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.
First pair goes to f(x) = 20x
Next pair goes to f(x) = 10x + 20
Third pair goes to f(x) = 10x + 25
Last pair goes to f(x) = 20x + 10
The answer is 880 ft cubed
Slope is 7/4 y intercept is -10
<span>The formula for the volume of a sphere is V=4/3πr^3. Find the radius of a sphere with each volume.
1. 10 in^3
Solve for r
r =~ 1.3365 inches</span>
