Answer :
To determine the density, we start with two given quantities: total mass and volume.
1. Total Mass: The total mass given is 600 pounds.
2. Volume: The volume given is 85 cubic feet.
Density (denoted as [tex]\( \rho \)[/tex]) is calculated using the formula:
[tex]\[ \rho = \frac{\text{mass}}{\text{volume}} \][/tex]
Plugging in the given values:
[tex]\[ \rho = \frac{600 \text{ pounds}}{85 \text{ cubic feet}} \][/tex]
Performing the division:
[tex]\[ \rho = 7.06 \text{ pounds per cubic foot} \][/tex]
Given the context, the units should be adjusted. The question mentions a unit of pounds per [tex]\( 600 t^2 \)[/tex]. However, typically, the unit of density derived in this type of problem would be pounds per cubic foot.
However, based on the constraint (place the correct numeric answer in the density box):
The result is:
[tex]\[ \boxed{7.06} \quad \frac{\text{ pounds }}{600 t ^2} \][/tex]
(Note: It is understood that there is a mistake in the unit [tex]\(600 t^2\)[/tex], and it should likely be [tex]\(\frac{\text{ pounds }}{\text{ cubic feet }}\)[/tex], or this might be a hypothetical unit given in the problem.)
1. Total Mass: The total mass given is 600 pounds.
2. Volume: The volume given is 85 cubic feet.
Density (denoted as [tex]\( \rho \)[/tex]) is calculated using the formula:
[tex]\[ \rho = \frac{\text{mass}}{\text{volume}} \][/tex]
Plugging in the given values:
[tex]\[ \rho = \frac{600 \text{ pounds}}{85 \text{ cubic feet}} \][/tex]
Performing the division:
[tex]\[ \rho = 7.06 \text{ pounds per cubic foot} \][/tex]
Given the context, the units should be adjusted. The question mentions a unit of pounds per [tex]\( 600 t^2 \)[/tex]. However, typically, the unit of density derived in this type of problem would be pounds per cubic foot.
However, based on the constraint (place the correct numeric answer in the density box):
The result is:
[tex]\[ \boxed{7.06} \quad \frac{\text{ pounds }}{600 t ^2} \][/tex]
(Note: It is understood that there is a mistake in the unit [tex]\(600 t^2\)[/tex], and it should likely be [tex]\(\frac{\text{ pounds }}{\text{ cubic feet }}\)[/tex], or this might be a hypothetical unit given in the problem.)