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Since n! is the product of all the numbers from 1 through n and (n+1)! is everything in n! multiplied by n+1, the quotient is n+1 ■
m = n/(n-1)
N is 98 and M is 1
Let the integers be m and n.∴ By the condition given in the question,m + n - mn = 72∴ m + n - mn = 71 + 1∴ m + n - mn - 1 = 71∴ (m - 1)(1 - n) = 71∴ (m - 1)(1 - n) = 1 × 71 = 71 × 1 = (-1) × (-71) = (-71) × (-1)Case 1:m - 1 = 1 and 1 - n = 71∴ (m, n) = (2, -70)Case 2:m - 1 = 71 and 1 - n = 1∴ (m , n) = (72, 0)This case is inadmissible as m, n are non zero integers.Case 3:m - 1 = -1 and 1 - n = -71∴ (m , n) = (0, 72)This case is inadmissible as m, n are non zero integers.Case 4:m - 1 = -71 and 1 - n = -1∴ (m, n) = (-70, 2)But as the given expression is symmetric, therefore, (2, -70) and (-70, 2) cannot be considered two different pairs.∴ We get only one solution.
Proof: P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n} (Bayes theorem) =P{T>n+m}/P{T>n} =((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m (1-p)^m = {T>m} So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.
5m-13n-n+4m= (5+4)m+(-13-1)n= 9m-14n
m-2+1-2m+1 When simplified: -m
if we have two negative numbers, -n and -m and we divide them: (-n)/(-m) then we can rewrite the quotient as : (-1)*n/(-1)*m and... (-1)/(-1) * n/m we know that any number divided by itself is 1 by some rule: so = (-n)/(-m) = n/m which is positive
Since n! is the product of all the numbers from 1 through n and (n+1)! is everything in n! multiplied by n+1, the quotient is n+1 ■
m = n/(n-1)
N is 98 and M is 1
The additiove opposite is -m+n or n-m. There is also a multiplicative opposite, which is 1/(m-n)
The equation for n layers is S(n) = n(n+1)(2n+1)/6It is simplest to prove it by induction.When n = 1,S(1) = 1*(1+1)(2*1+1)/6 = 1*2*3/6 = 1.Thus the formula is true for n = 1.Suppose it is true for n = m. That is, for a pyramid of m levels,S(m) = m*(m+1)*(2m+1)/6Then the (m+1)th level has (m+1)*(m+1) oranges and soS(m+1) = S(m) + (m+1)*(m+1)= m*(m+1)*(2m+1)/6 + (m+1)*(m+1)= (m+1)/6*[m*(2m+1) + 6(m+1)]= (m+1)/6*[2m^2 + m + 6m + 6]= (m+1)/6*[2m^2 + 7m + 6]= (m+1)/6*(m+2)*(2m+3)= (m+1)*(m+2)*(2m+3)/6= [(m+1)]*[(m+1)+1)]*[2*(m+1)+1]/6Thus, if the formula is true for n = m, then it is true for n = m+1.Therefore, since it is true for n =1 it is true for all positive integers.
Let the integers be m and n.∴ By the condition given in the question,m + n - mn = 72∴ m + n - mn = 71 + 1∴ m + n - mn - 1 = 71∴ (m - 1)(1 - n) = 71∴ (m - 1)(1 - n) = 1 × 71 = 71 × 1 = (-1) × (-71) = (-71) × (-1)Case 1:m - 1 = 1 and 1 - n = 71∴ (m, n) = (2, -70)Case 2:m - 1 = 71 and 1 - n = 1∴ (m , n) = (72, 0)This case is inadmissible as m, n are non zero integers.Case 3:m - 1 = -1 and 1 - n = -71∴ (m , n) = (0, 72)This case is inadmissible as m, n are non zero integers.Case 4:m - 1 = -71 and 1 - n = -1∴ (m, n) = (-70, 2)But as the given expression is symmetric, therefore, (2, -70) and (-70, 2) cannot be considered two different pairs.∴ We get only one solution.
Exponents are subject to many laws, just like other mathematical properties. These are X^1 = X, X^0 = 1, X^-1 = 1/X, X^m * X^n = X^m+n, X^m/X^n = X^m-n, (X^m)^n = X^(m*n), (XY)^n = X^n * Y^n, (X/Y)^n = X^n/Y^n, and X^-n = 1/X^n.
Proof: P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n} (Bayes theorem) =P{T>n+m}/P{T>n} =((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m (1-p)^m = {T>m} So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.
They are numbers of the form m, m+1 and m+2 where m is an integer. However, sometimes it can be easier - particularly with an odd number of consecutive integers - to write them as n-1, n and n+1 where n is an integer (= m+1).