Help me asnwer please

Suppose that an airline uses a seat width of 16.516.5 in. assume men have hip breadths that are normally distributed with a mean of 14.414.4 in. and a standard deviation of 0.90.9 in. complete parts (a) through (c) below. (a) find the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.516.5 in. the probability is nothing. (round to four decimal places as needed.)

If sec theta = 5/3 and the terminal point determined by theta is in quadrant 4, then

Consider this equation. |y + 6| = 2 what can be concluded of the equation? check all that apply. there will be one solution. there will be two solutions. the solution to –(y + 6) = 2 will be also be a solution to the given absolute value equation. the solution(s) will be the number(s) on the number line 2 units away from –6. the value of y must be positive since the variable is inside absolute value signs.