Let Ω be the set of all permutations of the numbers 1,2, ..., n. Let an almostfixed point be defined as follows: If we put the numbersi∈1,2, ..., naround acircle in clockwise order (such that 1 andnare next to each other) and thenassign another numberω(i)∈1,2, ..., nto it, if the numberω(i) is next toi(oris equal toi), we will say thatiis almost a fixed point. So, for the permutationω(1) = 5, ω(2) = 3, ω(3) = 1, ω(4) = 4, ω(5) = 2, we have that 1, 2, and 4 arealmost fixed points. Now, letX(ω) denote the number of almost fixed points inω∈Ω. FindE[X]and var(X). You may assume thatn≥5

the perimeter is increased by a factor of 2

38.5

step-by-step explanation:

lets say, first number = r

second number = 3r+7

total = 49

(r)+(3r+7)= 49

4r+7 = 49

4r = 42

r = 42/4

r= 10.5

3r+7 = 3(10.5)+7 = 38.5

hope this

215.92 dollars is the answer

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet