Well you have to find the GCF. The GCF in this problem would be 4. So divide 92 and 100 by 4. 92/4 is 23 and 100/4 is 25. So 92/100 is 23/25 in lowest terms.
Three times (3 times something) the difference of x and 2 (2-x) is greater than (>) -21 (-21)
3 times (2-x)>-21
3(2-x)>-21
distrobite using distributiver poerpoty
a(b+c)=ab+ac
3(2-x)=6+(-3x)=6-3x
6-3x>-21
add 3x to both sides
6>-21+3x
add 21 to both sides
27>3x
divide both sides by 3
9>x
x<9
so x= any number below 9 not including 9
11.22 is 33% of 34 if that is what youre asking
Answer:
4 games played = 1/56
5 games played = 5/56
6 games played = 15/56
7 games played = 35/56
Step-by-step explanation:
The probability of each time winning is 0.5
So there are a couple of ways the series could go.
- Team A could win all first 4 matches. We can depict this as A-A-A-A
- Team A could win 4, while having lost 1. Lets depict this a A-B-A-A-A
- Team A could win 4, while having lost 2. A-B-B-A-A-A
- Team A could win 4, while having lost 3. A-B-B-B-A-A-A
These 4 possibilities could be repeated with B winning as well. It should be noted that these are the only ways for the series to end. We find the number of permutations of each possibility above to find their probability.
I advise you to study 'how to permute identical objects' for this.
These permutations are stated below:
1. 4!/4! = 1
2.
= 5
3
= 15
4
= 35
These are they ways A or B could win. The total is 1+5+15+35 = 56. The answer given thus reflects the possibilities.
Answer:
<em>The probability is 0.89 or 89%</em>
Step-by-step explanation:
<u>Probability</u>
There are n=27 students in a certain Algebra 2 class. 18 out of them play basketball and 16 of them play baseball.
We also know 3 students play neither sport. This leaves 27-3=24 students playing sports.
The probability that a randomly chosen student plays basketball or baseball is:

Simplifying:

The probability is 0.89 or 89%