The population of Sasquatch in Salt Lake County was modeled by the function $$ P(t)=\frac{400 t}{t+40} $$, where $$ t=0 $$ represents the year 1803 . When were there fewer than 200 Sasquatch in Salt Lake County? Present your answer, to the nearest whole year, in interval notation. There were fewer than 200 Sasquatch in Salt Lake City between the years

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To determine when the population of Sasquatch in Salt Lake County was fewer than 200, we need to solve the inequality:

$$
P(t) < 200
$$

Given the population function:

$$
P(t) = \frac{400t}{t + 40}
$$

we set up the inequality:

$$
\frac{400t}{t + 40} < 200
$$

To eliminate the fraction, we can multiply both sides by $$ t + 40 $$ (noting that $$ t + 40 > 0 $$ for $$ t \geq 0 $$):

$$
400t < 200(t + 40)
$$

Expanding the right side:

$$
400t < 200t + 8000
$$

Now, we can isolate $$ t $$:

$$
400t - 200t < 8000
$$
$$
200t < 8000
$$
$$
t < 40
$$

This means that the population of Sasquatch was fewer than 200 for $$ t < 40 $$.

Since $$ t = 0 $$ corresponds to the year 1803, we can find the corresponding year for $$ t = 40 $$:

$$
1803 + 40 = 1843
$$

Thus, the population was fewer than 200 Sasquatch from $$ t = 0 $$ to $$ t = 40 $$, which corresponds to the years:

$$
1803 \text{ to } 1843
$$

In interval notation, this is represented as:

$$
(1803, 1843)
$$

Therefore, the final answer is:

$$
\text{There were fewer than 200 Sasquatch in Salt Lake County between the years } (1803, 1843).
$$
There were fewer than 200 Sasquatch in Salt Lake County between the years (1803, 1843).

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