Consider f (x) = StartRoot x squared minus 1 EndRoot and g (x) = StartRoot x squared + 1 EndRoot. What value(s) of x would make f(g(x)) and g(f(x)) commutative? x greater-than-or-equal-to 1 or x less-than-or-equal-to negative 1 Negative 1 less-than-or-equal-to x less-than-or-equal-to 1 any value of x no value of x

Any value of x

Step-by-step explanation:

Given

Required

What value of x is  

Solving for f(g(x))

Solve the inner square

Solving g(f(x))

Equate f(g(x)) and g(f(x))

This implies that at any value of x

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560

idk wbu

step-by-step explanation: