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3 years ago

What is the difference between a modular home and a manufactured home?

Engineering

1 answer:

Gennadij [26K]3 years ago

3 0

Answer:

Explanation:

modular homes are held to the same local, state and regional building codes required for on-site homes, while manufactured homes are held to a federal code set by the Department of Housing and Urban Development (HUD). I think this is right?

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3. (a) (5 points) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queue

Bezzdna [24]

Answer:

(N-1) × (L/2R) = (N-1)/2

Explanation:

let L is length of packet

R is rate

N is number of packets

then

first packet arrived with 0 delay

Second packet arrived at = L/R

Third packet arrived at = 2L/R

Nth packet arrived at = (n-1)L/R

Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R

Now

L / R = (1000) / (10^6 ) s = 1 ms

L/2R = 0.5 ms

average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2

the average queuing delay of a packet = 0 ( put N=1)

4 0

4 years ago

You are given a collection of n bottles of different widths and n lids of different widths and you need to find which lid goes w

Sladkaya [172]

Answer:

void bubble_sort( int A[ ], int n ) {

int temp;

for(int k = 0; k< n-1; k++) {

// (n-k-1) to ignore comparisons of already compared iterations

for(int i = 0; i < n-k-1; i++) {

if(A[ i ] > A[ i+1] ) {

// swapping occurs here

temp = A[ i ];

A[ i ] = A[ i+1 ];

A[ i + 1] = temp ;

}

5 0

3 years ago

Name the variable in this experiment

taurus [48]

You have to show the story or experiment so we know

7 0

3 years ago

A soil weighs 2,520 lbs/CY in its in situ condition, 1,970 lb/CY in its loose condition after excavation, and 3,025 lbs/CY in an

azamat

Answer:

load factor = 0.782

Shrink Factor = 0.833

no of truck is 62500

Explanation:

given data

soil weighs in situ condition = 2,520 lbs/CY

soil weighs in loose condition = 1,970 lb/CY

soil weighs in embanked state = 3,025 lbs/CY

average volume = 16 LCY

soil from a borrow pit = 1 million CCY

solution

first we get here Load Factor that is express as

load factor = $\frac{1,970}{2550}$

load factor = 0.782

and Shrink Factor will be as

Shrink Factor = $\frac{2520}{3025}$

Shrink Factor = 0.833

and

no of truck will be

no of truck = $\frac{1000000}{16}$

no of truck is 62500

6 0

4 years ago

Using the data in the photo write the complex waveform expression

UNO [17]

Answer:

1st Harmonic:

$v(t) = 50\cos(2000\pi t)$

3rd Harmonic:

$v(t) = 9\cos(6000\pi t)$

5th Harmonic:

$v(t) = 6\cos(10000\pi t)$

7th Harmonic:

$v(t) = 2\cos(14000\pi t)$

Explanation:

The general form to represent a complex sinusoidal waveform is given by

$v(t) = A\cos(2\pi f t + \phi)$

Where A is the amplitude in volts of the sinusoidal waveform

Where f is the frequency in cycles per second (Hz) of the sinusoidal waveform

Where $\phi$ is the phase angle in radians of the sinusoidal waveform.

1st Harmonic:

We have A = 50, f = 1000 and φ = 0

$v(t) = 50\cos(2\pi 1000 t + 0) \\\\v(t) = 50\cos(2000\pi t)$

3rd Harmonic:

We have A = 9, f = 3000 and φ = 0

$v(t) = 9\cos(2\pi 3000 t + 0) \\\\v(t) = 9\cos(6000\pi t)$

5th Harmonic:

We have A = 6, f = 5000 and φ = 0

$v(t) = 6\cos(2\pi 5000 t + 0) \\\\v(t) = 6\cos(10000\pi t)$

7th Harmonic:

We have A = 2, f = 7000 and φ = 0

$v(t) = 2\cos(2\pi 7000 t + 0) \\\\v(t) = 2\cos(14000\pi t)$

Note: The even-numbered harmonics have 0 amplitude that is why they are not shown here.

8 0

4 years ago

Read 2 more answers