Asked by Carlos
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:
Find an equation for the moose population, y,in terms of x, the years since 1994.
Find an equation for the moose population, y,in terms of x, the years since 1994.
Answers
Answered by
Bosnian
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
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
Answered by
carlos
thank you
Answered by
Carlos
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