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In 1994, the moose population in a park was measured to be 3340..

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4 Answers

This is similar to asking for the equation of a straight line given 2 points: (1994, 3340) and (1996,3200)

Slope m = (3340 - 3200) / (1994 - 1996) as given by the formula y2-y1 / x2-x1 which works out to -70.

Equation of the line then can be given by plugging this slope into y-y1 = m (m-x1) given by (y-3340) = -70 (x-1994). Note: You could use either point (1994, 3340) or (1996,3200) in the previous equation and you would still get the same answer.

This can be simplied into y = -70x +142920 (where y is the # Moose and x is the year)  or

P(t) = -70t + 142920 (to put this line equation in the form that has been asked for in the question).

Plugging in t=2009, P(t) = 2290 which is the moose population in the year 2009.

This seems plausible as we can see that the moose population seems to be decreasing with time.

 

 

 

 

 

Comments

isn't 142,920 a little too high for the population in 1990?

Ralph,

142,920 is the population at t=0 AD (when we started time in years)

If you wanted to find out how many moose are there in the year 1990, you would need to perform the following calculation: 142920 - 70*1990 = 3620.

Let me know if you have any additional questions!

 

but the rest of us got 3620 for t(0) or population in 1990?

 

Ralph, I am not sure I understand your question. In 1990 the population is 3620 which would be t(1990) not t(0).

Let me know what you think. 

Thanks,

Nandini

What I'm trying to say is that I used t=0 for 1990 so that each succeeding year from 1990 is t = 1, t =2, t=3, and so on. Since the problem stated: in terms of, "t", the years since 1990. P(t)= ????

Comment

So we know this from the problem:

Year |Moose Population

1994|3340

1996|3200

We are told to assume linear population change, so we can conclude from the data that the moose population will decrease by a steady 60 every year. Now, we should find the moose population for years since 1990. Since 1994 is 4 years later than that time and the moose population is 3340 then, the population for 1990 should be 3340 + 4 * 70 or 3620. 

So our y-intercept is 3620. With the decrease of 70 moose every year, the equation should then be P(t) = 3620 - 70t, where t is the number of years after 1990.

 

t is year since 1990

p(t) is population at time t

(t, p(t))

since we start at 1990, at 1994 t is equal to 1994 - 1990 which is 4

so, at 1996, t is equal to 1996 - 1990 which is 6

so first point is (4, 3340)

the second point (6, 3200)

so to get the slope we use  Δp(t)/Δt --->( 3200 - 3340)/(6 - 4) ---> -70

so, we have the equation p(t) = -70t + b

we use one point to figure out b:

3200 = -70(6) + b ---> b = 3200 + 70 * 6 = 3620

so we have the equation p(t) = -70t + 3620

so, at 2009, t will be 19. Plug in 19 to t to get:

p(19) = -70*(19) + 3620

so, p(19) = 2290

 

 

 

Formula for linear growth is f(x) = mx+b

m is the slope or in this case the population growth/decline.  x is the number of years and b is the original population.

in this case it will be p(t) = mt+b 

lets find the population decline m.

In 1996 the population was 3200.  This is a decline of 140 moose over 2 years. We need to know how many moose decline each year, so divide 140/2 and you get 70.

70 moose are declining each year.  So m = 70, but it's declining so we need a negative.  m = -70

So lets put it in the equation:

p(t) = -70t + b

So now it's easy to find b.  We just plug in the year and population and get our b.  Let's use 1996.  The time has been 6 years from 1990 and the population in 1996 is 3200.

use the formula p(t) = -70(t) + b

t = 6 

p(6) = 3200.  

So p(6) = -70(6) + b = 3200

p(6) = -420 + b = 3200

-420 + b = 3200

b = 3620

Now you can plug in the years for t and get the decline in population.

We need the year 2009.  So 2009-1990 = 19.  Our t=19.

p(19) = -70(19) + 3620 = -1330 + 3620 = 2290.

 

Comments

Hi there. But "3340" is not a y-intercept for this problem.

Comment