The following cryptogram has been received: ECGUCTUJKHV The encryption algorithm is as follows. Let x1, x2 be the roots of the polynomial x 2 + 3x + 1. Each letter of the plaintext is replaced with its position in the English alphabet (A is 1 and Z is 26). Then to every number the value of the polynomial f(x) = x 6 + 3x 5+ x 4 +x 3 + 4x 2 + 4x+ 3 either at x1 or at x2 is added. After that the numbers obtained are replaced with letters. Recover the plaintext. (Moral of the story: Repeating complex steps may lead to a very trivial cipher.)

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. the highest speed measured was 78.1 mbps. the complete list of 50 data speeds has a mean of x overbarequals16.11 mbps and a standard deviation of sequals18.65 mbps. a. what is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. how many standard deviations is that [the difference found in part (a)]? c. convert the carrier's highest data speed to a z score. d. if we consider data speeds that convert to z scores between minus2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant? a. the difference is nothing mbps.