Question

Since I can't delete my previous question, can someone help me with these problems? I couldn't edit my last question so I am putting it in a new post in hopes someone can explain this because I keep getting the answers wrong and have to generate a new question.

Answer 1

XDXD you should be able to edit it

Answer 2

Answer:

The equation is P=420t+6020

What happens in 2009?

15260

Step-by-step explanation:

So t represents the number of years after 1987.

So at t=1, the year is 1987+1=1988.

So at t=2, the year is 1987+2=1989.

And so on...

In year 1991, the moose population was 7700. In other words, at t=1991-1987=4, the population is 7700.

In year 1996, the moose population was 9800. In other words, at t=1996-1987=9, the moose population is 9800.

So we want to find an equation for the line going through (4,7700) and (9,9800).

First step, calculate the slope.

Line up points and subtract vertically.

Afterwards put 2nd difference on top of first difference.

(9,9800)

-(4,7700)

---------------

5, 2100

So the slope is 2100/5=420.

So using point-slope form [y-y1=m(x-x1)] the equation is:

y-7700=420(x-4)

Distribute 420 to terms in ( ):

y-7700=420x-1680

Add 7700 on both sides:

y=420x-1680+7700

Simplify:

y=420x+6020

_____________________

2009 is 2009-1987=22 years after 1987 so we are looking at t=22.

Replace x with 22 in y=420x+6020:

y=420(22)+6020

y=15260

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One last thing.

y=420x+6020

I'm going to write y as P and x as t:

P=420t+6020