An eager graduate student at Carleton, counts geese in the marshes off the Ottawa river. In 2004, ...

Sorry for the late reply, Katie b987

If you're still interested in the rest of the answer, let me know. I've gone ahead and done (a) for you.

If you need the rest, reply back:

Hi could you do the rest please! bio_man

I think you'd need to set-up an exponential equation:

y = initial population * growth factor ^ time
y = 2500 (1.126) ^ t

2 years:

y = 2500 (1.126) ^ 2 = 3169

4 years:

y = 2500 (1.126) ^ 4 = 4018

8 years:

y = 2500 (1.126) ^ 8 = 6460


I don't understand this one. I don't know know what "r" means. If I had to guess, it'd assume that r presents the 'growth rate'? In that case,

y = initial population * growth factor ^ time
5000 = 2500 (1.1725) ^ t

2 = (1.1725) ^ t
log [ 2 ] = log [ 1.1725 ^ t ]
t = 4.4 years

Again, I'm just speculating here. You might just be expected to use the formula:

td = ln(2) / r
td = ln(2) / 0.1725
td = 4 years


I think you're supposed to use the formula y = 2500 (1.126) ^ t again, and set 4235 as y to solve for t

4235 = 2500 (1.126) ^ t

4235/2500 = 1.126 ^ t

ln [ 4235/2500 ] = ln [ 1.126 ^ t ]

ln [ 4235/2500 ] / ln [ 1.126 ] = 4.46 years or 4.5 years

Are you sure about (a)?


In 2004, it was 2500

In 2005 = 2500 + 500 - 45 - 200 - 30 = 2725 geese.

Population change = 2725 - 2500= 225

% = (225/2500) x 100 = 9%

You did your calculation wrong:

In 2005 = 2500 + 500 + 45 - 200 - 30

That's why you got a different percentage than me.

Please review this resource to help you understand this a little bit more: https://biology-forums.com/index.php?action=downloads;sa=view;down=14509