Solve. Assume the exercise describes a linear relationship. When writing a linear equation, write the equation in
slope-intercept form.
-An investment is worth $3038 in 1991. By 1996 it has grown to $4908. Let y be the value of the investment in the year x, where x = 0 represents 1991. Write a linear equation that models the value of the investment in the
Year x.
A)y = $\frac { 1 } { 374 }$ 374x + 3038
B)y = -374x + 3038
C)y = -374x + 6778
D)y = 374x + 3038

Question

Quizplus · Accepted Answer

The value of the investment increases from $3038 in 1991 (x = 0) to $4908 in 1996 (x = 5). The increase in value is $4908 - $3038 = $1870 over 5 years. The slope (rate of change) is $1870 / 5 = $374 per year. Thus, the equation is y = 374x + 3038, where 374 is the slope and 3038 is the y-intercept, representing the value of the investment in the base year 1991.

Solve Assume the Exercise Describes a Linear Relationship 1374\frac { 1 } { 374 }3741​

Solve Assume the Exercise Describes a Linear Relationship $\frac { 1 } { 374 }$