Answer:
The probability that it will choose food #2 on the second trial after the initial trial = 0.3125
Step-by-step explanation:
Given - A lab animal may eat any one of three foods each day. Laboratory records show that if the animal chooses one food on one trial, it will choose the same food on the next trial with a probability of 50%, and it will choose the other foods on the next trial with equal probabilities of 25%.
To find - If the animal chooses food #1 on an initial trial, what is the probability that it will choose food #2 on the second trial after the initial trial?
Proof -
By the given information, we get the stohastic matrix
![H = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]](https://tex.z-dn.net/?f=H%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.5%260.25%260.25%5C%5C0.25%260.5%260.25%5C%5C0.25%260.25%260.5%5Cend%7Barray%7D%5Cright%5D)
As we know that,
The matrix is a Markov chain 
Let
The initial state vector be
![x_{0} = \left[\begin{array}{ccc}1\\0\\0\end{array}\right]](https://tex.z-dn.net/?f=x_%7B0%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
we choose this initial vector because given that If the animal chooses food #1 on an initial trial.
Now,
![x_{1} = Hx_{0} \\ = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]\left[\begin{array}{ccc}1\\0\\0\end{array}\right] \\= \left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right]](https://tex.z-dn.net/?f=x_%7B1%7D%20%3D%20Hx_%7B0%7D%20%5C%5C%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.5%260.25%260.25%5C%5C0.25%260.5%260.25%5C%5C0.25%260.25%260.5%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.5%5C%5C0.25%5C%5C0.25%5Cend%7Barray%7D%5Cright%5D)
∴ we get
![x_{1} = \left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right]](https://tex.z-dn.net/?f=x_%7B1%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.5%5C%5C0.25%5C%5C0.25%5Cend%7Barray%7D%5Cright%5D)
Now,
![x_{2} = Hx_{1} \\ = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]\left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right] \\= \left[\begin{array}{ccc}0.25+0.0625+0.0625\\0.125+0.125+0.0625\\0.125+0.0625+0.125\end{array}\right]\\= \left[\begin{array}{ccc}0.375\\0.3125\\0.3125\end{array}\right]](https://tex.z-dn.net/?f=x_%7B2%7D%20%3D%20Hx_%7B1%7D%20%5C%5C%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.5%260.25%260.25%5C%5C0.25%260.5%260.25%5C%5C0.25%260.25%260.5%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.5%5C%5C0.25%5C%5C0.25%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.25%2B0.0625%2B0.0625%5C%5C0.125%2B0.125%2B0.0625%5C%5C0.125%2B0.0625%2B0.125%5Cend%7Barray%7D%5Cright%5D%5C%5C%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.375%5C%5C0.3125%5C%5C0.3125%5Cend%7Barray%7D%5Cright%5D)
∴ we get
![x_{2} = \left[\begin{array}{ccc}0.375\\0.3125\\0.3125\end{array}\right]](https://tex.z-dn.net/?f=x_%7B2%7D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0.375%5C%5C0.3125%5C%5C0.3125%5Cend%7Barray%7D%5Cright%5D)
∴ we get
The probability that it will choose food #2 on the second trial after the initial trial = 0.3125
Question
Find the least common multiple of 8 and 15.
Answer= 120
Answer:
Rotate and upload.
Step-by-step explanation:
The magnify doesn't translate rotation.
No - there is a rule that state that the sum of two sides of a triangle must be greater than the 3rd side.
While 9+15 >4
and 4+15 >9
9+4 is not greater than 15 - so this triangle cannot be made.