Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum

Monthly Update sign up
Mailing List signup
Search
Follow The Annenberg Learner on LinkedIn Follow The Annenberg Learner on Facebook Follow Annenberg Learner on Twitter
MENU
Learning Math Home
Session 2: Solutions
 
Session 2 Part A Part B Part C Homework
 
Glossary
measurement Site Map
Session 2 Materials:
Notes
Solutions
 

A B C 
Homework

Video

Solutions for Session 2, Part C

See solutions for Problems: C1 | C2 | C3 | C4 | C5


Problem C1

The precision of a measurement is directly related to the partitioning of the measuring instrument. We can more precisely measure something when the partitions are closer together. For example, a beaker with milliliter partitions gives a more precise liquid measure than a measuring cup with 100-mL partitions.

<< back to Problem C1


 

Problem C2

a. 

The precision unit is 1 ft.

b. 

The precision unit is 1 mm.

c. 

The precision unit is 1 cm.

d. 

The precision unit is 1 in.

<< back to Problem C2


 

Problem C3

a. 

The maximum possible error is 0.5 ft.

b. 

The maximum possible error is 0.5 mm.

c. 

The maximum possible error is 0.5 cm.

d. 

The maximum possible error is 0.5 in.

<< back to Problem C3


 

Problem C4

The length of the hypotenuse mentioned in this problem is 4.2 cm, so the precision unit is 0.1 cm, because each centimeter is divided into tenths. The maximum possible error is 0.05 cm. The side length of the hypotenuse could be written as 4.2 cm 0.05 cm, or between 4.15 and 4.25 cm.

If you expressed the length of the hypotenuse as 42 mm, then the precision unit would be 1 mm, and the maximum possible error would be 0.5 mm (in either direction). The side length of the hypotenuse could be written as 42 mm 0.5 mm, or between 41.5 and 42.5 mm.

<< back to Problem C4


 

Problem C5

Since each centimeter is divided into tenths, the precision unit is 0.1 cm, and the absolute error for the calculations is 0.05 cm; therefore, the relative error is 0.05 divided by the measurement. For example, when the hypotenuse was measured as 4.2 cm, the relative error was 0.05/4.2, which gives you 0.012 or 1.2/100 (about 1.2%). Larger measurements with the same absolute error will have lower relative error.

<< back to Problem C5

<< back to Problem C5 non-interactive version


 

Learning Math Home | Measurement Home | Glossary | Map | ©

Session 2 | Notes | Solutions | Video

© Annenberg Foundation 2014. All rights reserved. Legal Policy