So all you have to do is multiply 400 times 400 which is 160,000 and divide 169,000 to 2 which is 80,000
Answer:
![f(x)=4\sqrt[3]{16}^{2x}](https://tex.z-dn.net/?f=f%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D)
Step-by-step explanation:
We believe you're wanting to find a function with an equivalent base of ...
![4\sqrt[3]{4}\approx 6.3496](https://tex.z-dn.net/?f=4%5Csqrt%5B3%5D%7B4%7D%5Capprox%206.3496)
The functions you're looking at seem to be ...
![f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x](https://tex.z-dn.net/?f=f%28x%29%3D2%5Csqrt%5B3%5D%7B16%7D%5Ex%5Capprox%202%5Ccdot2.5198%5Ex%5C%5C%5C%5Cf%28x%29%3D2%5Csqrt%5B3%5D%7B64%7D%5Ex%3D2%5Ccdot%204%5Ex%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D%5Capprox%204%5Ccdot%206.3496%5Ex%5C%20%5Cleftarrow%5Ctext%7B%20this%20one%7D%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B64%7D%5E%7B2x%7D%3D4%5Ccdot%2016%5Ex)
The third choice seems to be the one you're looking for.
When atleast one dice shows a 6 the possible outcomes will be:
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
(1,6), (2,6), (3,6), (4,6), (5,6)
Thus there are 11 total possible outcomes.
Among these outcomes, the sum of numbers greater than or equal to 9 can be obtained from:
(6,3), (6,4), (6,5), (6,6), (3,6), (4,6), (5,6)
This means there are 7 outcomes with sum greater than or equal to 9.
Thus, Probability of rolling a number greater than or equal to 9 with atleast one dice showing a 6 = 9/11
So, option A gives the correct answer
