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ololo11 [35]
3 years ago
10

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-ma

ils containing the same virus, after which the virus disables itself on that machine.(1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds.(2) Solve the recurrence relation.(3) How many e-mails are sent at the end of 20 seconds

Mathematics
2 answers:
pickupchik [31]3 years ago
8 0

Answer:

Answer explained below

Step-by-step explanation:

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-mails containing the same virus, after which the virus disables itself on that machine. (1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds. (2) Solve the recurrence relation. (3) How many e-mails are sent at the end of 20 seconds

1.START T=0.....

1000 EMAILS SENT AND RECEIVED BY 1000 M/CS.

T=1.....

EACH OF THE M/C SENDS 10 NEW MAILS ....

.................................

LET M[N] BE THE NUMBER OF MAILS SENT OUT AFTER N SECONDS.

SO , EACH OF THESE M/CS WILL SEND 10 MAILS IN NEXT 1 SECOND.

HENCE NUMBER OF MAILS SENT IN N+1 SECONDS=M[N+1]=

M[N+1]=10*M[N].......................1

THIS IS THE RECURRENCE RELATION.....

2.SOLUTION .....

M[N+1]=10M[N]=10*10M[N-1]=10*10*10M[N-2]=........

M(N+1)=[10^1][M(N)]=[10^2][M(N-1)]=[10^3][M(N-2)]=..........=[10^N][M(1)]=[10^(N+1)][M(0)]

M[N+1]=[10^(N+1)][1000]=[10^(N+1)][10^3]=[10^(N+4)]......................................2

THIS IS THE NUMBER OF MAILS SENT AFTER N+1 SECONDS .....OR ....

M[N]=[10^(N+3)].............................................3

..................IS THE SOLUTION FOR NUMBER OF MAILS SENT AFTER N SECONDS.....

3.AFTER N=20 SECONDS , THE ANSWER IS ....

M[20]=10^(20+3)=10^23

hoa [83]3 years ago
6 0

Answer:

The answer is attached below

Step-by-step explanation:

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kogti [31]

Answer:

1) 18 Quart of Sparkling water need to mix with 9 quart grape juice

2) 15/2  Quart of grape juice required for 15/4 quarts of sparkling water.

3) Quantity of grape juice  and  sparkling water in 100 quart of punch are 33 1/3 quart  and 66 2/3 quart respectively.

Step-by-step explanation:

1 1/2 quarts = 1 + (1/2) = 3/2 quarts

1)  water required for 3/4 quart of grape juice =  3/2 quarts

so water required for 1 quart of grape juice = (3/2) ÷ (3/4) = (3/2)× (4/3) = 2 quarts

so water required for 9 quart of grape juice = 9 * 2 = 18 quart

18 Quart of Sparkling water need to mix with 9 quart grape juice

2) From solution of 1 ,

   For 18 quart of Sparkling water , grape juice require = 9 quart

   So for 1 quart of Sparkling water , grape juice require = 9÷18 = 1/2 Quart

   so for 15/4 quart of Sparkling water ,  grape juice require = 1/2 × 15/4 =

15/2  Quart of grape juice required for 15/4 quarts of sparkling water.

3)

in 1 we calculated that 18 Quart of Sparkling water need to mix with 9 quart grape juice that is 2 quart of Sparkling water need to mix with 1 quart of grape juice.

In other words 1 quart of grape juice + 2 quart of sparkling water is 3 quart of punch.

Quantity of Grape juice in 3 quart of punch = 1 quart

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And quantity of grape juice in 100 quart of punch = 100×(1/3) = 100/3 =

33 1/3

Quantity of sparkling water in 3 quart of punch = 2 quart

so quantity of sparkling water  in 1 quart of punch = 2/3 quart

And quantity of sparkling water  in 100 quart of punch = 100 ×2/3 quart = 66 2/3


 


     


3 0
3 years ago
There are 50 students going on a class trip. 35
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The answer should be 74% probability

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Step-by-step explanation:

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12x-y-4

Step-by-step explanation:

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Step-by-step explanation:

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