Showing 1 - 20 results of 40,438 for search '(( 50 ((_ decrease) OR (a decrease)) ) OR ((( 12 nm decrease ) OR ( _ linear decrease ))))*', query time: 0.60s Refine Results
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    Time allocation may not decrease with price for a non-linear microscopic utility of leisure. by Ritwik K. Niyogi (427932)

    Published 2014
    “…Consequently, <i>TA</i> no longer decreases but may even increase with price (A, lower panel). …”
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    BMT decreased weight of different organs. by Saeed Katiraei (3956147)

    Published 2017
    “…<p>(A) BMT decreased the weight of liver, thymus, spleen and the pancreas both on LFD and HFD. …”
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    Decrease of synchronization measures during a pre-ictal interval. by Christian Meisel (46720)

    Published 2013
    “…<p>Left column: time series of maximum linear cross correlation during a pre-ictal (top) and an inter-ictal (middle) interval. …”
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    TUDCA decreases ER stress in HOX neonatal rat lungs. by Kirkwood A. Pritchard Jr. (13449794)

    Published 2022
    “…(<b>C</b>) In IHC stain, P-IRE1α levels are decreased (40.8±3.5 A.U. <i>vs</i> 53.1±5.0 A.U., p<0.001, n = 6, 3 for each sex) in chronic hyperoxia exposed neonatal rat lungs by TUDCA. …”
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    HFD induces LDs and decreases ER and mitochondria in nephrocytes. by Aleksandra Lubojemska (10746241)

    Published 2021
    “…(<b>E</b>) Low and high magnification views of pericardial nephrocytes (dotted outlines) from STD and HFD larvae, showing that HFD decreases mitochondria (marked with anti-ATP5A) and ER (marked with anti-KDEL) but increases LDs (marked with LipidTOX). …”
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    Linear regression analysis of decrease in viral load (VL) and trough concentrations of nevirapine (NVP C<sub>trough</sub>). by Jia Wang (51945)

    Published 2013
    “…<p>The log transformation of decrease in viral load (dlogVL) was significantly correlated with trough nevirapine concentrations (NVP Ctrough, µg/ml) among (A) all patients (r = 0.327, <i>p</i><0.05); and (B) patients with partial response (VL 50–400 copies/ml) and virologic failure (VL>400 copies/ml) (r = 0.619, <i>p</i><0.01).…”