Question

How to adjust this table ?

Answer 1

`y = -9/10x + 193`

Explanation:

Suppose that the `x`-axis represents age in years and the `y`-axis represents the pulse rate in beats/minute. From the table, we can see that when `x` increases by a constant amount (`10`), `y` decreases by a constant amount (`9`). This is a linear relationship between `x` and `y`.

As we are given a set of data points, we only need the slope `m` of the equation to represent it in point-slope form :

`y-y_1 = m(x-x_1)`

where `(x_1, y_1)` is a point on the graph.

The slope of a linear equation is given by `m = "change in y"/"change in x"`. Given two points on the graph `(x_1, y_1), (x_2, y_2)`, this translates to `m = (y_2-y_1)/(x_2-x_1)`.

Taking the first two points on the table, we get:

`m = (166-175)/(30-20) = -9/10`

Thus, together with the point `(20, 175)`, we get our equation in point slope form:

`y - 175 = -9/10(x - 20)`

We can also solve for `y` to put it in slope-intercept form :

`y = -9/10x + 193`

Answer 2

`y = -0.9 x + 193`

Explanation:

Given a list of values `{x_k,y_k}, k=1,2,cdots,n`

a line can be adjusted such that the accumulated deviation error is minimum.

Let the line be given by

`y = a x + b`

then the error at point `x_k` is `e_k = y_k - (a x_k + b)`

The accumulated quadratic error will be given by

`E(a,b) = sum_(k=1)^n e_k^2 =sum_(k=1)^n( y_k - (a x_k + b))^2`

`E(a,b)` have a minimum for `a_0,b_0` such that

`(partial E)/(partial a)(a_0,b_0) = 0-> b_0 sum x_k+a_0 sumx_k^2=sum x_ky_k`
`(partial E)/(partial b)(a_0,b_0) = 0 -> n b_0 + a_0 sum x_k =sum y_k`

solving for `a_0,b_0`

`a_0 = ( (sum x_k)( sum y_k)-n sum x_ky_k)/((sum x_k)^2-nsum x_k^2)`
`b_0 = ((sum x_k)(sum x_ky_k)-(sum x_k^2)(sum y_k))/((sum x_k)^2-nsum x_k^2`

appliying it to the table

`{{20, 175}, {30, 166}, {40, 157}, {50, 148}, {60, 139}, {70, 130}}`

we obtain:

`a_0 = -0.9, b_0 = 193`

so the adjusting line is

`y = -0.9 x + 193`