If the k-th entry is smaller than the (k+1)-th, make them swap places and remember that something changed. If not, don't change anything. Let's take the example from the beginning of the above section: 2/5 and 3/8. One way to compare them is to convert both to decimals. This would give: We know that 40 = 5 × 8, so to turn 2/5 into an equivalent fraction with denominator 40, we multiply its top and bottom by 8: Re-create the array with 1.0 metadata (ensure to specify all details, in case defaults have changed!)

To convert from percent to decimal, divide the percent by 100. Dividing by 100 will shift the decimal two places to the left. For fractions, the numerator or denominator and be calculated if desired. Clear both the numerator and denominator. Enter a value into the numerator or denominator. Enter a value into the percent box or decimal box. The blank box in the fraction either the numerator or denominator will be calculated.

circa early 2009, include init scripts that run any arrays that aren't started by auto-detect, which

If k < n – 1, increase k by 1, and repeat from point 2. Otherwise, recall if something changed in this run of the algorithm. If yes, repeat from point 1. If not, end the algorithm. As a workaround for the kernel-autodetection issue, several distributions, including Ubuntu and Fedora, If the k-th entry is larger than the (k + 1)-th, make them swap places and remember that something changed. If not, don't change anything. The brute-force method would be to multiply the two denominators. However, if you'd like to be a bit more subtle, you can choose their least common multiple. In fact, in our case, since the denominators are coprime (i.e., have no common factors), those two methods give the same result: 40. For instance, if you want to compare 23.1457 and 23.1349, then we begin with the left-most digits in the two numbers. Both of them are 2, so we move on to the next. Then we have 3-s, so again, we go one further to the left, and the same for 1-s. Finally, we arrive at two different digits: 4 and 3. Since 4 > 3, the first number is larger than the second (we don't have to look at the last digits).We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.