You just realized the turn signal on your vehicle is broken,

Part 1 of 4 parts for this set of problems: Given an 4777 byte IP datagram (including IP header and IP data, no options) which i

LekaFEV [45]

Answer:

Fragment 1: size (1332), offset value (0), flag (1)

Fragment 2: size (1332), offset value (164), flag (1)

Fragment 3: size (1332), offset value (328), flag (1)

Fragment 4: size (781), offset value (492), flag (1)

Explanation:

The maximum = 1333 B

the datagram contains a header of 20 bytes and a payload of 8 bits( that is 1 byte)

The data allowed = 1333 - 20 - 1 = 1312 B

The segment also has a header of 20 bytes

the data size = 4777 -20 = 4757 B

Therefore, the total datagram = 4757 / 1312 = 4

6 0

2 years ago

Based on the code you created in this Unit, propose a way to re-use most of the code (with different information in the variable

lisov135 [29]

Answer:

Is this a question or an answer

7 0

3 years ago

I need subscribers plz

melamori03 [73]

Answer:

No :) here this is my channel name jshdhebejdb

8 0

2 years ago

Read 2 more answers

In the second example with modulus math, why can't Eve find the solution?

Leno4ka [110]

C. She would have to resort to trial and errors to find a matching exponent

6 0

3 years ago

A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

$^nC_r = \frac{n!}{(n-r)!r!}$

Where $n = 15\ and\ r = 4$

$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

From the given parameters; 3 out of 15 is detective;

So;

$p = 3/15$

$p = 0.2$

$q = 1 - p$

$q = 1 - 0.2$

$q = 0.8$

Solving further using binomial;

$(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n$

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

$Probability = ^nC_3pq^3$

Substitute 4 for n, 0.2 for p and 0.8 for q

$Probability = ^4C_3 * 0.2 * 0.8^3$

$Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3$

$Probability = 4 * 0.2 * 0.8^3$

$Probability = 0.4096$

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

Probability that at least one is defective + Probability that at none is defective = 1

Probability that none is defective is calculated as thus;

$Probability = q^n$

Substitute 4 for n and 0.8 for q

$Probability = 0.8^4$

$Probability = 0.4096$

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

8 0

3 years ago