The sum from 1 to n is n*(n+1)/2In this case that mean 20*21/2 = 210The sum from 1 to n is n*(n+1)/2In this case that mean 20*21/2 = 210The sum from 1 to n is n*(n+1)/2In this case that mean 20*21/2 = 210The sum from 1 to n is n*(n+1)/2In this case that mean 20*21/2 = 210
If you mean: -6n+5 = 1 then the value of n is 2/3
If you mean -17 = 15n -20 then the value of n is 1/5
n is a variable. It could mean any number you want it to. Most of the time, it is not specified what it is. Sometimes they give you clues about it, like "5 times n is 5." In this case n is 1.
The mean of N numbers is the sum of those numbers divided by N. If you include 13 in that set and the new mean is 6, then the original N is 7. Let X be the original sum. You have a system of two simultaneous equations. 1.) X/N = 5 2.) X+13/N+1 = 6 To solve, substitute the value of X from eq. 1 into eq. 2 and then solve eq. 2 for N. (Some steps combined) X = 5N5N+13 = 6(N+1)5N = 6N-7-N = -7N = 7
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1 + 5/n
geometric mean is: (5 x 135)1/2 = √675 ≈ 25.98 To find the geometric mean of n numbers, multiply them together and take the nth root, so the geometric mean of x1, x2, ..., xn is: geometric mean = (Π xr)1/n for r = 1, 2, .., n
If you mean the sum of first n natural numbers then the formula is (n(n+1))/2 for example, 1+2+3+4+5=5*6/2=15
1. aveia n.f.2. calo n.3. calosidade n.f.4. milho n.5. trigo n.
The sum from 1 to n is n*(n+1)/2In this case that mean 20*21/2 = 210The sum from 1 to n is n*(n+1)/2In this case that mean 20*21/2 = 210The sum from 1 to n is n*(n+1)/2In this case that mean 20*21/2 = 210The sum from 1 to n is n*(n+1)/2In this case that mean 20*21/2 = 210
If you mean: -6n+5 = 1 then the value of n is 2/3
If you mean -17 = 15n -20 then the value of n is 1/5
n is a variable. It could mean any number you want it to. Most of the time, it is not specified what it is. Sometimes they give you clues about it, like "5 times n is 5." In this case n is 1.
If you mean: -17 = 15n -20 then the value of n works out as 1/5
The mean of N numbers is the sum of those numbers divided by N. If you include 13 in that set and the new mean is 6, then the original N is 7. Let X be the original sum. You have a system of two simultaneous equations. 1.) X/N = 5 2.) X+13/N+1 = 6 To solve, substitute the value of X from eq. 1 into eq. 2 and then solve eq. 2 for N. (Some steps combined) X = 5N5N+13 = 6(N+1)5N = 6N-7-N = -7N = 7
if n=4 then n+1 would be 4+1 which equals 5