Q: What is Min and Max Levels?

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The min and max is when the first derivative , or slope at any point, is zero. For f of x = 2x first derivative is 2, so this is constant slope with no min or max as this is not zero; min is thus negative infinity and max is infinity

Not necessarily.

the sixth thermometer

You take the integral of the sin function, -cos, and plug in the highest and lowest values. Then subtract the latter from the former. so if "min" is the low end of the series, and "max" is the high end of the series, the answer is -cos(max) - (-cos(min)), or cos(min) - cos(max).

max-- 0.2cm min-- 0.006cm

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Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. sonika aggarwal GNIIT

The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)

int sum (int min, int max) {return (max-min+1)*(max+min)/2;}

int a, b, c, d, max, min; scanf("%d%d%d%d",&a, &b, &c, &d); (a>b)?(max=a,min=b):(max=b,min=a); (c>d)?(a=c,b=d):(a=d,b=c); max=(a>max)?a:max; min=(b<min)?b:min; printf("%d %d\n", max, min);

min: 0.5 KVA MAX: 1.5 KVA

Usually, max 212 F and min 32 F It depends on for what the thermometer will be used. My greenhouse min/max thermometer has a min of -40°F and a max of 120°F A (non-digital) clinical thermometer has a min of 94°F and a max of 108°F A cook's (sugar/jam) thermometer has a min of 100°F and a max of 400°F

The min and max temp of mars i 1,299 degrees celcius and -1,000,000 degrees celcius

#include<stdio.h> int main() { int a[20],min,max; int n; printf("\nEnter the num of elements: "); scanf("%d",&n); printf("Enter the elements\n"); for(int i=0;i<n;i++) { scanf("%d",&a[i]); if(i==0) { min=max=a[i]; } if(a[i]<min) min=a[i]; else if(a[i]>max) max=a[i]; } printf("The largest element is %d. The smallest element is %d.", max, min); }

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.

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.

Min's and max's stand for minimums and maximum in U-Haul. These are terms that are used for inventory stocking purposes.