Common Core Performance Coach mathematics grade 7

Question

Julie estimated her towns population at 77500. The actual population of her town is 75000. Find the percent error.

Answer 1

Julie estimated her town's population at 77500. The actual population of her town is 75000. Find the percent error.

 

Percent difference, percent error, percent increase/decrease, etc. all have a base, or initial value, from which another value is compared.  This is why a 10% increase one year followed by a 10% decrease the next year does not result in the initial value.

 

The base value here is the actual population (75000).  The "error" is in Julie's estimate.  It is 2500 different from the base value of 75000, which is (2500/75000)*100%, or (0.03)*100% = 3%.

 

Now, PLZ, Julie made an "estimate," or Julie said it was "approximately 77500."   Now, even if Julie had said, "it is exactly 77501," [that is, (0.032)*100%] there is absolutely no justification for including additional digits (note: there must be whole people; this is an "estimate"].   Therefore, 3% is an "accurate" representation of the percent error her "estimate".

Answer 2

Hi Bree, 

Step One: Calculate the error (subtract one value form the other) ignore any minus sign.

77500-75000=2500

Step Two: Divide the error by the exact value (we get a decimal number)

2500/75000=0.33333333333 

Step Three: Convert that to a percentage (by multiplying by 100 and adding a "%" sign)

∴33%

Sincerely, 

Starlyte

Answer 3

Bree, take the difference and divide by the actual value.

 

2500/75000 = .333... = 33%

Answer 4

David was on the right track but got the decimal wrong. 2500/75000 is .0333333 which is 3.33% not 33%.