Question 1182816
the simplest thing you can do is apply the straight line equation of y = mx + b.
m is the slope
b is the y-intercept.


since your starting point is 345, then the y-intercept is 345.
each time you get a new data point, revise the equation as follows:


m = (y2 - y2) / (x2 - x1)


with the two data points that you currently have, the equation would become:
m = (302 - 345) / (49 - 0) = -43/49.


your equation becomes y = -43/49 * x + 345.


when x = 49, y becomes 302, as you already know.


to solve for when the rank becomes 0, replace y with 0 to get:


0 = -43/49 * x + 345


solve for x to get:


x = -345 / (-43/49) = 393.1395349.


this indicate a rank of 0 in approximately 393 days.


as you add data points, you revise the equation with the latest data point.


for example:


in 7 days, the new data point might be (56,295)


the equation becomes y = (295 - 345) / 56 * x + 345 which becomes y = -50 / 56 * x + 345.
set y = 0 and solve for x to get:
x = -345 / -(50/56) = 386.4 days.


after 14 more days, if your remaining days is 250, your data point becomes (63,250) and your equation becomes:
y = (250 - 345) / 63 * x + 345 which becomes y = -95/63 * x + 345.
set y = 0 and solve for x to get:
x = -345 / (-95/63) = approximately 228 days.


after 7 more days, if your remaining days is 245, your data point becomes (70,245) and your equation becomes:
y = (245 - 345) / 70 * x + 345 which becomes y = -100/70 * x + 345.
set y = 0 and solve for x to get:
x = -345 / (-100/70) = approximately 241 days.


each time you add a new data point, the equation will change and you will get a new estimate as to when your rank will become 0.


alternatively, you can use a linear regression formula.


with that, all you do is add the new data points when they become available and run the regression formula again to get a new equation.


for example, with the data points i assumed, you would have 5 data points.


they would be (0,345),(49,301),(56,295),(63,250),(70,245).


you would need a minimum of 5 points to run the regression formula.


that i have with my accumptions.


the data inputs to the regression calculator become:


0,49,56,63,70 on the first line (these are the x-values).
345,302,295,250,245 on the second line (these are the y-values).


the results are shown below:


<img src = "http://theo.x10hosting.com/2021/070804.jpg" >


the equation shown on the linear regression calculator was:


y=352.10828025478-1.3594176524113x


it's enough to round 352.... to 352 and to round -1.3594... to -1.40.


your equation becomes 352 - 1.40 * x


any greater accuracy is more then likely not requried.


with y = 352 -1.40 * x, you would set y = 0 and solve for x to get:


x = -352 / -1.40 = approximately 251 days.


each time you add another data point, the regression analysis will give you a new forecast.


the more data points, the more accurate the forecast from the regression formula.


the regression calculator can be found at <a href = "http://www.alcula.com/calculators/statistics/linear-regression/#gsc.tab=0" target = "_blank">http://www.alcula.com/calculators/statistics/linear-regression/#gsc.tab=0</a>