Question

The table lists the values for two parameters, x and y, of an experiment. What is the approximate value of y for x = 4.0?

Answer 1

Solving for the two unknowns using systems of linear equations (substitution or elimination method):
m= 11.9; b=-23.5

y=11.9x - 23.5
y=11.9*4-23.5
y=24.1
Therefore when x=4, the approximate value of y is 24.1

Answer 2

Answer:

The approximate value for x=4 is y=24.1

Explanation:

A practical method easy to use is the linear interpolation. In this procedure, the approximation is done using the secant line between the two nearest points. In this particular case those points are:

P1: (2.5,6.25)

P2:(9.4,88.36)

Where the first coordinate corresponds to the x coordinate and the second coordinate to the y coordinate. The expression to compute the secant line is:

$y-yo=m*(x-xo)$

Here m is the slope of the line and is calculated from:

$m=\frac{y2-y1}{x2-x1}$

And xo, yo could be the x and y coordinate of any of P1 or P2 points. Thus, for the present coordinates:

$m=\frac{88.36-6.25}{9.4-2.5}$

$m=11.9$

Choosing P1 coordinates as the xo and yo coordinates:

$y-6.25=11.9*(x-2.5)$

Them replacing the estimation value of x=4 and solving for y:

$y-6.25=11.9*(4-2.5)$

$y=11.9*(1.75)+6.25$

$y=24.1$