Answered

Sand spills out of a hole in a sandbag at a constant rate. The graph and table show how the weight of the sandbag changes with time.

Which function models this situation?

\begin{tabular}{|c|c|c|}
\hline
Time (minutes) & 4 & 5 \\
\hline
Weight (pounds) & 36 & 34 \\
\hline
\end{tabular}

A. [tex]\( a(n) = 42 - 2^{n-1} \)[/tex]
B. [tex]\( a(n) = 42 - 2^{n+1} \)[/tex]
C. [tex]\( a(n) = 42 - (n-1) \cdot 2 \)[/tex]
D. [tex]\( a(n) = 42 - n \cdot (2-1) \)[/tex]



Answer :

To find the function that models the situation, we need to analyze how the sandbag's weight changes over time. We are given a table with the following information:

[tex]\[ \begin{tabular}{|c|c|c|} \hline Time (minutes) & 4 & 5 \\ \hline Weight (pounds) & 36 & 34 \\ \hline \end{tabular} \][/tex]

From the table, we can see that:

- At 4 minutes, the weight of the sandbag is 36 pounds.
- At 5 minutes, the weight of the sandbag is 34 pounds.

Let's break down the steps to find a function that models the situation.

### Step 1: Calculate the Rate of Weight Loss

First, determine the change in weight and the corresponding time interval:

- Weight change: [tex]\(36 - 34 = 2\)[/tex] pounds.
- Time interval: [tex]\(5 - 4 = 1\)[/tex] minute.

Now we calculate the rate of weight loss:
[tex]\[ \text{Rate of weight loss} = \frac{\text{Weight change}}{\text{Time interval}} = \frac{2 \text{ pounds}}{1 \text{ minute}} = 2 \text{ pounds per minute}. \][/tex]

### Step 2: Find the Initial Weight

The weight of the sandbag decreases linearly at a constant rate. To model this with a linear function, we can use the formula:
[tex]\[ \text{Weight} = \text{Initial weight} - \text{Rate} \times \text{Time}.\][/tex]

We already know the weight of the sandbag at a specific time (4 minutes):
[tex]\[ 36 = \text{Initial weight} - 2 \times 4. \][/tex]

Rearranging to solve for the initial weight:
[tex]\[ \text{Initial weight} = 36 + 2 \times 4 = 36 + 8 = 44 \text{ pounds}. \][/tex]

Thus, the initial weight of the sandbag is 44 pounds.

### Step 3: Formulate the Function

Now we form the linear function using the initial weight and the rate. We have:
[tex]\[ \text{Weight} = 44 - 2 \times \text{Time}. \][/tex]

Rewriting this in terms of [tex]\(n\)[/tex], where [tex]\(n\)[/tex] represents the time in minutes:
[tex]\[ a(n) = 44 - 2n. \][/tex]

### Step 4: Check the Given Options

We check the provided options to see which function matches the one we derived:

A. [tex]\(a(n) = 42 - 2^{n-1}\)[/tex] - This does not match a linear form.

B. [tex]\(a(n) = 42 - 2^{n+1}\)[/tex] - This does not match a linear form.

C. [tex]\(a(n) = 42 - (n-1) \cdot 2\)[/tex] - Simplifying this:
[tex]\[ a(n) = 42 - 2(n-1) = 42 - 2n + 2 = 44 - 2n, \][/tex]
which matches the form we derived.

D. [tex]\(a(n) = 42 - n \cdot (2 - 1)\)[/tex] - This simplifies to:
[tex]\[ a(n) = 42 - n \cdot 1 = 42 - n, \][/tex]
which does not match.

### Conclusion

The correct function that models the weight of the sandbag over time is option C:
[tex]\[ a(n)=42-(n-1) \cdot 2. \][/tex]

This function accurately represents the initial weight and the constant rate of weight loss.