The base of a box is a rectangle. the width of the box is half its length. the height of the box is 0.5m. find the volume of the box if the area of the base is 1.08m² less than the combined area of the sides.

We assume the box is open-top.

If the width of the box is represented by w, then its length is 2w. The area of the base is then w·2w = 2w².

The combined area of the sides is the product of the perimeter of the base, 2(w+2w) = 6w and the height, 0.5, so is 3w.

Since the base area is 1.08 m² less than the lateral area, we have ...

... 2w² = 3w -1.08

... 2w² -3w +1.08 = 0 . . . . rearrange to standard form

... w² -1.5w +0.54 = 0 . . . make the leading coefficient 1 (divide by 2)

... (w -0.9)(w -0.6) = 0 . . . factor

The width of the box may be either 0.6 meters or 0.9 meters.

The volume is the product of width, length, and height, so is ...

... V = w·2w·0.5 = w²

... V = 0.36 m³ . . . or . . . 0.81 m³. . . . . . (there are two possible answers)

Check

For width 0.6 m, the length is 1.2 m and the perimeter is 2(0.6+1.2) = 3.6 m. The lateral area is then (0.5 m)(3.6 m) = 1.8 m². Subtracting 1.08 m² from this, we get 0.72 m². The area of the base is (0.6 m)(1.2 m) = 0.72 m², so this answer checks.

For width 0.9 m, the length is 1.8 m and the perimeter is 2(0.9+1.8) = 5.4 m. The lateral area is then (0.5m)(5.4 m) = 2.7 m². Subtracting 1.08 m² from this, we get 1.62 m². The area of the base is (0.9 m)(1.8 m) = 1.62 m², so this answer checks, too.

the correct answer is b) histograms can be used to exhibit the shape of distributions. this is because histograms count the frequency of each possible answer, so one is able to see which results are more common and which are less common.

answer: ouch. thats difficult

step-by-step explanation: