Can someone explain how to do this to me

there are 69 men and 59 women in a room. 26 of the men and 27 of the women earn more than $50,000 per year.

1) what is the probability of randomly selecting a man or a person that earns more than $50,000 from this group? round to the nearest 10th of a percent.

  75.0%

Step-by-step explanation:

There are a total of 69+59 = 128 people in the group. Of those, 69 are men and 27 more are people (women) who earn more than $50,000. So the group (man or person earning more than $50,000) consists of 96 people.

The probability of selecting one of those 96 from the group of 128 is ...

  96/128 = 3/4 = 75.0%

In a problem such as this, where the group of interest consists of parts of overlapping groups, the trick is always to find the easiest way to determine the size of the group of interest. Since the group "men" includes "high-earning men" and "low-earning men" and the group "women" includes "high-earning women" and "low-earning women", we could go to the trouble to figure the numbers in each of those four groups. In the end, we want ...

  man or high-earner = "low-earning man" + "high-earning man" + "high-earning woman"

As we saw above, the sum of the first two of these is just "men", so we can find the size of our group of interest by adding ...

  = "men" + "high-earning women"

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb