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And the question asks "Are you satisfied with your usual travel time for your most frequent trips?" The options are the same: "Not satisfied at all," "Not satisfied," "Slightly satisfied," "Moderately satisfied," and "Extremely satisfied".
The options are "Not satisfied at all," "Not satisfied," "Slightly satisfied," "Moderately satisfied," and "Extremely satisfied".
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21 The following scores indicate the levels of SWLS: 31 35 – Extremely satisfied; 26 30 – Satisfied; 21 25 – Slightly satisfied; 20 – Neutral; 15 19 – Slightly dissatisfied; 10 14 – Dissatisfied; 5 9 – Extremely dissatisfied.
Use this scale: 7: Very satisfied; 6: Moderately satisfied; 5: Slightly satisfied; 4: Neither satisfied nor dissatisfied; 3: Slightly dissatisfied; 2: Moderately dissatisfied; 1: Very dissatisfied.
The owner's assessment of the result of the treatment (1: not satisfied, 2: slightly satisfied, 3: satisfied and 4: very satisfied) was recorded at 1, 3, 6, 12 and 24 months.
The scoring was done on 5-point Likert scale (very dissatisfied = 1, slightly dissatisfied = 2, neutral = 3, slightly satisfied = 4 and very satisfied = 5).
Similarly, the majority of patients (90%%) were very satisfied (65%%) or slightly satisfied (25%%) with the level of support received while on treatment.
The co-primary efficacy end points were an improvement of one point or more from baseline rated by the clinician using the CR-SMFRS and a final score of 4 (slightly satisfied) to 6 (extremely satisfied) rated by the patient using the SSRS, indicating satisfaction with the appearance of the face and chin.
(*Sad trombone*) My takeaway is that I can only justify the minimum effort/expense needed to satisfy customers slightly more than our nearest competitor.
Remark 3 Notice that the lower and upper solutions obtained both in Propositions 1 and 2 satisfy a slightly stronger condition than the one required in Definition 1. Below are the links to the authors' original submitted files for images.
If (widetilde{varphi}:Lambda_{omega}^{p}(mathcal{M} rightarrow Lambda_{omega}^{p}(mathcal{M})) satisfies the slightly stronger condition (5.2) in Lemma 5.2, then (overline{varphi}:H^{q, omega}(mathcal{A} rightarrow H^{q, omega}(mathcal{A})) itself is a bounded everywhere defined operator.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com