Question

Q1 There are 170 deer on a reservation. The deer population is increasing at a rate of 30% each year. The table shows the number of deer, y, after x years.

Answer 1

To calculate the deer population over the years given that the population starts at 170 deer and increases at a rate of 30% per year, we can use the formula for exponential growth:

\[
y = y_0 \times (1 + r)^x
\]

Where:
- \( y \) is the population after \( x \) years,
- \( y_0 \) is the initial population (170 deer),
- \( r \) is the growth rate (30% or 0.30),
- \( x \) is the number of years.

For your scenario:

- Initial population \( y_0 = 170 \)
- Growth rate \( r = 0.30 \)

So the formula becomes:

\[
y = 170 \times (1 + 0.30)^x
\]
\[
y = 170 \times (1.30)^x
\]

Now we can calculate the population for different values of \( x \):

- For \( x = 0 \) (initial population):
\[
y = 170 \times (1.30)^0 = 170 \times 1 = 170
\]

- For \( x = 1 \) (after 1 year):
\[
y = 170 \times (1.30)^1 = 170 \times 1.30 = 221
\]

- For \( x = 2 \) (after 2 years):
\[
y = 170 \times (1.30)^2 \approx 170 \times 1.69 = 287.3 \approx 287
\]

- For \( x = 3 \) (after 3 years):
\[
y = 170 \times (1.30)^3 \approx 170 \times 2.197 = 373.49 \approx 373
\]

- For \( x = 4 \) (after 4 years):
\[
y = 170 \times (1.30)^4 \approx 170 \times 2.857 = 485.69 \approx 486
\]

- For \( x = 5 \) (after 5 years):
\[
y = 170 \times (1.30)^5 \approx 170 \times 3.7129 = 631.19 \approx 631
\]

In table format, you'll have:

| Year (x) | Deer Population (y) |
|----------|---------------------|
| 0 | 170 |
| 1 | 221 |
| 2 | 287 |
| 3 | 373 |
| 4 | 486 |
| 5 | 631 |

Feel free to ask if you have any other questions or need further calculations!