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2 years ago

Suppose that a 10-Mbps wireless station is transmitting 50-byte frames one immediately after the other.

2 answers:

7 0

8 bits = 1 byte

10 Mbps = 10/8 M bytes per second

1 frame = 50 bytes

Frame rate = 10 M / (8 x 50) = 10 M / 400 = 25,000 frames per second

========================

If each frame has an error probability of 0.004, then (0.004 x 25,000) = 100 frames per second are damaged.

1 hour = 3,600 seconds.

100 damaged frames per second = (100 x 3,600) = 360,000 damaged frames per hour.

LenaWriter [7]2 years ago

6 0

There are 1'000'000 Bytes in a Megabyte. In 10 Megabytes there are 10'000'000 bytes. So if it were transmitting 1-byte frames then the answer would be 10'000'000 frames. However this is not the case, because it is transmitting 50-byte frames, so we must divide 10'000'000 by 50.

Therefore, the answer to part (i) is: 200'000 Frames/second

If there are 200'000 Frames/second, then there are 12'000'000 Frames/hour (we get that by multiplying 200'000 by 60)

If the probability of a damaged frame is 0.004, then (multiplying 0.004 by 12'000'000) we get (ii) 48'000 Damaged Frames/hour

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RUDIKE [14]

If the VEHICLE(S) is/are out of balance, the driver will feel a pounding or shaking through the steering wheel.
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2 years ago

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Suppose that the data mining task is to cluster points (with (x, y) representing location) into three clusters, where the points

solong [7]

Answer:

Explanation:

K- is the working procedure:

It takes n no. of predefined cluster as input and data points.

It also randomly initiate n centers of the clusters.

In this case the initial centers are given.

Steps you can follow

Step 1. Find distance of each data points from each centers.

Step 2. Assign each data point to the cluster with whose center is nearest to this data point.

Step 3. After assigning all data points calculate center of the cluster by taking mean of data points in cluster.

repeat above steps until the center in previous iteration and next iteration become same.

A1(4,8), A2(2, 4), A3(1, 7), B1(5, 4), B2(5,7), B3(6, 6), C1(3, 7), C2(7,8)

Centers are X1=A1, X2=B1, X3=C1

A1 will be assigned to cluster1, B1 will be assigned to cluster2 ,C1 will be assigned to cluster3.

Go through the attachment for the solution.

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2 years ago

2. Because technology is always changing, there are new applications being developed constantly. (1 point)

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True because we need new tech

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2 years ago

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In this lab, you use the pseudocode in figure below to add code to a partially created Python program. When completed, college a

Elza [17]

Answer:

Python code is given below with appropriate comments

Explanation:

#Prompt the user to enter the test score and class rank.

testScore = input()

classRank = input()

#Convert test score and class rank to the integer values.

testScore = int(testScore)

classRank = int(classRank)

#If the test score is greater than or equal to 90.

if(testScore >= 90):

#If the class rank is greater than or equal to 25,

#then print accept message.

if(classRank >= 25):

print("Accept")

#Otherwise, display reject message.

else:

print("Reject")

#Otherwise,

else:

#If the test score is greater than or equal to 80.

if(testScore >= 80):

#If class rank is greater than or equal to 50,

#then display accept message.

if(classRank >= 50):

print("Accept")

#Otherwise, display reject message.

else:

print("Reject")

#Otherwise,

else:

#If the test score is greater than or equal to

#70.

if(testScore >= 70):

#If the class rank is greater than or equal

#to 75, then display accept message.

if(classRank >= 75):

print("Accept")

#Otherwise, display reject message.

else:

print("Reject")

#Otherwise, display reject message.

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print("Reject")

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2 years ago

Use the loop invariant (I) to show that the code below correctly computes the product of all elements in an array A of n integer

NeTakaya

Answer:

Given Loop Variant P = a[0], a[1] ... a[i]

It is product of n terms in array

Explanation:

The Basic Step: i = 0, loop invariant p=a[0], it is true because of 'p' initialized as a[0].

Induction Step: Assume that for i = n - 3, loop invariant p is product of a[0], a[1], a[2] .... a[n - 3].

So, after that multiply a[n - 2] with p, i.e P = a[0], a[1], a[2] .... a[n - 3].a[n - 2].

After execution of while loop, loop variant p becomes: P = a[0], a[1], a[2] .... a[n -3].a[n -2].

for the case i = n-2, invariant p is product of a[0], a[1], a[2] .... a[n-3].a[n-2]. It is the product of (n-1) terms. In while loop, incrementing the value of i, so i=n-1

And multiply a[n-1] with p, i.e P = a[0].a[1].a[2].... a[n-2].

a[n-1]. i.e. P=P.a[n-1]

By the assumption for i=n-3 loop invariant is true, therefore for i=n-2 also it is true.

By induction method proved that for all n > = 1 Code will return product of n array elements.

While loop check the condition i < n - 1. therefore the conditional statement is n - i > 1

If i = n , n - i = 0 , it will violate condition of while loop, so, the while loop will terminate at i = n at this time loop invariant P = a[0].a[1].a[2]....a[n-2].a[n-1]

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2 years ago