Very good. I have a doubt. I would love to hear your comment on it. In recent months, I have been reflecting on the apparent prevalence of certain predatory mega-journals, in particular MDPI's Sustainability, which stands out as the journal with the most publications on various topics, according to various tourism bibliometrics. However, this observation has led me to consider the need for further analysis. Specifically, it has caught my attention that when using the percentage of publications in relation to the specific research topic in percentage terms (number of articles on a topic divided by the total number of articles published), the magnitude of the contribution decreases drastically. To illustrate this point, let me present a hypothetical example: Journal A has published 10 articles on prospect theory in the last five years, but its total output is 600 articles. In comparison, Journal B has published 25 articles on prospect theory in the same period, but its total publication volume exceeds 49,000 articles. Some bibliometrics would say that Journal B is the one that publishes the most, however, it is just a matter of gaining by quantity. I gave the journals weights based on their percentages (Weight of journal = Percentage of Journal / Highest Percentage among journals) then I did the min-max normalisation (Normalised weight = (Weight of Journal−Min Weight) / (Max Weight−Min Weight)), Then I created a Weighted Metric with Normalisation (multiplying the normalised * their weight). The use of min-max normalisation in this one is correct? Do you think there is a better approach?
Something not mentions. Take that formula A = 2r+7 Instead of being given A, you are given that r is 4.5 instead of being given A, and let's be honest, usually you are measuring r and trying to find A When you sub in the 4.5 for the variable, you must group it with the exponent. A = 2r+7 becomes A = (2*4.5)+7 While not too important here, there are other math problems where if you don't group them like that, you will get the wrong answer when you start using orders of operations to solve.
A = r2+7 We know what A is, 16. So we pout 16 for A 16= r2+7 We use orders of Operations to solve problems. PEMDAS or BIDMAS, or whatever memory device you were taught. In this case, the problem is already solved, we are trying to reverse engineer it. So we have to do the order of operation backwards to find where this problem started. So PEMDAS becomes SADMEP (Which sounds like a depressed Muppet lol), and we do the opposite of whatever the sign says. Start with addition. 16= r2+7 The opposite of -7 is -7, so we subtract 7 from both sides. 16-7= r2+7-7 ----- 16-7 is 9, and 7-7 is 0 Now we have 9 =r2 r2 is (r times 2) The opposite of multiplying is dividing, and we have to do it to both sides so.... 9/2 =r2/2 9/2 is 4.5 and 2/2 is 1 So now we have 4.5 = r*1 Anything times 1stays itself so 4.5 = r I hope that helped.
The result You need is value of dy/dx. To do so, please apply the formula dy/dx=(dy/dt)/(dx/dt). It means for You dy/dx=y'/x'=3/4. Edit: If You need (d2y/dx2) please apply fromula: [(d2y/dt2)*(dx/dt)-(d2x/dt2)*(dy/dt)]/(dx/dt)^3 x and y are linear. 1st derivative will give a number. 2nd derivative will result as 0. It means that d2y/dx2 will be 0/number, so result is 0. Forther information are in ,,parametric functions".
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