A shipping box has the dimensions shown

A shipping box has the dimensions shown. What is the volume of the box in cubic yards? Not drawn to scale. (1)/(16)yd.^(3) (1)/(10)yd.^(3) (1)/(8)yd.^(3) (1)/(32)yd.^(3)

The correct answer and explanation is:

To calculate the volume of a shipping box in cubic yards, we need to multiply its three dimensions (length, width, and height). From the options you provided, we can see the dimensions are represented in fractional forms of cubic yards, which means we’re working with units in cubic yards.

However, there seems to be some missing information in your prompt. Normally, the dimensions (length, width, and height) would be given explicitly in linear units, such as yards, feet, or inches. In this case, you seem to be referencing fractional cubic yards, but without specifying the actual length, width, and height of the box, it is difficult to directly compute the volume.

Explanation:

When calculating the volume of a rectangular box, we use the formula: V=Length×Width×HeightV = \text{Length} \times \text{Width} \times \text{Height}

Where:

  • VV is the volume of the box.
  • Length, width, and height are the dimensions of the box.

If the dimensions are in yards, the resulting volume will be in cubic yards, since the volume of a three-dimensional object is measured in cubic units.

Let’s assume that the given fractions of cubic yards (like 1/16 yd³, 1/10 yd³, etc.) represent possible volumes based on different combinations of dimensions. To determine the correct volume, we would typically need the specific dimensions (length, width, and height) of the box to apply in the formula.

If we assume the box’s dimensions in yards are provided, we could calculate the volume by multiplying those dimensions. For example, if the box had dimensions like 1/2 yard by 1/4 yard by 1/8 yard, you would calculate: V=(12)×(14)×(18)=164 yd3V = \left(\frac{1}{2}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{8}\right) = \frac{1}{64} \, \text{yd}^3

Thus, the specific values for the dimensions and the correct answer would depend on this type of information being provided. If you have those, feel free to share, and I can walk you through the calculation!

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