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Let's go through the problem step by step.

The volume of the cube is given by \( s^3 = \frac{27}{1000} \, \text{ft}^3 \).

### 1. **The length of one side of the box, \( s \), is \( \frac{3}{10} \, \text{ft} \).**

The volume of a cube is calculated as \( s^3 \), where \( s \) is the side length. We are given that:
\[
s^3 = \frac{27}{1000}
\]
To find \( s \), we take the cube root of both sides:
\[
s = \sqrt[3]{\frac{27}{1000}} = \frac{\sqrt[3]{27}}{\sqrt[3]{1000}} = \frac{3}{10} \, \text{ft}
\]
This statement is **True**.

### 2. **The area of one face of the box, \( s^2 \), is \( \frac{6}{10} \, \text{ft}^2 \).**

The area of one face of the cube is \( s^2 \). Since we know \( s = \frac{3}{10} \), we calculate:
\[
s^2 = \left(\frac{3}{10}\right)^2 = \frac{9}{100} \, \text{ft}^2
\]
The statement says \( \frac{6}{10} \, \text{ft}^2 \), but \( \frac{6}{10} = 0.6 \), and \( \frac{9}{100} = 0.09 \). These are not equal, so this statement is **False**.

### 3. **The area of both labels on the box is \( \frac{9}{25} \, \text{ft}^2 \).**

If one face has an area of \( \frac{9}{100} \, \text{ft}^2 \), then the area of two faces would be:
\[
2 \times \frac{9}{100} = \frac{18}{100} = 0.18 \, \text{ft}^2
\]
The statement claims the area is \( \frac{9}{25} \, \text{ft}^2 \), which is:
\[
\frac{9}{25} = 0.36 \, \text{ft}^2
\]
These are not equal, so this statement is **False**.

### 4. **The perimeter of one face of the box, \( 4s \), is \( \frac{6}{5} \, \text{ft} \).**

The perimeter of a square face is \( 4s \). Since \( s = \frac{3}{10} \), we calculate:
\[
4s = 4 \times \frac{3}{10} = \frac{12}{10} = 1.2 \, \text{ft}
\]
The statement claims \( \frac{6}{5} \, \text{ft} \), and \( \frac{6}{5} = 1.2 \). So this statement is **True**.

### Summary of answers:
1. True
2. False
3. False
4. True

Would you like any additional details on these solutions, or do you have further questions?

### Here are 5 follow-up questions for practice:
1. How would the volume change if the side length \( s \) doubled?
2. What is the total surface area of the cube?
3. If each label covered exactly one face, how many square feet of labels are needed?
4. What would the perimeter be if the side length was increased by 50%?
5. How would the volume of the cube change if the side length was reduced by half?

### Tip:
To quickly check or manipulate powers of fractions, remember that both the numerator and denominator get raised to the respective power.

Sabrina is making a box in the shape of a cube with a side length of s. The volume of the box, s³, is 27/1000 ft³. Sabrina wants to completely cover two of the faces with labels. Determine whether each statement is true.

Solve a problem involving the volume and face properties of a cube-shaped box. Calculate the cube root, area of faces, and perimeters. Determine which statements are true for Sabrina’s box project. Suitable for Grades 6-8.

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