Solutions to Real-World Inequality Problems Practice%0D%0AComplete this assessment to review what you've learned. It will not count toward your grade.%0D%0A5 of 55 of 5 Items%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0A%0D%0AQuestion%0D%0AJerrica 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)%0D%0A1.3x+ 26 ≤65, where x is the number of books Jerrica can put in the box. Jerrica can put at most books in the box.

Answer 1

The inequality is 1.3x + 26 ≤ 65, where x is the number of books Jerrica can put in the box. The solution to this inequality is x ≤ (65 - 26)/1.3, which simplifies to x ≤ 29. Jerrica can put at most 29 books in the box.

Answer 2

that is wrong

Answer 3

Apologies for the incorrect response. Let's reevaluate the problem.

The weight of the container is 26 pounds. Let's denote the number of books as "x". Each book weighs 1.3 pounds. The total weight of the books will be 1.3x.

Jerrica can put at most 65 pounds of weight in the box. So the inequality to represent this situation is:

1.3x + 26 ≤ 65

To find the solution, we need to isolate "x" on one side of the inequality:

1.3x ≤ 65 - 26

1.3x ≤ 39

Now we divide both sides by 1.3:

x ≤ 39/1.3

x ≤ 30

Therefore, the correct inequality is 1.3x + 26 ≤ 65 and the solution is x ≤ 30. Jerrica can put at most 30 books in the box.