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Let's assume a variable $X$:

  • When we have a value of $100$ -> $X$ will be $50$.

  • When we have a value of $200$ -> $X$ will be $25$.

That means $X$ decreased with a steady value a total of $25$ from $100$ to $200$.

The question is if the value is $150$ (or in my case any value from $100$ to $200$), what is the value of $X$ based on these $2$ conditions.

cpiegore
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Makdous
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  • Do you want $X=75-\frac Y4$ or $X=\frac{5000}{Y}$ or something else? – Henry Mar 29 '22 at 16:15
  • I want the value of $X$ in a certain range (100 - 200), knowing the two conditions I mentioned above. – Makdous Mar 29 '22 at 16:27
  • It would help if you gave more information about how the value $X$ is calculated. The two examples in Henry's comment above both seem to work just fine. However, there are many others that will also work. – cpiegore Mar 29 '22 at 16:52
  • The $X$ always has a static value, in this case its 50 when the value is 100 & 25 when the value is 200. – Makdous Mar 29 '22 at 16:54
  • Right, but there are lots of ways $X$ could change as your variable $V$ does. It could follow a straight line, or could decay exponentially, or could move up and down like a piston over time. Knowing more about the system will let you decide what kind of interpolation to do to get the best approximation. – Mike Stay Mar 29 '22 at 17:00
  • Yeah you are right, but the problem assumes $X$ decline or incline as a constant increase/decrease. – Makdous Mar 29 '22 at 17:04

1 Answers1

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The general problem is called interpolation. Assuming linear interpolation, you have $(V_1, X_1) = (100, 50)$ and $(V_2, X_2) = (200, 25)$. The equation $X = \frac{-25}{100} V + 75$ is the line through those two points, and predicts $X = 37.5$ for $V=150$.

Mike Stay
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