written 8.8 years ago by
teamques10
★ 69k
|
•
modified 8.8 years ago
|
x |
$x^2$ |
$x^3$ |
$x^4$ |
y |
$xy$ |
$x^2y$ |
1 |
1 |
1 |
1 |
2.51 |
2.51 |
2.51 |
2 |
4 |
8 |
16 |
5.82 |
11.64 |
23.28 |
3 |
9 |
27 |
81 |
9.93 |
29.79 |
89.37 |
4 |
16 |
64 |
256 |
14.84 |
59.36 |
237.44 |
5 |
25 |
125 |
625 |
20.55 |
102.75 |
513.75 |
6 |
36 |
216 |
1296 |
27.06 |
62.36 |
974.16 |
$\sum x=21$ |
$ \sum x^2=91$ |
$\sum x^3=441$ |
$\sum x^4=2275$ |
$ \sum y=80.71$ |
$\sum xy=368.41$ |
$\sum x^2y=1840.51$ |
$
\\ \; \\ \; \\
\sum y \;=\; a\sum x + b \sum x^2 + c \sum x^2 + cN \; \; \;
\therefore 80.71=21a+91b+6c \; \; \; \; \ldots (i)
\\ \; \\ \; \\
\sum xy \;=\; a \sum x^2 + b \sum x^3 + c \sum x
\; \; \;
\therefore 91a+441b+21c \; \; \; \;
\ldots (ii)
\\ \; \\ \; \\
\sum x^2y \;=\;
a \sum x^3 + b \sum x^4 + c \sum x^2
\; \; \;
\therefore 1840.51 = 441a+2275b+91c \; \; \; \;
\ldots (iii)
$
Solving equations (i), (ii), (iii) simultaneously, we get
a=2.11 , b=0.4 , c=0
$
\therefore
y=2.11x+0.4x^2
$ This is an equation of parabola.
Note: This is another method where we didn’t convert into X=x-3.5 (in this case). We can solve by previous method also.