Hello DR. BOB,
the correct value in the problem posted at 1:17pm regarding Student A and Student B accuracy and random error is the:
MEAN of 2.63 g/mL. Could you take a look back at that post and let me know if i answered 2 and 3 correctly.
Hello! I'm not Dr. Bob, but I can definitely help you review the problem. Unfortunately, I cannot look back at the post you mentioned since I don't have access to past conversations. However, I can guide you through the process of finding the mean value of the data given.
To calculate the mean, you need a set of data values. In this case, you mentioned Student A and Student B accuracy and random error. If you have the values for these, you can sum them up and divide by the total number of values to find the mean.
For example, if Student A's accuracy and random error values are 3.2 g/mL, 1.8 g/mL, and 2.4 g/mL, you would add them together:
3.2 + 1.8 + 2.4 = 7.4 g/mL
Next, you divide by the total number of values, which is 3:
7.4 / 3 = 2.4667 g/mL
Rounding to the appropriate number of decimal places, you get the mean value of approximately 2.47 g/mL.
Please note that this explanation assumes you have access to the actual values for Student A and Student B accuracy and random error. If you have any additional details or specific calculations you would like me to review, please let me know!