Starting with an initial value of $$ P(0)=25 $$, the population of a prairie dog community grows at a rate of $$ P^{\prime}(t)=30-\frac{t}{5} $$ (in units of prairie dogs/month), for $$ 0 \leq t \leq 150 $$. a. What is the population 8 months later? b. Find the population $$ P(t) $$ for $$ 0 \leq t \leq 150 $$.

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To solve the problem, we will follow these steps:

### Part a: Find the population 8 months later

1. **Given Information:**
   - Initial population: $$ P(0) = 25 $$
   - Rate of change of population: $$ P'(t) = 30 - \frac{t}{5} $$

2. **Find $$ P(8) $$:**
   - We need to integrate $$ P'(t) $$ to find $$ P(t) $$.

### Part b: Find the population $$ P(t) $$ for $$ 0 \leq t \leq 150 $$

1. **Integrate $$ P'(t) $$:**
   $$
   P(t) = \int P'(t) \, dt = \int \left(30 - \frac{t}{5}\right) dt
   $$

2. **Calculate the integral:**
   - The integral of $$ 30 $$ is $$ 30t $$.
   - The integral of $$ -\frac{t}{5} $$ is $$ -\frac{t^2}{10} $$.
   - Therefore, we have:
   $$
   P(t) = 30t - \frac{t^2}{10} + C
   $$

3. **Determine the constant $$ C $$:**
   - We know that $$ P(0) = 25 $$:
   $$
   P(0) = 30(0) - \frac{(0)^2}{10} + C = 25 \implies C = 25
   $$

4. **Final expression for $$ P(t) $$:**
   $$
   P(t) = 30t - \frac{t^2}{10} + 25
   $$

Now, we will calculate $$ P(8) $$ and confirm the expression for $$ P(t) $$.

Let's perform the calculations.

### Step 1: Calculate $$ P(8) $$

We will substitute $$ t = 8 $$ into the expression for $$ P(t) $$:
$$
P(8) = 30(8) - \frac{(8)^2}{10} + 25
$$

### Step 2: Calculate $$ P(t) $$

We will also confirm the expression for $$ P(t) $$ as derived above.

Let's perform these calculations now.
Evaluate the expression by following steps:
- step0: Evaluate:
  $$30t-\frac{t^{2}}{10}+25$$
- step1: Substitute:
  $$30\times 8-\frac{8^{2}}{10}+25$$
- step2: Divide the terms:
  $$30\times 8-\frac{2^{5}}{5}+25$$
- step3: Multiply the numbers:
  $$240-\frac{2^{5}}{5}+25$$
- step4: Evaluate the power:
  $$240-\frac{32}{5}+25$$
- step5: Add the numbers:
  $$265-\frac{32}{5}$$
- step6: Reduce fractions to a common denominator:
  $$\frac{265\times 5}{5}-\frac{32}{5}$$
- step7: Transform the expression:
  $$\frac{265\times 5-32}{5}$$
- step8: Multiply the numbers:
  $$\frac{1325-32}{5}$$
- step9: Subtract the numbers:
  $$\frac{1293}{5}$$
Calculate or simplify the expression $$ 30t - \frac{t^2}{10} + 25 $$.
Simplify the expression by following steps:
- step0: Solution:
  $$30t-\frac{t^{2}}{10}+25$$
- step1: Reduce fractions to a common denominator:
  $$\frac{30t\times 10}{10}-\frac{t^{2}}{10}+\frac{25\times 10}{10}$$
- step2: Transform the expression:
  $$\frac{30t\times 10-t^{2}+25\times 10}{10}$$
- step3: Multiply the terms:
  $$\frac{300t-t^{2}+25\times 10}{10}$$
- step4: Multiply the numbers:
  $$\frac{300t-t^{2}+250}{10}$$
### Results

1. **Population 8 months later:**
   - The population $$ P(8) $$ is calculated as:
   $$
   P(8) = 30(8) - \frac{(8)^2}{10} + 25 = \frac{1293}{5} \approx 258.6
   $$
   Thus, the population 8 months later is approximately **258.6 prairie dogs**.

2. **Expression for $$ P(t) $$:**
   - The population $$ P(t) $$ for $$ 0 \leq t \leq 150 $$ is given by:
   $$
   P(t) = \frac{300t - t^2 + 250}{10}
   $$

### Summary of Answers
- **a.** The population 8 months later is approximately **258.6 prairie dogs**.
- **b.** The population function is $$ P(t) = \frac{300t - t^2 + 250}{10} $$ for $$ 0 \leq t \leq 150 $$.
a. The population 8 months later is approximately 258.6 prairie dogs. b. The population function is $$ P(t) = \frac{300t - t^2 + 250}{10} $$ for $$ 0 \leq t \leq 150 $$.

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