Question

The population of the Wallingfried neighborhood has been growing steadily since 1981. In 1988, the population was 36100 people. In 1994, it was 43900 people.

Find an equation in the form y=mx+b, where x is the number of years past 1981 and y is the population of Wallingfrie.

Answer 1

The equation that represents the population of Wallingfried neighborhood is y = 1300x + 1396000.

Step-by-step explanation:

To find the equation of the form y = mx + b that represents the population of the Wallingfried neighborhood, we need to determine the values of m (slope) and b (y-intercept). Since x represents the number of years past 1981 and y represents the population, we can use the two given data points to find the equation.

Let's use the point (1988, 36100) and (1994, 43900) to find m and b.

Step 1:

Calculate the change in population (Δy) and change in years (Δx).

Δy = 43900 - 36100 = 7800

Δx = 1994 - 1988 = 6

Step 2:

Find the value of m (slope).

m = Δy / Δx = 7800 / 6 = 1300

Step 3:

Substitute the slope and one of the data points into the equation y = mx + b to solve for b (y-intercept).

36100 = 1300 * 1988 + b

b = 36100 - (1300 * 1988)

Step 4:

Write the equation y = mx + b using the values of m and b.

y = 1300x + 1396000

Therefore, the equation that represents the population of Wallingfried neighborhood is y = 1300x + 1396000.

Answer 2

y=1300x+27000

Step-by-step explanation:

given that the population of the Wallingfried neighborhood has been growing steadily since 1981.

In 1988,the population was 36100 people. In 1994, it was 43900 people.

x is the number of years past 1981 and y , the population of Wallingfrie.

we have two points on the line as

when = 13, y = 43900

and when x = 7, y =36100

i.e. (13,43900) and (7,36100) lie on the line

Using two point formula we get the line as

`(y-36100)/(43900-36100) =(x-7)/(13-7) \\y-36100 =1300 (x-7)\\y =1300x +27000`

is the equation of the line