I did the usual analysis and found the first differences to be
145, 250, 358, and 465
and the second differences to be
105, 108, and 107
which appear to be relatively constant at appr 106, so I suspect a quadratic.
let N = ad^2 + bd + c, where d is the number of days, and N is the number of visitors.
we get:
52 = a + b + c
197 = 4a + 2b + c
447 = 9a + 3b + c
805 = 16a + 4b + c
1270 = 25a + 5b + c
Since we have more than 3 equations in only 3 unknowns, and our second differences were really not constant, our solutions could only be approximate.
taking the first 3 I got:
a = 5.5 , b=-12.5 , c = 12
taking the 1st, 3rd and 5th, I got
a = 53.5, b= -16.5, c = 15
approximating:
N = 53.5n^2 - 16.5n + 15
I know I get exact values for n = 1, 3, 5
how about n = 2 and n=4
N(2) = 196 instead of 197 , not bad
N(4) = 805 , right on!
I would say my equation is pretty good
The number of visitors at a Web site
over several days is shown in the table
at the right. What is an equation that models the data?
This is the given data:
Days: 1, 2, 3, 4, 5
Visitors: 52, 197, 447, 805, 1270
How do you write the equation?
3 answers
Just noticed a typo (sticking 3 key)
<taking the first 3 I got:
a = 5.5 , b=-12.5 , c = 12>
should say:
taking the first 3 I got:
a = 53.5 , b=-12.5 , c = 12
btw , I did not show how I solved those 3 equations in 3 unknowns, I used a webpage for that
http://www.1728.org/unknwn3.htm
<taking the first 3 I got:
a = 5.5 , b=-12.5 , c = 12>
should say:
taking the first 3 I got:
a = 53.5 , b=-12.5 , c = 12
btw , I did not show how I solved those 3 equations in 3 unknowns, I used a webpage for that
http://www.1728.org/unknwn3.htm
Thanks!