Respuesta :
From the question, we have the following information:
1. The weight limit for an 18-wheeler is 40 tons (weight of the truck + all material on board).
2. The 18-wheeler that Jack is driving weighs about 32,000 pounds (without any cargo onboard).
3. The average weight of a market pig is 250 pounds.
To solve this problem, we have that the conversion factor is 1T = 2000lbs, and we need to convert the given values into Tons to find the number of pigs that Jack can carry on the 18-wheeler.
Case: 18-wheeler ---> 32,000 pounds. We need to make the conversion into Tons. Then, we have:
[tex]32000lbs\cdot\frac{1T}{2000\text{lbs}}=16T[/tex]Case: the weight of the pigs ---> 250 pounds:
[tex]250\text{lbs}\cdot\frac{1T}{2000\text{lbs}}=0.125T[/tex]And now, we have that we have 40T - 16T = 24T (available to have as a cargo).
If we have that each pig weighs 0.125T, then, we can represent this situation as follows:
[tex]0.125x=24[/tex]To solve this equation, we can divide both sides of it by 0.125 (division property of equality):
[tex]\frac{0.125}{0.125}x=\frac{24}{0.125}\Rightarrow x=192[/tex]Then, we have that Jack can carry as many as 192 pigs (if they weigh 250 pounds on average) as cargo, and staying within the weight regulations.
