# How many significant digits does 110 have

## Significant places

#### What are significant digits / digits and what are they used for?

Significant digits are digits of a number that contribute to a measurement and that are useful as a rough method to round off a final calculation. For more complex systems such as the uncertainty of a dosimetry system or to determine the bacterial load of a product, more complex methods must be used, such as those described in NIST Technical Note 1297 (TN1297) with the name "Guidelines for Evaluating and Expressing the Uncertainty of NIST Measurement Results" (Guidelines for calculating and expressing the uncertainty of NIST measurement results) can be found.

#### What makes a number significant or insignificant?

All numbers that are not leading or trailing zeros are said to be significant, unless the trailing zero is a decimal (for example, 3.00 has 3 significant digits, whereas 300 has only one significant digit). In the case of a measuring instrument, if it has only been calibrated to a specific decimal place, any digit outside of this calibration range is not considered significant. For example, if a scale has only been calibrated to a tenth place (0.0) but also shows a hundredth place (0.00), only the tenth place may be documented using traditional rounding methods.

Example: A scale that has been calibrated to the tenths place shows a weight of 11.35 kg. The value is rounded to the tenth place and recorded as 11.4 kg.

#### Which rules regarding significant digits have to be considered when adding and subtracting?

With addition and subtraction, the final documented value may only have the same number of decimal places as the last precise measurement.

Example: The length of a building is 372.71 m. This value was measured with a measuring tape that is calibrated to the hundredth place. The width of the same building is 174.2 m, measured with a ruler calibrated to the tenth place. How big is the perimeter of the building?

The scope is:

P = 372.71 + 174.2 + 372.71 + 174.2

P = 1093.82 m

However, since the width of the building is only known up to the tenth place, our result can only be documented up to the tenth place. So the final result looks like this:

P = 1093.8 m

#### What rules regarding significant digits must be considered when multiplying and dividing?

With multiplication and division, the final result must only have the same number of significant digits as the last precise measurement.

Example: If the mass of a box is measured at 6.817 kg and the volume of the box is 18.39 cm3, what is the density of the box?

We calculate the density (ρ) by dividing the mass of the cardboard by its volume. So:

ρ =

6.817 kg /

18.39 cm3

ρ = 0.370 …… kg / cm3

Since the volume only has significant digits up to the hundredth place, but the mass has significant digits up to the thousandth place, we document the final density up to the hundredth place:

ρ = 0.37 kg / cm3

#### How do you handle constants when performing calculations with significant digits?

Recall the formula for calculating a circumference:

C = 2πr

In this equation, r represents a measurable quantity, the radius of the circle, and π is a constant. We know an infinite number of numbers after the decimal point for π, so we get the most imprecise value from our measurement of the radius. However, this is not the case with all constants.

In general, when doing calculations with constants, it is best to use one more decimal place than the least accurate measurement of decimal places has. So if we calculate the circumference of a circle with a radius of 4.2 cm, then we would substitute the value 3.14 for π (the radius is significant up to the tenth place, so we take one place more for π - up to the hundredth place).

#### If we calculate a value in several steps, when do we make the significant digits estimate?

The estimation of the significant digits takes place in the last step of the calculation. Let us recall our density example again: If our mass is now 5.312 kg and we have a box with the dimensions 2.54 x 2.54 x 2.54 cm, then we can calculate the volume as follows:

V = (2.54 cm) x ((2.54 cm)) x (2.54 cm)

V = 16.3871 ... cm3

And to calculate the density we use:

ρ = 0.3242 ... kg / cm3

And our final density is given to the hundredths place based on the accuracy of the height, width and depth of the box:

ρ = 0.32 kg / cm3

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