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Performance of Mussels, Mytilus edulis, Mytilus trossulus, and Their Hybrids in Cultivation at Three Scottish Lochs

PJ Dias, B Malgrange, M Snow… - Journal of the World …, 2011 - Wiley Online Library
... Me × Mt, 3, 5, 6, Mt, 1, 13, 14, Total N, 20, 20, 20, N correct, 16, 5, 14, Proportion, 0.800, 0.250,
0.700, N = 60, N correct = 35, Proportion correct = 0.583. Site B. Linear method for response:
genotype. Predictors: shell weight (g), shell length (mm), shell height (mm), shell width (mm ...
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[PDF] Assessment of drug interactions in elderly patients of a family health care unit in Aracaju (Brazil): A pilot study

[PDF] from academicjournals.orgRS Mendes-Netto, CQV Silva… - African Journal of …, 2011 - academicjournals.org
... Nonsteroidal antiinflammatory;**ASA, acetylsalicylic acid. Table 4. Drug administration according
to meal time of 35 elderly prescriptions assisted by a Family Health Care Unit, in Aracaju, Brazil,
2008. Drug Use Administration N % Correct Incorrect Captopril 18 51.4 16.7 83.3 ...
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[PDF] Vision Based Iterative Learning Control of a MEMS Micropositioning Stage with Intersample Estimation and Adaptive Model Correction

[PDF] from ntnu.noPJ White… - 2011 - nt.ntnu.no
... Depending on the time step one of two things comes next. If k is a multiple of N, correct the estimate
with a measurement using: 1 ( 1) ( 1) ( 1) T T d d d F k M k C CM k C R − ⎡ ⎤ + = 12) = + + )1
|1 (ˆ k kx [ ] )|1 (ˆ )1 ( )1 ( )|1 (ˆ k kxC ky kF k kx d m + − + + + 13) 4295 ...
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[PDF] simulateQuasif. fnc 111

[PDF] from psu.eduRH Baayen - Package 'languageR', 2011 - Citeseer
... x= data. frame (training. pca $ x) training. x= training. x [order (rownames (training. x)),] training.
pca. lda= lda (training [, 1: n], trainingAuthor) predictedClasses [i]= as. character (predict (training.
pca. lda, spanish. t [, 1: n]) $ class [i]) } ncorrect= sum (predictedClasses== as. ...
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Survey and management of mussel Mytilus species in Scotland

[PDF] from hcmr.grPJ Dias, SB Piertney, M Snow… - Hydrobiologia, 2011 - Springer
... Me 16 2 0 Me 9 Mt 3 5 6 Mt 1 13 14 Total N 20 20 20 N correct 16 5 14 Proportion 0.8 0.25 0.70
Summary N = 60 N correct = 35 Proportion correct = 0.583 Site B Put into group Me Me 9 Mt Mt
Me 17 6 0 Me 9 Mt 3 5 7 Mt 0 9 13 Total N 20 20 20 N correct 17 5 13 ...
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