Beth measured the distance from home plate to the pitcher’s mound on a baseball field. She measured the distance to be 69 ft. The actual distance from home plate to the pitcher’s mound is 60.5 ft.

What is the percent error in Beth’s measurement? Round to the nearest tenth of a percent, if necessary.

A.
8.5%

B.
12.3%

C.
14%

D.
85%

Question

Beth measured the distance from home plate to the pitcher’s mound on a baseball field. She measured the distance to be 69 ft. The actual distance from home plate to the pitcher’s mound is 60.5 ft.
 
What is the percent error in Beth’s measurement? Round to the nearest tenth of a percent, if necessary.
 
 A.
8.5%
 
 B.
12.3%
 
 C.
14%
 
 D.
85%

anonymous · Answer

I Think D

I Dont Know How I Got It Though

anonymous · Answer

For every 69 feet she measured, there is an error of 8.5 feet(69-60.5), 
percentage error = (8.5/69)*100 = 12.3%

anonymous · Answer

So B ?_?

anonymous · Answer

yep

phi · Answer

you should figure out the "error"
and then the fraction  
$$ \frac{error}{truth}$$
and then change that to a decimal and then to percent

phi · Answer

in this problem, the error is the difference between measured and truth
(to find a difference between two numbers, you subtract them)

also, truth is 60.5  (not 69 , which is the measured)