The salary of an employee in 1985 is $1200,in 1987 it will be $1350. Express salary as a linear function of time and estimate his salary in 1988.
y = 1200 + (1350-1200)/2 x
Now just plug in x=3 (3 years since 1985)
To express the salary as a linear function of time, we need to find the equation of a line that represents the relationship between the salary and time.
Let's assign the year 1985 as "0" and the year 1987 as "2".
Using the two given data points, we can set up a system of equations to find the slope and y-intercept of the line:
Point 1: (0, 1200)
Point 2: (2, 1350)
The slope of the line is given by the formula:
slope = (change in y) / (change in x)
slope = (1350 - 1200) / (2 - 0)
slope = 150 / 2
slope = 75
Now, we can use the point-slope form of a line to find the y-intercept:
y - y1 = m(x - x1)
Using point 1:
y - 1200 = 75(x - 0)
y - 1200 = 75x
y = 75x + 1200
Therefore, the linear function expressing the salary as a function of time is:
Salary = 75x + 1200
To estimate the employee's salary in 1988, we need to assign 1988 as "3" and substitute it into the equation:
Salary = 75 * 3 + 1200
Salary = 225 + 1200
Salary = $1425
Therefore, the estimated salary of the employee in 1988 is $1425.
To express the salary as a linear function of time, we can use the formula for a linear equation: y = mx + b. In this case, y represents the salary, x represents the time, m represents the rate of change, and b represents the initial value.
We are given two data points: in 1985, the salary is $1200, and in 1987, the salary will be $1350. We can use these points to find the values of m and b.
Let's start by finding the rate of change (m). The rate of change represents how much the salary increases per year. We can calculate it using the formula:
m = (y2 - y1) / (x2 - x1), where (x1, y1) represents the first data point and (x2, y2) represents the second data point.
Using the given data:
x1 = 1985, y1 = 1200
x2 = 1987, y2 = 1350
m = (1350 - 1200) / (1987 - 1985)
= 150 / 2
= 75
Now that we have the value of m, we can use it to find the initial value (b). We can substitute the values of one of the data points into the equation and solve for b:
y = mx + b
1200 = 75 * 1985 + b
Simplifying this equation:
1200 = 149625 + b
b = 1200 - 149625
b = -148425
Now we have the values of m and b, and we can write the linear function for the salary:
y = 75x - 148425
To estimate the salary in 1988 (one year after 1987), we substitute x = 1988 into the equation:
y = 75 * 1988 - 148425
Simplifying this expression:
y = 149100 - 148425
y ≈ 675
Therefore, the estimated salary in 1988 is $675.