Explanation:
Any calibration scale consists of markings indicating calibrated values. The ‘space’ between the marks (lines) is the area of uncertainty with respect to the calibration.
Thus, the possible error is always one-half of the value between the markings, because ON either one you have a calibrated value. In between, no matter how close you think you can “judge” the distance, there is no calibrated reference point, so the ‘error’ of stating a value is +/- the value of half of the calibration accuracy. 0.991 is accurate (assuming that is the calibration limit), and 0.992 or 0.990 would also be “accurate”. The possible error is the +/- 0.0005 beyond that third digit that might be more to one side or the other.
That means the measured value of 0.991g could be between 0.9905g and 0.9915g.
Hope this helps. Not positive if it’s correct, however.
Answer:
impossible?
Step-by-step explanation:
Explanation is in a file
bit.
ly/3a8Nt8n
Answer:
The answer to your question is below
Step-by-step explanation:
4.-
4a. 72
_<u>x 9</u>
638
4b. The error this student is doing is that he multiplies the first number but he must add the decenes of this result to the result of the multiplication of the second number and he is not doing that.
4c.- I told him the correct way to solve multiplications.
a) Multiply the first number and only write the units.
b) If the are decenes add then to the units of the result of the multiplication of the second number.
Example
72 9 x 2 = 18
<u>x9 </u> Just write the 8
8
72
<u>x 9</u>
648 9 x 7 = 63 + 1 from the previous operation
= 64
