Question #2c114
1 Answer
Explanation:
For starters, the fact that the percent error carries a negative sign tells you that the mass Walter measured is smaller than the actual mass of the object.
#"% error" = color(red)(-)2.75color(blue)(%)# This means that for every
#color(blue)("100 g")# that you have for the actual mass of the object, Walter measured#"2.75 g"# #color(red)("less")# .
In other words, for every
#"100 g " - " 2.75 g" = "97.25 g"#
Now, you know that Walter determined that the object has a mass of
#100.7 color(red)(cancel(color(black)("g measured"))) * "100 g actual"/(97.25color(red)(cancel(color(black)("g measured")))) = color(darkgreen)(ul(color(black)("104 g")))#
The answer is rounded to three sig figs, the number of sig figs you have for the percent error.
ALTERNATIVE METHOD
Alternatively, you can determine the actual mass of the object by using the equation
#"% error" = (|"measured value " - " actual value"|)/"actual value" xx 100%#
Notice that the equation uses the absolute value of the difference between the measured value and the actual value
However, in your case, you already know that the percent error carries a negative sign, i.e. the measured values is smaller than the actual value, so you can ditch the absolute value sign to say that
#"% error" = ("measured value " - " actual value")/"actual value" xx 100%#
This will get you
#"actual value" xx ("% error" + 100%) = "measured value" xx 100%#
which s equivalent to
#"actual value" = ("measured value" xx 100%)/(("% error" + 100%))#
Plug in your values to find
#"actual value" = ("100.7 g" xx 100 color(red)(cancel(color(black)(%))))/((-2.75 + 100)color(red)(cancel(color(black)(%)))) = color(darkgreen)(ul(color(black)("104 g")))#