Problem
The percent of births to teenage mothers that are out-of-wedlock can be approximated by a linear function of the number of years after 1953. The percent was 16 in 1970 and 72 in 2006 Complete parts (a) through (c) (a) What is the slope of the line joining the points $(17,16)$ and $(53,72)$ ? The slope of the line is 1.56. (Simplify your answer: Round to two decimal places as needed) (b) What is the average rate of change in the percent of teenage out-of-wedlock births over this period? The average rate of change in the percent of teenage out-of-wedlock births over this period is 1.56 . (Simplify your answer. Round to two decimal places as needed) (c) Use the slope from part (a) and the number of teenage mothers in 2006 to write the equation of the line The equation is $p=$ (Do not factor. Type an expression using $x$ as the variable)
Solution
Step 1 ：Given two points (17,16) and (53,72), we can find the slope of the line joining these points using the formula (y2 - y1) / (x2 - x1).
Step 2 ：Substitute the given points into the formula: slope = (72 - 16) / (53 - 17).
Step 3 ：Calculate the slope: slope = 1.5555555555555556.
Step 4 ：Round the slope to two decimal places: slope = \(\boxed{1.56}\).
