If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers. Round to six decimal places. (Hint: these are dependent events)

P = 0.008908

Step-by-step explanation:

The complete question is:

The table below describes the smoking habits of a group of asthma sufferers

               Nonsmokers      Light Smoker     Heavy smoker      Total

Men                        303                     35                           37        375

Women                   413                     31                            45        489

Total                        716                     66                           82        864

If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers.

The number of ways in which we can select x subjects from a group of n subject is given by the combination and it is calculated as:

Now, there are 82C2 ways to select subjects that are both heavy smokers. Because we are going to select 2 subjects from a group of 82 heavy smokers. So, it is calculated as:

At the same way, there are 864C2 ways to select 2 different people from the 864 subjects. It is equal to:

Then, the probability P that two different people from the 864 subjects are both heavy smokers is:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

you probably already have the answer but anyways it's b. 15.0

step-by-step explanation: