Question

2. Suppose the population of deer in a region was 3500 in the year 2000. Since then the population has grown by 3.5% annually. What will the approximate population be in the year 2020?

Answer 1

We can use the formula for exponential growth:

\[y=a(1+b)^{x}\]


Y is the answer to your question.

A is the starting population of deer.
(In this case, we would 3500.)

B is the percent change, in decimal form.
(If the deer grow 3.5% every year,
then the decimal form would be 0.035.)

X is the amount of time passed.
(In this case, that would be 20 years,
because that's how long it is, from 2000 to 2020.)

~~~~~~~~~~~~~

You see?

It's a very simple equation,
and all we have to do is plug in the numbers into the right places,
which I've explained there.

If we plug in all the numbers,
and solve for Y,
we will find out how many deer there would be, by the year 2020.

Figure out the answer, and show me.
I'll check it for you.

Answer 2

i dont get it

Answer 3

like i tried solving but it dident come out right

Answer 4

Can you show me the equation that you tried to solve?

Answer 5

y=3,500(1+0.035)^20

Answer 6

Right. That's the right equation.

Remember PEMDAS?

Parentheses
Exponents
Multiplication or Division
Addition or Subtraction

We have to make the equation smaller and smaller,
in that order.

So first, let's take care of the parentheses.

What's 1 + 0.035 = ?

Answer 7

1.035

Answer 8

so now its y=3,500(1.035)^20

Answer 9

Right.

Now, let's take care of the EXPONENT.

The EXPONENT is 20.

What's 1.035 to the 20th power?

Do you have a scientific calculator?

Answer 10

no

Answer 11

1.9898

Answer 12

Well that's the right answer.

So now, its...

y = 3500(1.9898)

Answer 13

That's 3,500 times 1.9898.

Answer 14

6964.3

Answer 15

Yep.

There's our answer.

If the deer population increases by 3.5% every year,
then there should be about 6964 deer,
by the year 2020.

Answer 16

i need help with 2 more

Answer 17

If you'd like, you can make a separate post.

I'll come along and take a look. I'll help you if I think I can.

And if not, then the question will be there for someone else to come along and help.

Answer 18

ok