Answer:
Your annswer is b
Step-by-step explanation: because it is least than all of the others
<em>Greetings from Brasil</em>
From radiciation properties:
![\large{A^{\frac{P}{Q}}=\sqrt[Q]{A^P}}](https://tex.z-dn.net/?f=%5Clarge%7BA%5E%7B%5Cfrac%7BP%7D%7BQ%7D%7D%3D%5Csqrt%5BQ%5D%7BA%5EP%7D%7D)
bringing to our problem
![\large{6^{\frac{1}{3}}=\sqrt[3]{6^1}}](https://tex.z-dn.net/?f=%5Clarge%7B6%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7B6%5E1%7D%7D)
<h2>∛6</h2>
Answer:
a. 24
b. 2
c. 0.0833 = 8.33%
Step-by-step explanation:
a.
The first "slot" of person to arrive has 4 possibilities, then the second "slot" will have 3 possibilities, as one has already arrived, then the third "slot" has 2 possibilities, and the fourth "slot" has just 1 possibility.
So, multiplying all these combinations, we have 4*3*2*1 = 24 possible ways they can arrive
b.
If the first and the last person are already "locked", we just have possibilities for the second and third person. The second will have 2 possibilities (Sergio or Tyrone), and the third will have only 1 (the person that wasn't the second between Sergio and Tyrone). So, the number of possibilities is 2*1 = 2
c.
If we have 2 cases where Dawn is first and Jim is last, from a total of 24 possible cases, the probability is 2/24 = 1/12 = 0.0833 = 8.33%
I think that would be 9 I am not sure.
The answer is 300 divided by 20 which you would get 15.