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BlackZzzverrR [31]
3 years ago
9

If I could make a Short Film on any topic it would be...how could that film change the world?

Computers and Technology
1 answer:
tensa zangetsu [6.8K]3 years ago
6 0
Your Decision but a good one will be exploring the ocean or the underground it will be very shocking if you find something no one ever has found.
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What is the answer for 2.8.10 word games? This is what I have so far, but I can’t seem to be able to figure out the bananaSplit
Feliz [49]

Answer:

 public String bananaSplit(int insertIdx, String insertText) {

     return word.substring(0, insertIdx) + insertText + word.substring(insertIdx);

 }

Explanation:

Do you have the other parts of the WordGames class?

8 0
2 years ago
Nj hj hjkbh hj g7yubuyiycrtdryfrrcctcftt
Dafna1 [17]

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nice letters

Explanation:

6 0
3 years ago
Read 2 more answers
1. The process of creating workable systems in a very short period of time is called
12345 [234]

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okay i know only 1st question and answer

answer:rapid application development

6 0
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
How many channels can the 2.4 GHz band be divided into?
wlad13 [49]

Answer:Fourteen channels

Explanation:

8 0
3 years ago
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