The answer is -2 for the csc and it’s also a rational denominator
0% because there is no chance it can happen
There are 4! = 24 poosible permutations of the four letters. Let the letters be A, B, C and D. Two permutations will have only letter A in the correct envelope, two more permutations will have only letter B in the correct envelope, two more will have only letter C in the correct envelope and two more will have only letter D in the correct envelope. Therefore 8 out of the 24 possible permutations will have only one letter with the correct address. The required probability is 8/24 = 1/3.
Answer:
![\sqrt[3]{4(x-3)} /2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B4%28x-3%29%7D%20%20%20%20%2F2)
Step-by-step explanation:
