Answer:
the first one
Step-by-step explanation:
![2(\sqrt[3]{16x^{3}y })+4(\sqrt[3]{54x^{6}y^{5} } )= 2(2x\sqrt[3]{2y })+4(3x^{2} y\sqrt[3]{2y^{2} } ) = 4x(\sqrt[3]{2y })+12x^{2} y(\sqrt[3]{2y^{2} } )](https://tex.z-dn.net/?f=2%28%5Csqrt%5B3%5D%7B16x%5E%7B3%7Dy%20%7D%29%2B4%28%5Csqrt%5B3%5D%7B54x%5E%7B6%7Dy%5E%7B5%7D%20%20%7D%20%29%3D%202%282x%5Csqrt%5B3%5D%7B2y%20%7D%29%2B4%283x%5E%7B2%7D%20y%5Csqrt%5B3%5D%7B2y%5E%7B2%7D%20%7D%20%29%20%3D%204x%28%5Csqrt%5B3%5D%7B2y%20%7D%29%2B12x%5E%7B2%7D%20y%28%5Csqrt%5B3%5D%7B2y%5E%7B2%7D%20%7D%20%29)
Answer:

The probability is 3/12. The third option is correct.
Step-by-step explanation:
The sample space is

Note that this sample space is not equally probable.
The probability of getting a given number followed is the probability of getting an even number from the 6 numbers (3/6) multiplied by the probability of getting a head after getting that even number, that is 1/2, because is equally probable to get heads or tails from one single coin toss (note that we are assuming that the dice was even, thats why there is a single coin toss).
Therefore, the probability of getting an even number and a head is
P( D in {2,4,6} , H = 1) = P(D in {2,4,6}) * P(H=1 | D in {2,4,6}) = 3/6 * 1/2 = 3/12.
A. -89
B. -6
C. -17
hope this helped