In order to find the probability of drawing a marble that is NOT BLUE, we must find the ratio of non-blue marbles to the total amount of marbles
Let's count how many marbles total there are first
8 red + 3 blue + 1 green = 12 marbles total
Now, how many marbles are there that <u>ARE NOT BLUE </u>
8 red + 1 green = 9 marbles <u>THAT ARE NOT BLUE </u>
Now<em> let's create a ratio</em> to find a probility
9/12
This can be simplified to 3/4 by taking out the common factor of 3
Your answer is D! "three-fourths"
Answer:
x=35
'x-10'=25 degrees
'x'=35 degrees
The last one is 120 degrees
Step-by-step explanation:
A straight line is 180 degrees
180-60=120
2x+110=180
2x=70
x=35
Answer:

Step-by-step explanation:
1) Find the Greatest Common Factor (GCF).
1 - What is the largest number that divides evenly into
and
?
It is 
2 - What is the highest degree of
that divides evenly into
and
?
It is 1, since
is not in every term.
3 - Multiplying the results above,
The GCF is 4.
2) Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)

3) Simplify each term in parentheses.

4) Factor out common terms in the first two terms, then in the last two terms.

5) Factor out the common term
.
