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lions [1.4K]
3 years ago
13

A chef writing up her famed recipe for beef stew realizes she has switched parsley and oregano everywhere in the recipe. The che

f wants to change all occurrences of “parsley" to “oregano" and all occurrences of “oregano" to “parsley." The chef will use the fact that the word “thyme" does not appear anywhere in the recipe. Which of the following algorithms can be used to fix the chef’s recipe? First, change all occurrences of “parsley" to “oregano." Then, change all occurrences of “oregano" to “parsley." First, change all occurrences of “parsley" to “oregano." Then, change all occurrences of “oregano" to “parsley." Last, change all occurrences of “thyme" to “oregano." First, change all occurrences of “parsley" to “thyme." Then, change all occurrences of “thyme" to “oregano." Last, change all occurrences of “oregano" to “parsley." First, change all occurrences of “parsley" to “thyme." Then, change all occurrences of “oregano" to “parsley." Last, change all occurrences of “thyme" to “oregano."
Computers and Technology
1 answer:
kkurt [141]3 years ago
8 0

Answer:

the last three sentences

Explanation:

first change all of the occurrences of parsley to thyme,

and then all of oregano to parsley

and then all of thyme to oregano

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A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co
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Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k  ...(1)

probability of 2 hops = k(1-k)  ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

                                                       = (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

                                                       = k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops                = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

     ∞                             n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

    n-1

   = 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet =  k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet =  k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

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