There are 12 marbles in total in the bowl.
From this bowl, we will choose 4 marbles at random.
That's 12C4 or (12*11*10*9) or 11,880 ways
Out of this, what is the probability that there is exactly 1 blue marble.
There are 5 blue balls but only 1 is selected, 5C1 ways
The rest except blue calls are 7, but only 3 slots are left, 7C3 ways
The probably of exactly 1 blue of 4 marbles is then 5C1*7C3/12C4 = 1050/11,880 = 0.0884 or 8.84%
Answer:
look at the picture i have sent
Answer:
21
Step-by-step explanation:
Given,
- Number of red marble = 4
- number of green marble = 3
- number of transparent marble = 1
- number of yellow marble = 3
- number of orange marble = 3
Total number of marble except red and green = 3 +3+1
= 7
So, the total number of possible sets of five marbles such that none of them are green or red can be given by
![n\ =\ ^7C_5](https://tex.z-dn.net/?f=n%5C%20%3D%5C%20%5E7C_5)
![=\ \dfrac{7!}{(7-5)!.5!}](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B7%21%7D%7B%287-5%29%21.5%21%7D)
![=\ \dfrac{7!}{5!.2!}](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B7%21%7D%7B5%21.2%21%7D)
![=\ \dfrac{42}{2}](https://tex.z-dn.net/?f=%3D%5C%20%5Cdfrac%7B42%7D%7B2%7D)
= 21
So, the required number of possible sets are 21.
Common factor is an expression that is common. Her we can see 2x+5 is a factor for both of the terms so it is the common factor.
1. B(4,0) and D(8,10)
2. m=rise/run=5/4
y=mx+b
b=0
y=(5/4)x
point (16/20) , 20=5/4*16, 20=20 true , so point (16/20) is on the line
3.point(20,24) , y=(5/4)x, 24=(5/4)*20, 24=25 false, so point (20,24) is not on the line
4. point(80,100) , y=(5/4)x, 100=(5/4)*80, 100=100 true, so point (80, 100) is on the line
5. you need substitute values of x and y into equation y=(5/4)x, if it will be true, then the point is on the line