Question
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Jerrica is packing a box that can hold at most 65 pounds. She starts by putting a container in the box that weighs 26 pounds. She plans to fill the rest of the box with books. Each book she places in the box weighs 1.3 pounds. Write an inequality for the number of books Jerrica can put in the box, and describe the solutions. (2 points) Check answer Remaining Attemots. 3 bost

Ask by Ruiz Flynn. in the United States
Jan 23,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

Jerrica can put up to 30 books in the box without exceeding the 65-pound limit.

Solution

To determine the number of books \( x \) that Jerrica can put in the box without exceeding the weight limit, we can set up the following inequality: \[ 26 + 1.3x \leq 65 \] **Explanation:** - **26 pounds** is the weight of the container. - **\( 1.3x \)** represents the total weight of \( x \) books, with each book weighing **1.3 pounds**. - The sum of the container and books should be **less than or equal to 65 pounds**. **Solving the Inequality:** \[ \begin{align*} 26 + 1.3x &\leq 65 \\ 1.3x &\leq 65 - 26 \\ 1.3x &\leq 39 \\ x &\leq \frac{39}{1.3} \\ x &\leq 30 \end{align*} \] **Conclusion:** Jerrica can place **up to 30 books** in the box without exceeding the 65-pound limit. **Final Inequality:** \[ 26 + 1.3x \leq 65 \quad \text{where} \quad x \leq 30 \]

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First, let’s set up the inequality. Jerrica has already filled the box with a 26-pound container, which means she can use up to \(65 - 26 = 39\) pounds for books. If each book weighs 1.3 pounds, we can express this as: \(1.3b \leq 39\), where \(b\) is the number of books. Now let's solve for \(b\): \[b \leq \frac{39}{1.3}\] Calculating this gives us \(b \leq 30\). This means that Jerrica can put a maximum of 30 books in the box, and the solution set represents all possible whole numbers of books from 0 to 30, inclusive! Now, imagine Jerrica stacking those books like a game of Jenga—careful precision is crucial to make sure they don't topple over or exceed that magical 65-pound limit!

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