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3 years ago

5

Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1

Gbps. What is the queueing delay of packet number 48 ?(in milliseconds, rounded to one decimal place, e.g. 0.01234 seconds would be entered as "12.3")

1 answer:

tia_tia [17]3 years ago

4 0

Answer:

938.7 milliseconds

Explanation:

Since the transmission rate is in bits, we will need to convert the packet size to Bits.

1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

5 MiB = 5 × 8 × 2^20 bits.

The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R

where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.

Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9

queueing delay for 48th packet = 0.938725181 seconds

queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds

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Answer:

222.5 Gpa

Explanation:

From definition of engineering stress, $\sigma=\frac {F}{A}$

where F is applied force and A is original area

Also, engineering strain, $\epsilon=\frac {\triangle l}{l}$ where l is original area and $\triangle l$ is elongation

We also know that Hooke's law states that $E=\frac {\sigma}{\epsilon}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}$

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F= 89000 N

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By substitution we obtain

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3 years ago

In multi-grade oil what is W means?

irga5000 [103]

Answer:

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Explanation:

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3 years ago

A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

diamong [38]

Answer:

Explanation:

Given

$q_1=11.5\ nC$ charge is placed at $x=0\ cm$

another charge of $q_2=-1.2\ nC$ is at $x=3\ cm$

We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to $q_2$ at some distance r from it

$E_2=\frac{kq_2}{r^2}$

Now Electric Field due to $q_1$ is

$E_1=\frac{kq_1}{(3+r)^2}$

Now $E_1+E_2=0$

$\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0$

$\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}$

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thus $r=1.43\ cm$

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3 years ago

Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of poun

Alex17521 [72]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.

Data Given:

x = 27 , 44 , 32 , 47, 23 , 40, 34, 52

y = 30, 19, 24, 13 , 29, 19, 21, 14

It is given that,

∑x = 299

∑y = 167

∑ $x^{2}$ = 11887

∑ $y^{2}$ = 3773

We are asked to verify the above values manually in this question.

So,

1. ∑x = 299

Let's verify it:

∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52

∑x = 299

Yes, it is equal to the given value. Hence, verified.

2. ∑y = 167

Let's verify it:

∑y = 30 + 19 + 24 + 13 + 29 + 19 + 21 + 14

∑y = 169

No, it is not equal to the given value.

3. ∑ $x^{2}$ = 11887

Let's verify it:

For this to find, first we need to square all the value of x individually and then add them together to verify.

∑ $x^{2}$ = $27^{2}$ + $44^{2}$ + $32^{2}$ + $47^{2}$ + $23^{2}$ + $40^{2}$ + $34^{2}$ + $52^{2}$

∑ $x^{2}$ = 11,887

Yes, it is equal to the given value. Hence, verified.

4. ∑ $y^{2}$ = 3773

Let's verify it:

Again, for this we need to find the squares of all the y values and then add them together to verify it.

∑ $y^{2}$ = $30^{2}$ + $19^{2}$ + $24^{2}$ + $13^{2}$ + $29^{2}$ + $19^{2}$ + $21^{2}$ + $14^{2}$

∑ $y^{2}$ = 3,845

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3 years ago

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Radda [10]

Answer:

<u>0.59</u>

Explanation:

The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.

The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.

By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

$\frac{214}{225}\cdot\frac{213}{224}\cdot\frac{212}{223}\cdot\frac{211}{222}\cdot\frac{210}{221}\cdot\frac{209}{220}\cdot\frac{208}{219}\cdot\frac{207}{218}\cdot\frac{206}{217}\cdot\frac{205}{216}\cdot\frac{204}{215}\cdot\frac{203}{214}\cdot\frac{202}{213}\cdot\frac{201}{212}\cdot\frac{200}{211}\cdot\frac{199}{210}$

The result is: 0.41278 ≈ 0.41

The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.

Therefore, <em>the probability that the batch is rejected</em> is 1 - 0.41 = 0.59.

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3 years ago