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What is the probability the number contains at least one digit 2?
The probability is 1. And the same is true for any other digit.This counter-intuitive result can be proved as follows:Ignore numbers with leading zeros.There are 9 1-digit numbers of which 8 do not contain any 2s;There are 90 2-digit numbers of which 8*9 do not contain any 2s;There are 900 3-digit numbers of which 8*9*9 do not contain any 2s;and so on.The total number of k digit numbers is 9 + 90 + 900 + ... + 9*10^(k-1)= 9*[10^k-1]/[10 - 1] = 10^k - 1Of these, the count of numbers which do not contain any 2s = 8 + 8*9 + 8*9^2 + ... + 8*9^(k-1)= 8*[9^k-1]/[9 - 1] = 9^k - 1So the probability that the numbers do not contain any 2s is [9^k - 1]/[10^k - 1] which, as k gets larger, approaches 9^k/10^k = (0.9)^k which tends to 0.So, the probability that the number DOES contain at least one 2 = 1 - Pr(It contains none) = 1 - 0 = 1.
What is the sum of k squared and k?
k^2 + k = k^2 + k k^2 x k = k^3
What does 2 less than k is?
The formula is 2-k. The answer depends on the value of k.
How do you find the two square roots of i imaginary number Answer in rectangular form?
The two square roots of i are (k, k) and (-k, -k) where k = sqrt(2)/2 = 1/sqrt(2).
What is the value of k when the line y equals 3x plus 1 is a tangent to the curve of y2 plus x2 equals k?
If: y = 3x +1 then y^2 = 9x^2 +6x +1 If: y^2 +x^2 = k then y^2 = k -x^2 So: 9x^2 +6x +1 = k -x^2 Transposing terms: 10x^2 +6x +(1 -k) = 0 Using the discriminant: 6^2 -4*10*(1 -k) = 0 Solving the discriminant: k = 1/10
