In 1994 the moose population in a park was measured to be 6400. By 1996, the population was measured again to be 6800. If the population continues to change linearly:

Question

Find an equation for the moose population, y,in terms of x, the years since 1994.

Answer 1

The two-point form of a straight line passing through the points ( x1, y1 ) and ( x2 , y2 ) is given by:

y − y1 = ( y2 − y1 ) * ( x − x 1 ) / ( x2 − x1 )

Ini this case:

x1 = 1994 , x2 = 1996

y1 = 6400 , y2 = 6800

So:

y − y1 = ( y2 − y1 ) * ( x − x 1 ) / ( x2 − x1 )

y − 6400 = ( 6800 − 6400 ) * ( x − 1994 ) / ( 1996 − 1994 )

y − 6400 = 400 * ( x − 1994 ) / 2

y − 6400 = 200 * ( x − 1994 )

y − 6400 = 200 * x − 200 * 1994

y − 6400 = 200 x − 398800 Add 6400 to both sides

y − 6400 + 6400 = 200 x − 398800 + 6400

y = 200 x - 392400

Proof:

x = 1994

y = 200 x - 392400 = 200 * 1994 - 392400 = 398800 - 392400 = 6400

x = 1996

y = 200 x - 392400 = 200 * 1996 - 392400 = 399200 - 392400 = 6800

Answer 2

thank you

Answer 3

Now it is asking what will be the prediction for 2006?