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Taya2010 [7]
3 years ago
13

Write a design brief for the given senario.Answer the following questions.​

Mathematics
1 answer:
alisha [4.7K]3 years ago
5 0

He should design a crane which can have its magnet turned on and off. That magnet is called an electromagnet. An electromagnet can be switched on and off when you decide which helps helps the car dealer sort and remove the pieces of magnetic material easily.

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School starts at 8:55. one day katie was a quarter of an hour late. what time did she arrive​
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Step-by-step explanation:

15 minutes is  a quarter of an hour. Add that to the time school starts and it gives you 9:10. Let me know if you have any questions and I'll try my best to help. :)

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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]

∴ we get

x_{2} = \left[\begin{array}{ccc}0.375\\0.3125\\0.3125\end{array}\right]

∴ we get

The probability that it will choose food #2 on the second trial after the initial trial = 0.3125

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3 years ago
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