Answer:
Barbara's horse can add
Step-by-step explanation:
To test if the horse could add, Eric he tried to calculate experimentally the probability that the horse had guessed the 8 answers.
If we call "trial" the action of throwing 8 coins and count the number of tails obtained, then eric made 53 trials.
During each trial, 8 events could have occurred:
Get 1, 2, 3, 4, 5, 6, 7 or 8 tails.
The plot of points obtained by Eric shows the frequency in which each of these 8 events occurred during the 53 trials.
The fact that the point diagram shows only one point on event 8, means that it occurred only once during the 53 trials, and compared to the other 7 events, this was the one with a lower frequency (the one that obtained the the highest frequency was the 4-tailed one, with 11 points).
This means that it is very unlikely to get 8 queues when casting 8 coins.
Therefore it also means that it is unlikely that Barbara's horse has guessed the 8 answers.
So, curiously, Barbara's horse maybe knows how to add
![(\sqrt[4]{7})^5 = (7^{\frac 1 4})^5 = 7^{\frac 5 4} ](https://tex.z-dn.net/?f=%20%28%5Csqrt%5B4%5D%7B7%7D%29%5E5%20%3D%20%287%5E%7B%5Cfrac%201%204%7D%29%5E5%20%3D%207%5E%7B%5Cfrac%205%204%7D%0A)
![(3^{2/3})^{1/6} = 3^{ \frac 2 3 \times \frac 1 6 } = 3^{\frac 1 9} = \sqrt[9]{ 3} ](https://tex.z-dn.net/?f=%283%5E%7B2%2F3%7D%29%5E%7B1%2F6%7D%20%3D%203%5E%7B%20%5Cfrac%202%203%20%5Ctimes%20%5Cfrac%201%206%20%7D%20%3D%203%5E%7B%5Cfrac%201%209%7D%20%3D%20%5Csqrt%5B9%5D%7B%203%7D%20%0A%0A%0A)
![\dfrac {2^{3/4} } {2 ^{1/2} } = 2^{3/4 - 1/2} = 2^{1/4} = \sqrt[4] {2} ](https://tex.z-dn.net/?f=%5Cdfrac%20%7B2%5E%7B3%2F4%7D%20%7D%20%7B2%20%5E%7B1%2F2%7D%20%7D%20%20%3D%202%5E%7B3%2F4%20-%201%2F2%7D%20%3D%202%5E%7B1%2F4%7D%20%3D%20%5Csqrt%5B4%5D%20%7B2%7D%0A)