Celina measured the height of a window frame as 5.55 feet, but the actual height was 6 feet. What is the percentage of error in Celina's mea

not quite Jbow

Celina measured the height of a window frame as 5.55 feet, but the actual height was 6 feet. What is the percentage of error in Celina's measurement?

she is  6-5.55 = 0.45 feet off

 $${\frac{{\mathtt{0.45}}}{{\mathtt{6}}}}{\mathtt{\,\times\,}}{\mathtt{100}} = {\frac{{\mathtt{15}}}{{\mathtt{2}}}} = {\mathtt{7.5}}$$  

error is  7.5% 

She was 0.05 inches off. That in decimal form is 5%