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lesya [120]
2 years ago
8

What is a drawback to being in Slide Show mode? a- Being able to review each slide in order

Computers and Technology
2 answers:
pishuonlain [190]2 years ago
8 0

Answer:

b-Inability to make changes to slides

Explanation:

Slide Show mode allows you to see the presentation as the audience will view it and let you practise your speech. This mode helps you to see the final or current result of your presentation so that you can check if your work is satisfactory. It doesn't allow you to make changes to the slides, so if you want to make them, you have to exit from this mode, make changes and come back to slide show mode to see the corrections.

Ilya [14]2 years ago
4 0
B. I'm slideshow mode you are unable to edit slides.
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B. emotional

Explanation:

An emotional argument. An argument does not always have to be made in words.

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Which answer would it be?
tigry1 [53]
<h2>My Answer:</h2>

FTP: The definition on Wikipedia is; "The File Transfer Protocol (FTP) is a standard network protocol used for the transfer of computer files between a client and server on a computer network."

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~TheExpertNerd

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3 years ago
Show the ERD with relational notation with crowfoot. Drexel University Financial Office has made contracts with several local ba
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3 years ago
A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co
tresset_1 [31]

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} }

6 0
3 years ago
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