Consider the following: cos(x) = x^3
The equation cos(x) = x^3 is equivalent to the equation f(x) = cos(x) - x^3 = 0. f(x) is continuous on the interval [0, 1], f(0) = 1 and f(1) = Since there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos(x) = x^3, n the interval (0, 1).
(a) Prove that the equation has at least one real root.
(b) Use a calculator to find an interval of length 0.01 that contains a solution.

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

i think the answer would be c

step-by-step explanation:

f(gx)=f(1/x+2)

=(1/x+2)+2

=(1+2x)/2+2

=(1+2x+4)/2

= (2x+5)/2