What is the solution to −1.2b−5.3≥1.9

Mathematics

2 answers:

katen-ka-za [31]3 years ago

5 0

B is less than or equal to -6

wel3 years ago

3 0

Answer:

If you want me to solve for b then

b ≤ − 6

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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,

Tresset [83]

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

$H = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]$

As we know that,

The matrix is a Markov chain $x_{k+1} = Hx_{k}$

Let

The initial state vector be

$x_{0} = \left[\begin{array}{ccc}1\\0\\0\end{array}\right]$

we choose this initial vector because given that If the animal chooses food #1 on an initial trial.

Now,

$x_{1} = Hx_{0} \\ = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]\left[\begin{array}{ccc}1\\0\\0\end{array}\right] \\= \left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right]$

∴ we get

$x_{1} = \left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right]$

Now,

$x_{2} = Hx_{1} \\ = \left[\begin{array}{ccc}0.5&0.25&0.25\\0.25&0.5&0.25\\0.25&0.25&0.5\end{array}\right]\left[\begin{array}{ccc}0.5\\0.25\\0.25\end{array}\right] \\= \left[\begin{array}{ccc}0.25+0.0625+0.0625\\0.125+0.125+0.0625\\0.125+0.0625+0.125\end{array}\right]\\= \left[\begin{array}{ccc}0.375\\0.3125\\0.3125\end{array}\right]$