Question

I’ve tried to do the first one and i still dont get it… u dont have to show work ill do that just fill in the boxes and answer the letter question at the bottom pls

Answer 1

Answer: A % error = 0.6878%, accuracy = 3, precision = 3
B % error = 0.2751%, accuracy = 2, precision = 2
C % error = 0.1376%, accuracy = 1, precision = 1

Step-by-step explanation:

For calculating the percent error for each student, I took the average of each of the student's measurements, and used that as my "measured" value. Then, I used the percent error formula to find each percent error.

A: 14.8cm + 14.1cm + 14.5cm + 14.6cm + 14.2cm = 72.2cm
72.2cm/5 = 14.44cm

`(|14.44-14.54|)/(14.54)` x 100% = 0.006878 x 100%
= 0.6878% error for student A

B: 14.8cm + 14.2cm + 14.6cm + 14.5cm + 14.8cm = 72.9cm
72.9cm/5 = 14.58cm

`(|14.58-14.54|)/(14.54)` x 100% = 0.002751 x 100%
= 0.2751% error for student B

C: 14.6cm + 14.5cm + 14.5cm + 14.4cm +14.6cm = 72.6cm
72.6cm/5 = 14.52cm

`(|14.52-14.54|)/(14.54)` x 100% = 0.001376 x 100%
= 0.1376% error for student C

For part B, I'm unsure what it's exactly asking, but how I would record each student's accuracy and precision is by doing a scale of 1 to 3, with 1 being the most accurate or precise and 3 being the least accurate or precise.

For both parts B and C, accuracy is how close the measured values are to the actual values, while precision is how close the measured values are to each other.

For student A, 2 out of 5 measurements were close to the actual value, but A's measurements were the furthest from each other, so I would put 3 for A's accuracy and precision.

For student B, 2 out of 5 measurements were close to the actual value, but B's measurements were closer to each other than A's measurements, so I would put 2 for B's accuracy and precision.

For student C, 5 out of 5 measurements were close to the actual value, and C's measurements were the closest to each other, so I would put 1 for C's accuracy and precision.

Student Accuracy Precision

A 3 3
B 2 2
C 1 1

For part C:

Student A and student B both had 2 measurements that were close to the actual value (14.5cm and 14.6cm, compared to 14.54cm), and student C had all 5 measurements be close to the actual value, ranging from 14.4cm to 14.6cm. Student C's measurements are closer to the actual value of 14.54cm, so student C is the most accurate.

Student A's measurements varied from 14.1cm to 14.8cm, student B's measurements varied from 14.2cm to 14.8cm, and student C's measurements varied from 14.4cm to 14.6cm. Student C's measurements are closer to each other's than student A or B, so student C is the most precise.

So, your answer for the last question will be "a. Student C has done the most precise and accurate work."

I hope this helps! :)