I need help with #4 plzz

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

pickupchik [31]

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

8 0

3 years ago

Read 2 more answers

Let B be the basis of P3 consisting of the Hermite polynomials in Exercise 21, and let p.t / D 7 ! 12t ! 8t 2 C 12t 3. Find the

Anastasy [175]

To calculate the relative vector of B we have to:

$P_B=\left[\begin{array}{ccc}3\\3\\-2\\3/2\end{array}\right]$

The coordenates of:

$p(t)= 12t^3-8t^2-12t+7$ , with respect to B satisfy:

$C_1(1)+C_2(2t)+C_3(-2+4t^2)+C_4(-12t+8t^3)= 7-12t-8t^2+12t^3$

Equating coefficients of like powers of t produces the system of equation:

$\left \{ {C_1-2C_3=7} \atop {2C_2-12C_4=-12} \right. \\\left \{ {{4C_3=-8} \atop {8C_4=12}} \right.$

After solving this system, we have to:

$C_1=3\\C_2= 3\\C_3= -2\\C_4= \frac{3}{2}$

And the result is:

$P_B=\left[\begin{array}{ccc}3\\3\\-2\\3/2\end{array}\right]$

Learn more: brainly.com/question/16850761

4 0

2 years ago

Evaluate t-u/v when t=2 u=(-8) v(-1/2)

Dovator [93]

Simple, plug n chug...

$t- \frac{u}{v}$

t=2

u=-8

v=-0.5

Now, plug in what you have,

$2- \frac{-8}{-0.5}$

$\frac{-8}{-0.5} =16$

Now,

2-16=-14

Thus, your answer, -14.

5 0

2 years ago

Will give brainliest for right answer!

Annette [7]

This is a system of equations. We solve it by setting it up so that when we add the two equations together, one of the variables will cancel out. We can do this by multiplying the bottom equation by 3. This will make our system of equations equal:

3a + b + 225
-3a + 3b = 75

Now we add these two equations together, because the a terms will cancel out.

4b = 300.

We can find what b is by dividing both sides by 4.

b = 75

Next, we plug in b in one of the equations and solve for a. You can use either equation, but I will use the second.

75 - a = 25

Subtract 25 from both sides and add a to both sides:

a = 50

So, the first option is correct. A = 50 and b = 75.

8 0

3 years ago

Tyler used 1 1/2 gallons of paint to paint 5/7 of a fence. How many gallons will it take to paint the whole fence?

Mademuasel [1]

9514 1404 393

Answer:

1) (1 1/2)(7/5) = g

2) (1 1/2)/(5/7) = g

Step-by-step explanation:

1) Tyler has painted 5/7 of the 7/7 of the fence. So the total amount of the fence is (7/7)/(5/7) = 7/5 as much as what was already painted.

The total amount of paint needed is then 7/5 the amount already used:

g = (7/5)(1 1/2) . . . . . gallons of paint needed for the whole fence

__

2) If g gallons are required for the whole fence, the amount used so far for 5/7 of the fence is 5/7 of the quantity required:

(5/7)g = 1 1/2

g = (1 1/2)/(5/7) . . . . division expression for the paint needed

3 0

2 years ago