A researcher collects data on the relationship between the amount of daily exercise an individual gets and the percent body fat of the individual. The following scores are recorded.

Individual 1 2 3 4 5
Exercise (min) 10 18 26 33 44
% Fat 30 25 18 17 14
Individual: 1, 2, 3, 4, 5

Excercise (min): 10, 18, 26, 33, 44
% Fat: 30, 25, 18, 17, 14
a) Based on the above data, if an individual exercises 20 minutes daily, his predicted % body fat would be

a. 21.63
b. 27.74
c. 27.88
d. 23.75

Option D is correct.

If exercise time = 20 mins,

Body fat = 23.75%

Step-by-step explanation:

After plotting a graph of the variables given, with body fat on the x-axis and exercise time on the y-axis.

The equation of the graph was obtained to be

Y = -1.931X + 65.936

If exercise time = 20 mins,

Y = 20

20 = -1.98X + 66.936

-1.98X = -46.936

X = 23.7 = 23.75% From the options.

The problem can also be solved by doing the least squares regression line method, but this is more straight forward.

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form