يعرض 21 - 40 نتائج من 80,148 نتيجة بحث عن '(((( 50 a decrease ) OR ( 10 ((ng decrease) OR (we decrease)) ))) OR ( 50 mean decrease ))', وقت الاستعلام: 1.08s تنقيح النتائج
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    Flow chart of the study procedures. حسب Kazuhiko Ikeuchi (9234268)

    منشور في 2023
    "…In the multivariable analysis, higher age (predicted monthly RPR relative change on the log2 scale 0.23/10 years [95% confidence interval [CI], 0.090–0.37]), history of syphilis (0.36 [95% CI, 0.07–0.65]), and a lower ratio of RPR at treatment initiation/diagnosis (−0.52/every 10-fold increase [95% CI, −0.81 to −0.22]) were associated with a slower RPR decrease after treatment.…"
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    Datasheet1_Impact of lockdown on children with type-1 diabetes: returning to the community was associated with a decrease in HbA1c.pdf حسب Benjamin Morat (17677977)

    منشور في 2023
    "…No significant difference was observed between medians of HbA1c_mean and HbA1c_after values (8.37% [7.88; 9.32%] vs. 8.50% [7.70; 9.50%], respectively; p = 0.391). …"
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    Datasheet2_Impact of lockdown on children with type-1 diabetes: returning to the community was associated with a decrease in HbA1c.pdf حسب Benjamin Morat (17677977)

    منشور في 2023
    "…No significant difference was observed between medians of HbA1c_mean and HbA1c_after values (8.37% [7.88; 9.32%] vs. 8.50% [7.70; 9.50%], respectively; p = 0.391). …"
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    Predicting pattern diversity decreases as a function of and . حسب Selim Haj Ali (21222613)

    منشور في 2025
    "…(a) and (b): Average negative predictive value for and , respectively for and . …"
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    Estimated mean values for light interception. حسب Torsten Schober (20485754)

    منشور في 2024
    "…Empirical models for the relationships between the investigated plant traits and PD/DVP were created using linear regression analysis preceded by a lack-of-fit test. An increase in PD led to a linear decrease in inflorescence yield per plant (<i>p</i> = 0.02), whereas a positive linear relationship was found for inflorescence yield (<i>p</i> = 0.0001) and CBD yield (<i>p</i> = 0.0002) per m<sup>2</sup>. …"
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