The answer would be 11
2 goes into 22, 11 times
So 11 would be your answer!
Hope this helps!!
C and b , both can be a number multiplied by 10 to the power of -6
<u>Answer-</u>
<em>The probability that in the box there are 1 red and 8 blue markers is</em><em> 0.111 or 11.1%</em>
<u>Solution-</u>
In the box all markers are red or blue. There are total of 9 markers.
So the number of possible combination number of red or blue marker is,
![S=[(0,9),(1, 8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1),(9,0)}]](https://tex.z-dn.net/?f=S%3D%5B%280%2C9%29%2C%281%2C%208%29%2C%282%2C7%29%2C%283%2C6%29%2C%284%2C5%29%2C%285%2C4%29%2C%286%2C3%29%2C%287%2C2%29%2C%288%2C1%29%2C%289%2C0%29%7D%5D)
As at the random drawn, there was a Blue marker, so the condition of (9,0) i.e 9 Red marker and 0 Blue marker is not a case.
So the sample space becomes,
![S=[(0,9),(1, 8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1)}]](https://tex.z-dn.net/?f=S%3D%5B%280%2C9%29%2C%281%2C%208%29%2C%282%2C7%29%2C%283%2C6%29%2C%284%2C5%29%2C%285%2C4%29%2C%286%2C3%29%2C%287%2C2%29%2C%288%2C1%29%7D%5D)
![|S|=9](https://tex.z-dn.net/?f=%7CS%7C%3D9)
Let us assume that E is the event that in the box there are 1 red and 8 blue markers. So
![E=[(1,8)]](https://tex.z-dn.net/?f=E%3D%5B%281%2C8%29%5D)
![|E|=9](https://tex.z-dn.net/?f=%7CE%7C%3D9)
The probability that in the box there are 1 red and 8 blue markers is,
![P(\text{1 Red and 8 Blue})=\dfrac{|E|}{|S|}=\dfrac{1}{9}=0.111=11.1\%](https://tex.z-dn.net/?f=P%28%5Ctext%7B1%20Red%20and%208%20Blue%7D%29%3D%5Cdfrac%7B%7CE%7C%7D%7B%7CS%7C%7D%3D%5Cdfrac%7B1%7D%7B9%7D%3D0.111%3D11.1%5C%25)
Answer:
I’m not sure but maybe 8?
Step-by-step explanation: