The discussion revolves around completing a table of values related to an equation of the form y=(k/x)+b, where some values of x and y are unknown. Participants are tasked with determining the constants k and b based on given data points.
There is ongoing exploration of the relationships between the known and unknown values, with some participants suggesting that the first three entries are sufficient to determine k and b. Guidance has been offered regarding the calculation of unknowns based on known values, but no consensus has been reached on the correctness of the table or the values of k and b.
Participants are working under the constraints of an assignment that requires them to find specific values without complete information. There is uncertainty about the assumptions made regarding the unknown values and their relationships.
Gustavs1337 said:Homework Statement
there's a table x - 1;2;3;u;u;u;u;8
y - 370;160;90;u;24;u;u;u u - unknown value
i have to complete this table with this equation y=(k/x)+b you have to find k and b values
Homework Equations
y=(k/x)+b x is relevant to y x -1 is y -370 and x-2 is y-160 and so on
The Attempt at a Solution
I have tried doing this witha calculator adding all kinds of numbers but with no progress :(.
the ones in y are different, but i believe that the ones in x might be the same, because 1;2;3 right and the last is 8 not any other number. oh and yea i guess its exact fit if yu mean like 2 or 3 not 2,5 or 3,5Ray Vickson said:Are the 4 u-values in x all the same---a single unknown number---or are they 4 unknown numbers? Are the 3 u-values in y the same or different? Are the u's for x the same or different from the u's for y? Do you want an exact fit, or do you want a "best" fit, such a least-squares fit or something similar?
What makes you think x is 5 there?Gustavs1337 said:I just figured out something but I am not sure, if k is 420 and b is -50 then it goes correctly by the table but it gets wrong at the 24, because 420/5 -50 is 34 not 24. Can it be a mistake in the table or is there smth I am mistaken in ?
Well its mostly because the x-8 is in the end so i thought that the missing x'es were [4;7], but now i think it might not be true because the results don't match. I don't want to believe that there's a mistake in the table tho. I also thought that why are the x'es unknown if it were from 1 to 8, so i think there is some kind of secret or something that i do not knowharuspex said:What makes you think x is 5 there?
Just drop the assumption that the missing x values are in arithmetic progression. The first three entries in the table are enough to find k and b. Thereafter it is just a matter of calculating the unknown number (x or y) from the known one.Gustavs1337 said:Well its mostly because the x-8 is in the end so i thought that the missing x'es were [4;7], but now i think it might not be true because the results don't match. I don't want to believe that there's a mistake in the table tho. I also thought that why are the x'es unknown if it were from 1 to 8, so i think there is some kind of secret or something that i do not know
haruspex said:Just drop the assumption that the missing x values are in arithmetic progression. The first three entries in the table are enough to find k and b. Thereafter it is just a matter of calculating the unknown number (x or y) from the known one.
By the way, the easiest way to find k and b is to consider the product xy. This should equal k+bx, so you can do a straight line fit.
Yes.Gustavs1337 said:do you mean xy=k+bx
Yes.Gustavs1337 said:do you think the number 24 fits in the table
haruspex said:Just drop the assumption that the missing x values are in arithmetic progression. The first three entries in the table are enough to find k and b. Thereafter it is just a matter of calculating the unknown number (x or y) from the known one.
By the way, the easiest way to find k and b is to consider the product xy. This should equal k+bx, so you can do a straight line fit.
so, by my calculations x5 = 5.68 ?haruspex said:Yes.
Yes.
Yes.Gustavs1337 said:so, by my calculations x5 = 5.68 ?
Well that's close. ##\ \frac{210}{37}=5\frac{25}{37}\ .##Gustavs1337 said:so, by my calculations x5 = 5.68 ?