Find the solution set to each inequality. Express the solution in set notation.
6+2 < 5−4
a/5 +8 ≤ 13
−3 −7 > −27
8(−6) > 4(−4)

First, solve each inequality. I'll solve the first one first.

7

≥

2

x

−

5

12

≥

2

x

6

≥

x

Therefore, x could be any number less than or equal to 6. In interval notation, this looks like:

(

−

∞

,

6

]

The parenthesis means that the lower end is not a solution, but every number above it is. (In this case, the lower end is infinity, so a parenthesis must be used, since infinity is not a real number and so it cannot be a solution.) The bracket means that the upper end is a solution. In this case, it indicates that not only could

x

be any number less than 6, but it could also be 6.

Let's try the second example:

3

x

−

2

4

>

4

3

x

−

2

>

16

3

x

>

18

x

>

6

Therefore, x could be any number greater than 6, but x couldn't be 6, since that would make the two sides of the inequality equal. In interval notation, this looks like:

(

6

,

∞

)

The parentheses mean that neither end of this range is included in the solution set. In this case, it indicates that neither 6 nor infinity are solutions, but every number in between 6 and infinity is a solution (that is, every real number greater than 6 is a solution).

Now, the problem used the word "OR", meaning that either of these equations could be true. That means that either

x

is on the interval

(

−

∞

,

6

]

or the interval

(

6

,

∞

)

. In other words,

x

is either less than or equal to 6, or it is greater than 6. When you combine these two statements, it becomes clear that

x

could be any real number, since no matter what number

x

is, it will fall in one of these intervals. The interval "all real numbers" is written like this:

(

−

∞

,

∞

)

Final Answer

a) one student is chosen from classroom a, then a student is chosen from classroom b.  

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b.   this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes.   this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes.   this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes.   this makes them dependent events.