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solving real - world volume problems a type of cracker, rectangular in …

Question

solving real - world volume problems
a type of cracker, rectangular in shape, is stored in a vertical column with all of the crackers stacked directly on top of each other. each cracker measures 2 inches in length by (1\frac{1}{2}) inches in width. the volume of the column is 15 inches cubed. if there are 40 crackers in the column, what is the height of each individual cracker?
(\bigcirc) (\frac{3}{40}) inch
(\bigcirc) (\frac{1}{8}) inch
(\bigcirc) (\frac{1}{5}) inch
(\bigcirc) (\frac{3}{8}) inch

Explanation:

Step1: Calculate the volume of one cracker

The volume of the column is \(15\) cubic inches and there are \(40\) crackers. So the volume of one cracker \(V=\frac{15}{40}=\frac{3}{8}\) cubic inches.

Step2: Use the volume formula for a rectangular prism

The volume formula for a rectangular prism is \(V = l\times w\times h\). Given \(l = 2\) inches and \(w=1\frac{1}{2}=\frac{3}{2}\) inches. Substitute \(V=\frac{3}{8}\), \(l = 2\), and \(w=\frac{3}{2}\) into the formula: \(\frac{3}{8}=2\times\frac{3}{2}\times h\).

Step3: Simplify the equation

First, \(2\times\frac{3}{2}=3\). So the equation becomes \(\frac{3}{8}=3h\).

Step4: Solve for \(h\)

Divide both sides of the equation by \(3\): \(h=\frac{3}{8}\div3=\frac{3}{8}\times\frac{1}{3}=\frac{1}{8}\) inch.

Answer:

\(\frac{1}{8}\text{ inch}\)