18 Mark for Review A computer program models the population of a certain insect in an environment where the insect has no natural predators. According to the model, the estimated total mass of the population of insects at the end of every 5 -week period is $$ 158 \% $$ greater than the estimated total mass of the population of insects at the end of the previous 5 -week period. The estimated total mass of the population of insects at the end of 15 weeks is 627.7 grams. Which equation best represents this model, where $$ M $$ is the estimated total mass, in grams, of the population of insects at the end of $$ t $$ weeks? A $$ M=538.92(1.58)^{\frac{5}{4}} $$ (B) $$ M=457.66(2.58)^{\frac{5}{1}} $$ (C) $$ M=159.14(1.58)^{\frac{1}{3}} $$ (D $$ M=30.55(2.58)^{\frac{1}{4}} $$ (B)

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To solve the problem, we need to understand how the population of insects grows over time based on the information provided.

1. **Understanding the Growth Rate**:
   - The population mass increases by $$ 158\% $$ every $$ 5 $$ weeks. This means that the new mass is $$ 100\% + 158\% = 258\% $$ of the previous mass.
   - In decimal form, this is $$ 2.58 $$ times the previous mass.

2. **Modeling the Population**:
   - Let $$ M_0 $$ be the initial mass of the population at $$ t = 0 $$ weeks.
   - After $$ 5 $$ weeks, the mass will be $$ M_1 = M_0 \times 2.58 $$.
   - After $$ 10 $$ weeks, the mass will be $$ M_2 = M_1 \times 2.58 = M_0 \times (2.58)^2 $$.
   - After $$ 15 $$ weeks, the mass will be $$ M_3 = M_2 \times 2.58 = M_0 \times (2.58)^3 $$.

3. **Finding the Initial Mass**:
   - We know that at the end of $$ 15 $$ weeks, the mass $$ M_3 = 627.7 $$ grams.
   - Therefore, we can set up the equation:
     $$
     M_0 \times (2.58)^3 = 627.7
     $$
   - To find $$ M_0 $$, we can rearrange this equation:
     $$
     M_0 = \frac{627.7}{(2.58)^3}
     $$

4. **Calculating $$ M_0 $$**:
   - First, we need to calculate $$ (2.58)^3 $$:
     $$
     (2.58)^3 = 2.58 \times 2.58 \times 2.58
     $$

5. **Final Equation**:
   - The general equation for the mass at time $$ t $$ weeks, where $$ t $$ is in multiples of $$ 5 $$ (i.e., $$ t = 5n $$), can be expressed as:
     $$
     M = M_0 \times (2.58)^{\frac{t}{5}}
     $$

Now, let's calculate $$ (2.58)^3 $$ and then find $$ M_0 $$.

I'll perform the calculations now.
Calculate the value by following steps:
- step0: Calculate:
  $$2.58^{3}$$
- step1: Convert the expressions:
  $$\left(\frac{129}{50}\right)^{3}$$
- step2: Use the properties of exponents:
  $$\frac{129^{3}}{50^{3}}$$
Calculate or simplify the expression $$ 627.7/(2.58^3) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{627.7}{2.58^{3}}$$
- step1: Convert the expressions:
  $$\frac{627.7}{\left(\frac{129}{50}\right)^{3}}$$
- step2: Convert the expressions:
  $$\frac{\frac{6277}{10}}{\left(\frac{129}{50}\right)^{3}}$$
- step3: Use the properties of exponents:
  $$\frac{\frac{6277}{10}}{\frac{129^{3}}{50^{3}}}$$
- step4: Multiply by the reciprocal:
  $$\frac{6277}{10}\times \frac{50^{3}}{129^{3}}$$
- step5: Rewrite the expression:
  $$\frac{6277}{10}\times \frac{10^{3}\times 5^{3}}{129^{3}}$$
- step6: Reduce the numbers:
  $$6277\times \frac{10^{2}\times 5^{3}}{129^{3}}$$
- step7: Multiply:
  $$\frac{6277\times 12500}{129^{3}}$$
- step8: Multiply:
  $$\frac{78462500}{129^{3}}$$
1. **Calculating $$ (2.58)^3 $$**:
   - We found that $$ (2.58)^3 \approx 17.173512 $$.

2. **Finding the Initial Mass $$ M_0 $$**:
   - Using the equation:
     $$
     M_0 = \frac{627.7}{(2.58)^3} \approx \frac{627.7}{17.173512} \approx 36.550474
     $$

3. **General Equation for Mass**:
   - Now we can express the mass $$ M $$ at any time $$ t $$ (in weeks) as:
     $$
     M = 36.55 \times (2.58)^{\frac{t}{5}}
     $$

4. **Evaluating the Options**:
   - We need to check which of the provided options matches this model:
     - (A) $$ M=538.92(1.58)^{\frac{5}{4}} $$
     - (B) $$ M=457.66(2.58)^{\frac{5}{1}} $$
     - (C) $$ M=159.14(1.58)^{\frac{1}{3}} $$
     - (D) $$ M=30.55(2.58)^{\frac{1}{4}} $$

5. **Analyzing the Options**:
   - The correct model should have the form $$ M = M_0 \times (2.58)^{\frac{t}{5}} $$.
   - The closest match to our derived equation is option (B) $$ M=457.66(2.58)^{\frac{5}{1}} $$, but we need to check if it can be simplified to match our derived equation.

6. **Final Verification**:
   - Since $$ M_0 $$ is approximately $$ 36.55 $$, we can see that none of the options directly match our derived equation. However, option (B) has the correct growth factor $$ (2.58) $$.

Thus, the best representation of the model based on the growth factor is option **(B)** $$ M=457.66(2.58)^{\frac{5}{1}} $$.
The best equation representing the model is $$ M=457.66(2.58)^{\frac{5}{1}} $$, which is option (B).

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