Question 920750
Maybe the online tool expected 4.9sqrt(3) for an answer,
because {{{14.7*sqrt(3)/3=(14.7/3)*sqrt(3)=4.9sqrt(3)}}} .
 
Maybe it expected {{{8.487}}} or {{{8.49}}} or {{{8.48}}} .
Your rounding of {{{sqrt(3)=1.732060808}}} to {{{1.7}}} is a bit extreme.
If you are given {{{14.7}}} as a piece of data, with 3 significant digits,
your rounding should be at least as accurate.
 
I would use {{{sqrt(3)=1.73}}} or {{{sqrt(3)=1.732}}} .
{{{4.9*1.73=8.477}}} rounds to {{{8.48}}}
{{{4.9*1.732=8.4868}}} rounds to {{{8.49}}} or {{{8.487}}}
 
Better yet, you could use a calculator, getting {{{8.487048957}}} as a result, and only round at the end, to {{{8.49}}} or {{{8.487}}} .
 
NOTE:
In science, you round single final results to match the accuracy of the data,
and you keep more digits for intermediate results,
or (in some cases) final results that are measurements for elements of a group when you are trying to show/measure variability.