00:01
So in this problem, we're given this function for p of t, where it represents the number of fish in terms of thousands, and t is measured in terms of years.
00:09
So in part a of this problem, what we want to do is find the fish population after three years.
00:14
Well, that would mean that t is equal to three.
00:16
So all we're doing is substituting three in place of t.
00:19
So you have 10 divided by 1 plus 4e to the negative 0 .8 times 3 power.
00:26
And now this is something we do in our calculator.
00:29
Don't forget.
00:30
To make sure your entire denominator is in the parentheses.
00:33
So we'll have p of 3 equal to, and this is how you'll type it in.
00:37
10 divided by, open parentheses, 1 plus 4, e, and that should put you with an exponent, and then you're going to do negative 0 .8 times 3.
00:47
And then you want to make sure that you close your parentheses.
00:50
And in this case, they don't tell us what two round two, so i'm just going to round to nearest whole number because we're talking about fish.
00:56
Or actually, i'll round three places after the decimal.
00:59
So that would be 7.
01:00
3, 3, 7.
01:02
So that's how many fish in terms of thousands.
01:04
I should put that.
01:06
That's how many fish in terms of thousands you would expect after three years.
01:11
All right.
01:11
Next, let's go to part b.
01:14
So in part b, we want to know after how many years will the fish population reach 5 ,000.
01:20
Well, that would mean that's our p of t value.
01:22
So we're going to have 5 ,000 equal to 10 divided by 1 plus 4e to the negative 0 .8 t -p and now what we need to do is solve this equation for t.
01:35
So first, we have to multiply both sides of our equation by our denominator.
01:40
So 1 plus 4e to the negative 0 .8t power.
01:44
And again, we're going to do it on the left -hand side as well.
01:49
Because on the right -hand side, they'll cancel, so we'll just have 10...