Question

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 1954. The percent was 13 in 1974 and 79 in 2010.a. what is the slope of the line joining points (18,19) and (54,72)b. what is the average rate of change in the percent of teenage out of wedlock births over this period?c. Use the slope from part a and the number of teenage mothers in 2010 to write the equation of the line.

Answer 1

Step-by-step explanation:

a) The slope of the line joining points (18,19) and (54,72):

$(x_1,y_1)(x_2,y_2)=(18,19) and (54,72)$

Slope of the line , m= $\frac{y_2-y_1}{x_2-x_1}$

The slope line passing through these points is :

$m=\frac{72-19}{54-18}=1.472$

The slope of the line joining points (18,19) and (54,72) is 1.472.

b) The average rate of change in the percent of teenage out of wedlock births over this period will be equal to the slope of the line which represent change of percentage of with respect to time period.

The average rate of change in the percent of teenage out of wedlock births over this period is 1.472.

c)The equation of the line:

$(y-y_1)=m\times (x-x_1)$

we have :

Slope of the line: $m = \frac{53}{36}$

(year, percentage)

$(x_1,y_1)=(56,79)$

$(y-79)=\frac{53}{36}\times (x-56)$

$y-79=\frac{53}{36}x-\frac{2,968}{36}$

$y=\frac{53}{36}x-\frac{2,968}{36}+79$

$y=\frac{53}{36}x-\frac{124}{36}$

The equation of the line is : $y=\frac{53}{36}x-\frac{124}{36}$