# Question #2c114

##### 1 Answer

#### 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 **actual mass** of the object is

#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 *smaller* or *bigger* than the actual value, all that matters is **how small** or **how big** this difference is.

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")))#