0.006, 0.008, 0.081, 0.091, 0.095, 0.0.8, 0.09, 0.1
0.8 0.04 0.53
I believe the answer is: 7,7,11
1.6, 6.1, 6.166, 6.61, and 6.66.
123478
the mean is where you add all the numbers then divide by the number of numbers the median is when u write all the numbers in order then find the one in the middle and the mode is the most common one. the range is the smallest number subtracted from the largest number.
just go the reverse of how they are listed now they are currently largest to smallest
0.8 0.04 0.53
Use the following algorithm (written in pseudocode). Let largest be the lowest possible real number. Let smallest be the greatest possible real number. Repeat while there is input... { Read real number r from input. If r is greater than largest then let largest be r. If r is less than smallest then let smallest be r. } End repeat. Let range be largest minus smallest. Output range.
I believe the answer is: 7,7,11
0.001 0.05 0.2 0.5 3.3 5.2
1.6, 6.1, 6.166, 6.61, and 6.66.
smallest: zero largest: five thousand
#include #include #include int main(int argc, char *argv[]){int n, smallest, largest, sum, temp;if(argc < 2){printf("Syntax: foo val1[val2 [val3 [...]]]\n");exit(1);}smallest = largest = sum = atoi(argv[1]);for(n = 2; n < argc; n++){temp = atoi(argv[n]);if(temp < smallest) smallest = temp;if(temp > largest) largest = temp;sum += temp;}printf("Smallest: %i\nLargest: %i\nAverage: %i\n", smallest, largest, sum / (argc - 1));return 0;}
use the numbers 5,6,7,8 to write an equation with the largest possible equation
write these numbers in order of size smallest first 2177 914 941 944 909
Asia and Australia are continents.
final double[] ns = new double[10]; final Random rnd = new Random(System.currentTimeMillis()); // Fill... for (int i = 0; i < ns.length; ++i) { ns[i] = rnd.nextDouble(); } // Get largest/smallest... double largest = Double.MIN_VALUE; double smallest = Double.MAX_VALUE; for (double n : ns) { if (n > largest) { largest = n; } if (n < smallest) { smallest = n; } } // largest and smallest are now the proper values.