Every year, people in the US buy 2.7 billion boxes of breakfast cereal. A “typical” cereal box has dimensions of 2.5 inches by 7.75 inches by 11.75 inches. Imagine a warehouse that has a rectangular floor and that contains all the boxes of breakfast cereal bought in the US in one year. If the warehouse is 10 feet tall, what could the side lengths be? Is is possible to design a different sized cereal box to increase the amount of cereal in this warehouse?

A possible pair of values for length and width of the warehouse are:

Length = 10^4 ft, Width = 3.5571 * 10^3 ft.

It is not possible to redesign the cereal box to increase the amount of cereal in this warehouse, because the total volume occupied inside the warehouse will be the same (limited by the warehouse volume).

Step-by-step explanation:

First we need to find the volume of one box. The volume of the box can be calculated multiplying all the three dimensions:

V_box = 2.5 * 7.75 * 11.75 = 227.6563 in3

We know that the warehouse can store 2.7 billion boxes, so we can find the volume of the warehouse by multiplying this amount by the volume of each box:

V_warehouse = 227.6563 * 2.7 * 10^9 = 614.6719 * 10^9 in3

Converting from cubic inches to cubic feet (1 ft3 = 12^3 in3), we have:

V_warehouse = 614.6719 * 10^9 / 12^3 = 3.5571 * 10^8 ft3

If we know that the height of the warehouse is 10 feet, we have that the base area is:

Area = V_warehouse / height = 3.5571 * 10^7 ft3

A possible pair of values for length and width of the warehouse are:

Length = 10^4 ft, Width = 3.5571 * 10^3 ft

(We just need to choose value that makes the relation Length * Width = Area be true)

It is not possible to redesign the cereal box to increase the amount of cereal in this warehouse, because the total volume occupied inside the warehouse will be the same (limited by the warehouse volume)

29 mil

step-by-step explanation:

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~