the population of deer in a certain area of cabarrus county grows proportional to itself. the population of deer in 2000 was found to be 45, and by 2008 the population had grown to 65. Find the growth constant k(rounded to 6 decimal places), and the expected deer population in 2015. (Round to 2 decimal places)

Just need to know the formula and what to plug in where.

You know the growth function will look like

g(t) = a*e^(kt)
So, just plug in your numbers:

g(0) = a = 45
g(8) = 45e^(8k) = 65, so
k = ln(65/45)/8 = 0.046

Now just figure g(15)

wow, I actually did it right, I got .045 for k, and p(t) = 89.67.

I can't believe I actually did this problem correctly, I hope.(crosses fingers)

I meant .046 not .045

I got 89.67 by using 45e^(ln(65/45)/8*15)

To solve this problem, you can use the exponential growth formula:

P(t) = P(0) * e^(kt)

Where:
- P(t) represents the population at a given time t
- P(0) represents the initial population
- e is the base of the natural logarithm
- k is the growth constant
- t represents the time in years

To find the value of the growth constant, we can use the information given in the problem.

Given:
- In 2000, the population of deer, P(0), was 45.
- In 2008, the population of deer, P(8), was 65.

We can now substitute these values into the exponential growth formula:

65 = 45 * e^(k * 8)

To solve for k, divide both sides of the equation by 45:

65/45 = e^(k * 8)

1.44444 ≈ e^(k * 8)

Now, take the natural logarithm (ln) of both sides:

ln(1.44444) ≈ ln(e^(k * 8))

ln(1.44444) ≈ k * 8

Now divide both sides by 8 to solve for k:

ln(1.44444)/8 ≈ k

Using a calculator, you can find the value of ln(1.44444)/8 ≈ 0.018327.

Rounded to 6 decimal places, the growth constant k ≈ 0.018327.

To find the expected deer population in 2015, we can use the formula again:

P(15) = P(0) * e^(k * 15)

Substituting the known values:

P(15) = 45 * e^(0.018327 * 15)

Using a calculator to calculate e^(0.018327 * 15), the expected deer population in 2015 is approximately 96.27.

Therefore, rounded to 2 decimal places, the expected deer population in 2015 is 96.27.