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Svetradugi [14.3K]
3 years ago
7

If you embed a multimedia file, you reduce the bandwidth your Web site must manage.

Engineering
1 answer:
slega [8]3 years ago
7 0
The answer is true hope this helps
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What is Applied Science?​
andreev551 [17]

Answer:

Applied science is the application of existing scientific knowledge to practical applications, like technology or inventions. Within natural science, disciplines that are basic science develop basic information to predict and perhaps explain and understand phenomena in the natural world.

6 0
3 years ago
1. Implement the k-means clustering algorithm either in Java or Python. • The program should be executable with at least 3 param
givi [52]

Answer:

The code for this Question in Python is as follows:

matplotlib inline

from copy import deepcopy

import numpy as np

import pandas as pd

from matplotlib import pyplot as plt

plt.rcParams['figure.figsize'] = (16, 9)

plt.style.use('ggplot')

# Importing the dataset

data = pd.read_csv('xclara.csv')

print(data.shape)

data.head()

# Getting the values and plotting it

f1 = data['V1'].values

f2 = data['V2'].values

X = np.array(list(zip(f1, f2)))

plt.scatter(f1, f2, c='black', s=7)

# Number of clusters

k = 3

# X coordinates of random centroids

C_x = np.random.randint(0, np.max(X)-20, size=k)

# Y coordinates of random centroids

C_y = np.random.randint(0, np.max(X)-20, size=k)

C = np.array(list(zip(C_x, C_y)), dtype=np.float32)

print(C)

# To store the value of centroids when it updates

C_old = np.zeros(C.shape)

# Cluster Lables(0, 1, 2)

clusters = np.zeros(len(X))

# Error func. - Distance between new centroids and old centroids

error = dist(C, C_old, None)

# Loop will run till the error becomes zero

while error != 0:

   # Assigning each value to its closest cluster

   for i in range(len(X)):

       distances = dist(X[i], C)

       cluster = np.argmin(distances)

       clusters[i] = cluster

   # Storing the old centroid values

   C_old = deepcopy(C)

   # Finding the new centroids by taking the average value

   for i in range(k):

       points = [X[j] for j in range(len(X)) if clusters[j] == i]

       C[i] = np.mean(points, axis=0)

   error = dist(C, C_old, None)

# Initializing KMeans

kmeans = KMeans(n_clusters=4)

# Fitting with inputs

kmeans = kmeans.fit(X)

# Predicting the clusters

labels = kmeans.predict(X)

# Getting the cluster centers

C = kmeans.cluster_centers_

fig = plt.figure()

ax = Axes3D(fig)

ax.scatter(X[:, 0], X[:, 1], X[:, 2], c=y)

ax.scatter(C[:, 0], C[:, 1], C[:, 2], marker='*', c='#050505', s=1000)

4 0
4 years ago
A direct-coupled amplifier has a low-frequency gain of 40 dB, poles at 2 MHz and 20 MHz, a zero on the negative real axis at 200
rewona [7]

Answer:

Explanation:

Low frequency gain is= 40db= 20logK=>100 poles at 2MHz,20MHz

Zero at -200MHz, zero at infinity.

A) A(s) = 100FH(s)

B) Poles (1): 2 pi × 2 × 10^6= 4pi × 10^6MHz

              (2): 2pi × 20 × 10^6= 4pi × 10^6 MHz

 Zeroed: 2pi × 10^6 × 200= 400pi × 10^6, at infinity.

T/(S) = (1 + S/400π × 10^6)/S(1 + S/4π × 10^6)(1 + S/4π × 10^6)

8 0
3 years ago
A system consists initially of nA moles of gas A at pressure p and temperature T and nB moles of gas B separate from gas A but a
Volgvan

Answer:

A) б = - R ( nA In Ya - nB In Yb )

B) s2 = ( nA + nB ) s( T,P )

C) No entropy will be produced

Explanation:

A) assuming ideal gas behavior the expression for entropy produced

for a closed system : s2 - s1 = б

where : s1 ( initial entropy ) = nA sA ( T, P ) + nB sB ( T, P )

s2 ( final entropy ) = nA sA ( T, YaP ) + nB sB ( T, YbP )

∴ б = - R ( nA In Ya - nB In Yb )

B) Given that

Ya and Yb are less than 1  respectively, hence the value of б  = positive

also assuming the gases are identical

s2 = ( nA + nB ) s( T,P )

C) No entropy will be produced when same gas at same temperature and same pressure are mixed

5 0
3 years ago
Can somebody please help me
astraxan [27]

Answer:

C. 420 x 594 millimeters

6 0
3 years ago
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