The equation $$ N(m)=65(Q)^{\frac{n}{4}} $$ gives the predicted population $$ N(m) $$, in thousands, of a certain bacteria colony $$ m $$ minures after the initial measurement, where $$ Q $$ is a constant greater than 1 . The predicted population increases by $$ p \% $$ every 120 seconds. What is the value of $$ p $$ in terms of $$ Q $$ ? (A) $$ 100\left(Q^{\frac{1}{2}}+1\right) $$ (B) $$ 100\left(Q^{30}+1\right) $$ (C) $$ 100\left(Q^{\frac{1}{2}}-1\right) $$ (D) $$ 100\left(Q^{30}-1\right) $$ (DCO (C)

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To determine the value of $$ p $$ in terms of $$ Q $$, let's analyze the given information step by step.

**Given:**
- The population of the bacteria colony is modeled by:
  $$
  N(m) = 65 \cdot Q^{\frac{m}{4}}
  $$
  where $$ N(m) $$ is in thousands and $$ m $$ is the time in minutes.
- The population increases by $$ p\% $$ every 120 seconds (which is equivalent to 2 minutes).

**Objective:**
Find the expression for $$ p $$ in terms of $$ Q $$.

**Solution:**

1. **Understand the Growth Over 2 Minutes:**
   The population increases by $$ p\% $$ every 2 minutes. This means that after 2 minutes, the population becomes:
   $$
   N(m + 2) = N(m) \cdot \left(1 + \frac{p}{100}\right)
   $$

2. **Use the Given Population Model:**
   Substitute $$ m + 2 $$ into the population model:
   $$
   N(m + 2) = 65 \cdot Q^{\frac{m + 2}{4}} = 65 \cdot Q^{\frac{m}{4}} \cdot Q^{\frac{2}{4}} = N(m) \cdot Q^{\frac{1}{2}}
   $$

3. **Set the Two Expressions for $$ N(m + 2) $$ Equal:**
   $$
   N(m) \cdot \left(1 + \frac{p}{100}\right) = N(m) \cdot Q^{\frac{1}{2}}
   $$
   Divide both sides by $$ N(m) $$:
   $$
   1 + \frac{p}{100} = Q^{\frac{1}{2}}
   $$

4. **Solve for $$ p $$:**
   $$
   \frac{p}{100} = Q^{\frac{1}{2}} - 1
   $$
   $$
   p = 100 \left(Q^{\frac{1}{2}} - 1\right)
   $$

**Answer:**
(C) $$ 100\left(Q^{\frac{1}{2}}-1\right) $$
$$ p = 100\left(Q^{\frac{1}{2}} - 1\right) $$

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