Solve the problem.
-Assume that the situation described can be modeled by a straight line graph. Use the given information to find the y = mx + b form of the equation of the line. The number of births in County A has been increasing in recent
Years. Use the information given on the bar graph for the years 1960 and 1990. Let x = 0 represent the year 1960
And y represent the number of births.
A) $y = \frac { 30 } { 497 } x + \frac { 930,140 } { 497 }$
B) $y = \frac { 1990 } { 497 } x - 2190$
C) $y = \frac { 497 } { 1990 } x + 1693$
D) $y = \frac { 497 } { 30 } x + 1693$

Question

Solve the problem.
-Assume that the situation described can be modeled by a straight line graph. Use the given information to find the y = mx + b form of the equation of the line. The number of births in County A has been increasing in recent
Years. Use the information given on the bar graph for the years 1960 and 1990. Let x = 0 represent the year 1960
And y represent the number of births. 
A) $y = \frac { 30 } { 497 } x + \frac { 930,140 } { 497 }$ 
B) $y = \frac { 1990 } { 497 } x - 2190$ 
C) $y = \frac { 497 } { 1990 } x + 1693$ 
D) $y = \frac { 497 } { 30 } x + 1693$

Quizplus · Accepted Answer

The Answer of Solve the problem.
-Assume that the situation described can be modeled by a straight line graph. Use the given information to find the y = mx + b form of the equation of the line. The number of births in County A has been increasing in recent
Years. Use the information given on the bar graph for the years 1960 and 1990. Let x = 0 represent the year 1960
And y represent the number of births. 
A) $y = \frac { 30 } { 497 } x + \frac { 930,140 } { 497 }$ 
B) $y = \frac { 1990 } { 497 } x - 2190$ 
C) $y = \frac { 497 } { 1990 } x + 1693$ 
D) $y = \frac { 497 } { 30 } x + 1693$

Solve the Problem y=30497x+930,140497y = \frac { 30 } { 497 } x + \frac { 930,140 } { 497 }y=49730​x+497930,140​

Solve the Problem $y = \frac { 30 } { 497 } x + \frac { 930,140 } { 497 }$