:pairwise independence. suppose we want to generate n pairwise independent random variables in the range {0, 1, 2, . . , m − 1}. we will assume that n and m are powers of 2 and let n = {0, 1} n and m = {0, 1} m (hence n = log n and m = log m). we saw a scheme in the lecture using mn bits. here we will revisit that scheme in a different way and then see how it can be made more randomness-efficient. pick a uniformly random matrix a ∈ {0, 1} m×n and a random vector b ∈ {0, 1} m . then for a vector v ∈ {0, 1} n , set xv = av + b mod 2 (by this we mean component wise mod 2). (a) suppose we pick a and b uniformly at random. show that under this scheme, for all w ∈ {0, 1} n where w 6= 0 and for all γ ∈ {0, 1} m , pa[aw = γ mod 2] = 1 2m . why does this guarantee that xu and xv are independent for u 6= v and u 6= 0, v 6= 0?

$18 was the price of the t-shirt

step-by-step explanation:

i think it’s c i believe

the speed is   miles per hour > is false

step-by-step explanation:

we know that  

a relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form or

in a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

the graph shown in the figure represent a direct variation

the slope of the line represent the speed

for the point

substitute the values of x and y

the linear equation is equal to

statements

case a) it takes hours to go miles

the statement is true

because

substitute

for

find the value of y in the linear equation

> is correct

case b) the speed is   miles per hour

the statement is false

because the speed is

case c) the unit rate for the trip is

the statement is true

because the speed is equal to the slope of the linear equation

case d) it took minutes to go the first miles

the statement is true

because

substitute

for

find the value of y in the linear equation

> is correct