Algorithm
Step1: Read A, B, C
Step2: If A > B is True, then check whether A > C, if yes then A is greatest otherwise C is greatest
Step3: If A > B is False, then check whether B > C, if yes then B is greatest otherwise C is greatest
Following these steps flowchart can be made.
int secondmax (int a[], int n){int i, max, second;for (i=0; imax) second= max, max= a[i];else if (i==1 a[i]>second) second= a[1];}return second;}int middle (int a[3]){return secondmax (a, 3);}
An algorithm is a set of instructions that tells something (in this case, a computer) how to complete a specific task. An algorithm can be as simple as the following: 1. Move hand to wrench.2. Pick up wrench.3. Move wrench over to bolts.4. Turn wrench clockwise five times.5. Move wrench to original location.6. Let go of wrench.However, algorithms can be extremely complex and resource-intensive depending on the task you are trying to accomplish.
Cls input "enter two no.s ",a,b sum=a+b print "sum = ";sum end
You basically multiply 3 with 8 and 3 X 8 = 24
1. High accuracy. Comparing to Basic Incremental algorithm (especially if the slope were > 1.) 2. High speed. Comparing to Digital Differenmtial algorithm. 3. Draws the line between any two points. Comparing to Basic Incremental algorithm which can't draw if x0 > x1 ( the format is: (x0, y0), (x1, y1). )
Calculating the average of a set of data is the same as calculating the mean, or the value that is in the middle of the data. For example, if you were given the numbers 3, 7, 10, 4, and 5, and was asked to find the average, this is how it would be done:First, add up all the numbers. 3 + 7 + 10 + 4 + 5 = 29Then, divide the added numbers by how many numbers there are, total, which is 5. 29/5 = 5.8The average of 3, 4, 5, 7, and 10, is 5.8.
1.Start Algorithm 2.Enter first number 3.Enter second number 4.Enter third number 5.Enter fourth number 6.Enter fifth number 7.Add five number 8.display five number / 2 9.Display result 10.End Algorithm
The time complexity of the Karatsuba algorithm for multiplying two numbers is O(nlog2(3)), which is approximately O(n1.585).
To find six numbers that average out to 4, we can use the formula for calculating averages: Average = (Sum of all numbers) / (Total number of numbers). Since we have six numbers, the sum of all numbers must be 6 * 4 = 24. To evenly distribute this sum, we can choose the numbers 3, 3, 4, 4, 5, and 5. Their sum is 24, and when divided by 6, the average is indeed 4.
int secondmax (int a[], int n){int i, max, second;for (i=0; imax) second= max, max= a[i];else if (i==1 a[i]>second) second= a[1];}return second;}int middle (int a[3]){return secondmax (a, 3);}
add all the outcome / example 15, 64, 21, 7, 10, 3 = 120. Then divide the sum of all the numbers by the number of numbers/ example, take 120 divided by 6=20. So the average number out of all the numbers above is 20.
To get the simple average of a series of numbers, you add the numbers up and divide by however many numbers there were. For example, the average of 2, 3 and 4 is the sum of the numbers (which adds up to 9) divided by 3 (because there are three numbers). So the average is 9/3 = 3.
The average of 3, 3.5, and 6 is 4.17. Add the three numbers together and divide by 3 to find the average.
2
Yes. For example, the average of the numbers 1, 2, and 3 is 2. 1+2+3=6 6/3=2
What exactly do you mean "yields only prime numbers"? If you mean a formula that when given the numbers n=1, 2, 3, ... and so on generates the nth prime number (or a different prime number for each n) then no. If you mean an algorithm whereby a number can be tested to be a prime number then yes. (Using this prime_test algorithm, a simple algorithm can be written that would supply numbers one at a time to it and use its result to decide whether to yield the tested number or not, only yielding those numbers which pass the test.)
Add the 3 numbers together and divide that answer by 3. Example: The average of 2, 4, and 9. Average = (2+4+9)/3 = 15/3 =5.