Zoom with me because I need help with a question on math please. Meeting ID: 523 121 0760
Passcode: 0CHTAZ

(a) (8%) compute the probability of an even integer among the 100 integers 1! , 2! , 3! , until 100! (here n! is n factorial or n*(n-1)*(n-2) *… 1) (b) (16%) compute the probability of an even integer among the 100 integers: 1, 1+2, 1+2+3, 1+2+3+4, …., 1+2+3+… + 99, and 1+2+3+… + 100

Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). solution (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) we start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .

The scatterplot below shows the number of hours that students read weekly and the number of hours that they spend on chores weekly. which statement about the scatterplot is true?