均值化,equalization
1)equalization[英][,i:kw?lai'zei??n][美][,ikw?l?'ze??n]均值化
1.With the help of the data processing software DPS,we make the non-dimensional processing of the raw data by means of the equalization method.用均值化方法借助DPS数据处理软件,对原始数据进行无量纲化处理,不仅有效消除了量纲与数量级的影响,而且使得处理后的数据包含了原始数据的全部信息。
2)Winsorized meanWinsor化均值
1.The results showed that the trimmed mean and Winsorized mean could describe the baseline values accurately.结果表明通过稳健估计方法中的切尾均值和Winsor化均值可较准确地表达环境的基线值。
3)mean quantization均值量化
1.Mean Quantization Audio Watermarking Algorithm Based on Wavelet Transform;基于小波域的均值量化数字音频水印算法
2.A blind watermarking scheme based on fuzzy entropy and mean quantization;基于模糊熵和均值量化的盲文水印方案
3.Audio watermarking scheme based on DWT and mean quantization基于DWT和均值量化的音频水印算法
英文短句/例句

1.Audio watermarking scheme based on DWT and mean quantization基于DWT和均值量化的音频水印算法
2.An Audio Watermarking Algorithm Based on Wavelet Domain Quantization Average;一种基于小波域均值量化的音频数字水印算法
3.Robust Digital Watermarking Based on Audio Statistics Characteristics and Synchronization Code一种基于均值量化的抗去同步攻击数字水印算法
4.A Blind Color Picture Watermarking Algorithm Based on Adaptive Mean Quantization DWT基于DWT域自适应均值量化的彩色图像盲水印算法
5.average sample number平均抽查数;平均检查量;平均取样量;平均值抽样数
6.tiny deviations from the average.任何偏离平均值的变化。
7.The Study on Some Problems Concerning Dynamic Coherent Risk Measures and Mean-Variance Optimizations;动态Coherent风险度量和均值方差优化若干问题的研究
8.The Adjustment of Mean Estimation with Extreme Value;极端值影响下总体均值估计量的调整
9.In 2001, the annual average of NO2 concentration in the air of 341 cities attained the Grade II national standards for air quality.2001年,341个城市空气中二氧化氮浓度年均值均达到国家环境空气质量二级标准。
10.The amount by which the average of a set of values departs from a reference value.一组数值的平均值对某参考值的偏离量。
11.The ratio of the average power loss to the power loss under peak loading.平均能量衰减与峰值负荷时能量衰减的比值。
12.The standard error of estimate, on the other hand, measures the variability, or scatter, of the observed values around the regression line.而估计值的平均误差,却是度量观察值围绕着回归直线的变化程度或分散程度。
13.The square voltage was used to describe thermal noise voltage in determination.热噪声电压通常采用均方值定量描述。
14.standard deviation of the arithmetic mean of a series of measurements测量列算术平均值的标准(偏)差
15.______% more or less both in amount and quantity allowed at the Sellers' option.数量及总值均得有___%的增减,由卖方决定。
16.With% more or less both in amount and quantity allowed at the Sellers option.数量及总值均得有%增减,由卖方决定。
17.The norm of a quaternion is similar to that of a vector.四元数的平均值与矢量的有点相似。
18.The Mathematical Mean of the Physical Quantity and the Physical Experimental Error;物理量的数学平均值和物理实验误差
相关短句/例句

Winsorized meanWinsor化均值
1.The results showed that the trimmed mean and Winsorized mean could describe the baseline values accurately.结果表明通过稳健估计方法中的切尾均值和Winsor化均值可较准确地表达环境的基线值。
3)mean quantization均值量化
1.Mean Quantization Audio Watermarking Algorithm Based on Wavelet Transform;基于小波域的均值量化数字音频水印算法
2.A blind watermarking scheme based on fuzzy entropy and mean quantization;基于模糊熵和均值量化的盲文水印方案
3.Audio watermarking scheme based on DWT and mean quantization基于DWT和均值量化的音频水印算法
4)mean-quantization均值量化
1.Digital audio fragile watermarking algorithm based on mean-quantization;基于均值量化的音频脆弱水印算法研究
5)Zero mean normalization零均值化
6)normalized mean泛化均值
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