simulation with Effective Permeability Tensor;天然裂缝性油藏等效介质法数值模拟研究
5.In- situ hollow floor hollow cylinder transverse direction equivalent thickness simplified numerical solution现浇空心楼板横筒方向等效厚度简化数值解法
6.equivalent binary digit等价的二进制数字;等效二进位数;等值二进数
7.Numerical Simulation of Air Spring Shock Based on Control Volume Method and Equivalent Viscous Damping Method基于控制体积法及等效粘性阻尼法的空气弹簧冲击数值仿真
8.We get a better result comparing with mean filter、 reciprocal gradient filter、 linear filter and median filter.和领域平均法、数梯度加权法、线性插值法、中值滤波法等相比较,其效果改善明显。
9.A TOF Measurement Method Based on Equivalent Sampling and Data Average等效采样与数值平均相结合的激光飞行时间测量方法
10.The artical introduces elementary equation, boundary and initial conditions of model of equivalent conduction heat transfer, physical property prarameters processing and the method of value solving.介绍了等效传导传热模型的基本方程、界条件和初始条件、性参数的处理以及数值求解方法等。
11.Study on Equivalent Characteristic and Numerical Simulation of Seepage in Fractured Rock;裂隙岩体渗流等效性及数值模拟研究
12.Numberical Simulation Analysis of Equivalent Bolt Rock Mass and Its Application;锚固岩体等效参数数值模拟研究及应用
13.An Efficient Method for Parameter Pass Evaluation:Equivalence Transform Method有效的参数传递求解方法—等价变换法
14.{An argument name is missing a value after its equal sign.}{参数名称在等号后无法找到值。}
15.LOGCONTRAST METHOD FOR EQUATING TEST BASED ON IRT;项目反应理论框架下的新等值方法——对数对比等值法
16.The Damage Model of Jointed Rock Mass Equivalent Behavior and Numerical Simulation;节理岩体等效性能的损伤模型及其数值模拟
17.Failure Analysis and Stress Simulation on Plasma Sprayed Al_2O_3 Ceramic Coating;等离子弧喷涂Al_2O_3陶瓷涂层失效分析及数值模拟
18.Equivalent Prameters Computation of Periodic Composites for Dynamic Coupled Thermoelasticity Problems周期复合材料等效热弹性性能的数值计算
相关短句/例句
mean equivalent均值等效
1.Sometimes the comparison between the drug effects of the two kinds is needed with themean equivalent of two sided t-test;then the dose of a drug can be gradually extended.用均值等效二次单侧检验方法,并对其逐步推广,从而对处理组与对照组进行检验,确定是否等效。
3)advanced mean value method高等均值法
4)average equivalence method平均等效法
5)extreme equivalent method极限等效法
6)equivalent numerical method等效数值法
1.Introducing the concept and character of entropy and entropy weight,theequivalent numerical method based on entropy weights was established,and applied the model for evaluating the eutrophication of Shanzi Reservoir comprehensively.介绍了熵和熵权的概念与性质,结合等效数值评价方法,对福州山仔水库富营养化进行综合评价,评价结果与物元分析模型、属性识别理论方法、模糊隶属函数法进行对比,结果表明,利用熵权等效数值法进行富营养化综合评价是可行的。
延伸阅读
均值不等式几个重要不等式(一)一、平均值不等式设a1,a2,…, an是n个正实数,则,当且仅当a1=a2=…=an时取等号1.二维平均值不等式的变形(1)对实数a,b有a2+b2³2ab (2)对正实数a,b有(3)对b>0,有, (4)对ab2>0有,(5)对实数a,b有a(a-b)³b(a-b) (6)对a>0,有(7) 对a>0,有 (8)对实数a,b有a2³2ab-b2(9) 对实数a,b及l¹0,有二、例题选讲例1.证明柯西不等式证明:法一、若或命题显然成立,对¹0且¹0,取代入(9)得有两边平方得法二、,即二次式不等式恒成立则判别式例2.已知a>0,b>0,c>0,abc=1,试证明:(1)(2)证明:(1)左=[]=³(2)由知同理:相加得:左³例3.求证:证明:法一、取,有a1(a1-b)³b(a1-b), a2(a2-b)³b(a2-b),…, an(an-b)³b(an-b)相加得(a12+ a22+…+ an2)-( a1+ a2+…+ an)b³b[(a1+ a2+…+ an)-nb]³0所以法二、由柯西不等式得: (a1+ a2+…+ an)2=((a1×1+ a2×1+…+ an×1)2£(a12+ a22+…+ an2)(12+12+…+12)=(a12+ a22+…+ an2)n,所以原不等式成立例4.已知a1, a2,…,an是正实数,且a1+ a2+…+ an<1,证明:证明:设1-(a1+ a2+…+ an)=an+1>0,则原不等式即nn+1a1a2…an+1£(1-a1)(1-a2)…(1-an)1-a1=a2+a3+…+an+1³n1-a2=a1+a3+…+an+1³n…………………………………………1-an+1=a1+a1+…+an³n相乘得(1-a1)(1-a2)…(1-an)³nn+1例5.对于正整数n,求证:证明:法一、>法二、左==例6.已知a1,a2,a3,…,an为正数,且,求证:(1)(2)证明:(1)相乘左边³=(n2+1)n证明(2)左边= -n+2(= -n+2×[(2-a1)+(2-a2)+…+(2-an)](³ -n+2×n