The reason we use max for union is because the union of two sets should include all elements from both sets, so the max operation ensures that the larger of the two elements is included. Conversely, we use min for intersection because we want to find the common elements shared by both sets, so the min operation ensures we only include elements that are in both sets.
You can use the SEQUENCE function in Excel to create a series of numbers from max to min or min to max. For example, to create a series from 10 to 1 in descending order, you could use =SEQUENCE(10,1,-1). This formula generates a range of numbers from 10 to 1 in descending order.
It is recommended to use enamelled copper wire min. 99,9 %.
Minimum oxidation number: -1 Maximum oxidation number: 1 Min. common oxidation no.: 0 Max. common oxidation no.: 1 I got this of a really cool website that has alot of infromation and useful facts about each element and trust me you'll use it a lot in chemistry. www.chemicool.com
At an intersection.
To calculate lambda max (λ max), generally used in spectrophotometry, you can use the formula: λ max = 1 / εmax, where εmax is the molar absorptivity (extinction coefficient) of the compound at its maximum absorbance wavelength. The λ max represents the wavelength at which the compound absorbs light most strongly.
in simple calculations we use + - × ÷ in computer like sum average min and max
You can use the SEQUENCE function in Excel to create a series of numbers from max to min or min to max. For example, to create a series from 10 to 1 in descending order, you could use =SEQUENCE(10,1,-1). This formula generates a range of numbers from 10 to 1 in descending order.
For two sets, the Venn diagram will consist of two overlapping ovals. The area of the overlap is the intersection. The entire area of both ovals is the union.
The Demorgans Law includes the union, intersection, and complement in mathematics. Examples are A intersection B and B union A. Those are the basic examples.
The MAX function gets the highest value in a range and the MIN function gets the lowest. If there were values in all the cells from A2 to A20, you could use the two functions to get the highest and lowest values in the range:=MAX(A2:A20)=MIN(A2:A20)
You can use MAX to evaluate a list of thousands of sales transactions to determine quickly which day had the largest single transaction.You can use the MIN function to compare hundreds of wholesale costs to determine the lowest cost for a particular item.
Use the median-of-three algorithm: int min (int a, int b) { return a<b?a:b; } int max (int a, int b) { return a<b?b:a; } int median_of_three (int a, int b, int c) { return max (min (a, b), min (max (a, b), c)); } Note that the algorithm does not cater for equal values which creates a problem when any two values are equal, because there are only two values to play with, neither of which can be regarded as being the middle value. If the equal value is the lower of the two values, the largest value is returned if and only if it is the last of the three values, otherwise the lowest value is returned. But when the equal value is the larger of the two values, the largest value is always returned. Lowest value is equal: Input: 0, 0, 1 = max (min (0, 0), min (max (0, 0), 1)) = max (0, min (0, 1)) = max (0, 1) = 1 Input: 0, 1, 0 = max (min (0, 1), min (max (0, 1), 0)) = max (0, min (1, 0)) = max (0, 0) = 0 Input: 1, 0, 0 = max (min (1, 0), min (max (1, 0), 0)) = max (0, min (1, 0)) = max (0, 0) = 0 Highest value is equal: Input: 0, 1, 1 = max (min (0, 1), min (max (0, 1), 1)) = max (0, min (1, 1)) = max (0, 1) = 1 Input: 1, 0, 1 = max (min (1, 0), min (max (1, 0), 1)) = max (0, min (1, 1)) = max (0, 1) = 1 Input: 1, 1, 0 = max (min (1, 1), min (max (1, 1), 0)) = max (1, min (1, 0)) = max (1, 0) = 1 The only way to resolve this problem and produce a consistent result is to sum all three inputs then subtract the minimum and maximum values: int median_of_three (int a, int b, int c) { return a + b + c - min (min (a, b), c) - max (max (a, b), c)); } Lowest value is equal: Input: 0, 0, 1 = 0 + 0 + 1 - min (min (0, 0), 1) - max (max (0, 0), 1) = 1 - 0 - 1 = 0 Input: 0, 1, 0 = 0 + 1 + 0 - min (min (0, 1), 0) - max (max (0, 1), 0) = 1 - 0 - 1 = 0 Input: 1, 0, 0 = 1 + 0 + 0 - min (min (1, 0), 0) - max (max (1, 0), 0) = 1 - 0 - 1 = 0 Highest value is equal: Input: 0, 1, 1 = 0 + 1 + 1 - min (min (0, 1), 1) - max (max (0, 1), 1) = 2 - 0 - 1 = 1 Input: 1, 0, 1 = 1 + 0 + 1 - min (min (1, 0), 1) - max (max (1, 0), 1) = 2 - 0 - 1 = 1 Input: 1, 1, 0 = 1 + 1 + 0 - min (min (1, 1), 0) - max (max (1, 1), 0) = 2 - 0 - 1 = 1 This makes sense because when we sort 0, 0, 1 in ascending order, 0 is in the middle, while 0, 1, 1 puts 1 in the middle.
MAX gives you the highest value in a specified range. MIN gives you the lowest value in a specified range. COUNT will count how many values there are in a specified range, ignoring cells that do not have numbers in them. You could use them like this: =MAX(A2:A20) =MIN(A2:A20) =COUNT(A2:A20)
#include <stdio.h> #include <stdlib.h> #include <time.h> int GetRand( int min, int max) { // Swap min and max if wrong way around. if(max<min)min^=max^=min^=max; return(rand()%(max-min+1)+min); } int main() { // Seed the random number generator with current time to ensure // random numbers are different on each run. // Note: use srand(1) to generate the same sequence on every run. srand((unsigned)time(NULL)); printf("Number from 0 to 10 : %d\n", GetRand(0,10)); printf("Number from 1 to 100: %d\n", GetRand(1,100)); return(0); }
If you are specifying a component to a manufacturer, which has to fit into some assembly made up of components from various places, it is essential to give max-min tolerances to ensure it all fits together.
using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace printnumber { class Program { static void Main(string[] args) { int i; for (i = 0; i < 20; i++) { Console.Write(i); Console.Read(); } } } }
MIN is used to find the lowest number in a range. You would use MAX or LARGE to find the highest number in a range.