WILL MARK BRANLIEST IF GOTTEN RIGHT!
2 answers:
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The complete question in the attached figure
Part A) How much sand is currently in the container?
[sand currently in the container]=(5)*(4 1/2)*(2.25)-----> (5)*(4.5)*(2.25)
[sand currently in the container]=50.625 in³
the answer Part a) is 50.625 in³
Part B) How much more sand could the container hold before?
[sand could the container hold before]=[5*4.5*3]-[50.625]
[sand could the container hold before]=[67.5]-[50.625]------> 16.875 in³
the answer Part B) is 16.875 in³
Part C) What percent of the container is filled with sand?
the volume of container is [5*4.5*3]=67.5 in³
the volume filled with sand=50.625 in³
therefore
if 100%----------------> 67.5 in³
X---------------------> 50.625 in³
X=(50.625*100)/67.5=75 %
the answer Part C) is 75%
Part A) How much sand is currently in the container?
[sand currently in the container]=(5)*(4 1/2)*(2.25)-----> (5)*(4.5)*(2.25)
[sand currently in the container]=50.625 in³
the answer Part a) is 50.625 in³
Part B) How much more sand could the container hold before?
[sand could the container hold before]=[5*4.5*3]-[50.625]
[sand could the container hold before]=[67.5]-[50.625]------> 16.875 in³
the answer Part B) is 16.875 in³
Part C) What percent of the container is filled with sand?
the volume of container is [5*4.5*3]=67.5 in³
the volume filled with sand=50.625 in³
therefore
if 100%----------------> 67.5 in³
X---------------------> 50.625 in³
X=(50.625*100)/67.5=75 %
the answer Part C) is 75%