just go the reverse of how they are listed now they are currently largest to smallest
0.001 0.05 0.2 0.5 3.3 5.2
smallest: zero largest: five thousand
k
Fourteen and two hundred fifty-one thousandths.
Write the place value of 5 in the following decimal numbers: 5.21
0.8 0.04 0.53
0.006, 0.008, 0.081, 0.091, 0.095, 0.0.8, 0.09, 0.1
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.
#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;}
smallest: zero largest: five thousand
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.