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ss7ja [257]
3 years ago
8

Problem 16-03 (Algorithmic) The computer center at Rockbottom University has been experiencing computer downtime. Let us assume

that the trials of an associated Markov process are defined as one-hour periods and that the probability of the system being in a running state or a down state is based on the state of the system in the previous period. Historical data show the following transition probabilities: To From Running Down Running 0.70 0.30 Down 0.20 0.80 If the system is initially running, what is the probability of the system being down in the next hour of operation

Mathematics
1 answer:
ozzi3 years ago
6 0

Find the attachments for complete solution

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A regular hexagon has an apothem measuring 14 cm and an approximate perimeter of 96 cm.
Mariulka [41]

ANSWER

672cm²

EXPLANATION

The area of a regular hexagon is given by the formula,

Area =  \frac{1}{2} ap

where a=14cm is the apothem and p=96cm is the perimeter of the regular hexagon.

We substitute the values into the formula,

Area =  \frac{1}{2}  \times 14 \times 96 {cm}^{2}

Area =672 {cm}^{2}

Therefore the approximate area of the hexagon is 672cm²

8 0
3 years ago
Read 2 more answers
A study of the effects of color on easing anxiety compared anxiety test scores of participants who completed the test printed on
NikAS [45]

Answer:

Step-by-step explanation:

To solve this, we would follow these simple steps. We have

unvrs :

The arithmetic mean, x-bar for the yellow paper group (Y) = 20.6

The arithmetic mean, x-bar for the green paper group (G) = 21.75

Recall that, H0: µY = µG

And from the data we have, we can see that

H0: µY< µG

We proceed to say that the

T-Test-statistic = -0.404

Also, the p-value: 0.349

From our calculations, we can see that the p-value > 0.05, and as such, we conclude that we will not reject H0. This is because there is not enough evidence to show that test that is printed on the yellow paper decreases anxiety at a 0.05 significance level.

6 0
3 years ago
Little Snail is going to visit his friend over at the next pond, 3 miles away. He can crawl ( 1/2. 1/3, 1/4, 3/4, 2/3 ) of a mil
KIM [24]

Solution :

It is given that Little Snail is going to visit one of his friend at the pond which is 3 miles away.

When the snail crawls 1/2 of a mile per day, it will take him, $1 \times \frac{2}{1} \times 3$

   =  6 days to get to the next pond.

When the snail crawls 1/3 of a mile per day, it will take him, $1 \times \frac{3}{1} \times 3$

   =  9 days to get to the next pond.

When the snail crawls 1/4 of a mile per day, it will take him, $1 \times \frac{4}{1} \times 3$

   = 12 days to get to the next pond.

When the snail crawls 3/4 of a mile per day, it will take him, $1 \times \frac{4}{3} \times 3$

   =  4 days to get to the next pond.

When the snail crawls 2/3 of a mile per day, it will take him, $1 \times \frac{3}{2} \times 3$

   =  4.5 days to get to the next pond.

3 0
3 years ago
The number 27 is <br> please help!
alina1380 [7]

Answer:

27 is a composite number

5 0
3 years ago
The table below shows the results of flipping two coins. how does the experimental probability of getting HH compare to the theo
marusya05 [52]
The correct answer is option B. i.e. the experimental probability is 3% greater than the theoretical probability<span>

The </span>theoretical Outcomes are: HH HT TH TT
 Then, the probability of getting HH = 1/4 = 0.25 = 25%

Now, Experimental Outcomes : <span>HH=28 HT=22 TH=34 TT=16
Total number of outcomes = 28+22+34+16 = 100
</span>Then, the probability of getting HH = 28/100 = 0.28 = 28%

Thus, <span>the experimental probability is 3% greater than the theoretical probability</span>
8 0
3 years ago
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