when you are performing an advanced filter, in which argument box do you enter the range for the dataset?

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

What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?

nasty-shy [4]

Answer:

$0.14\ lb/in^{3}$

Explanation:

Density is defined as mass ler unit volume, expressed as

$\rho=\frac {m}{v}$

Where m is mass, $\rho$ is density and v is the volume. For a sphere, volume is given as

$v=\frac {4\pi r^{3}}{3}$

Replacing this into the formula of density then

$\rho=\frac {m}{\frac {4\pi r^{3}}{3}}$

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

$\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}$

4 0

4 years ago

2.) A fluid moves in a steady manner between two sections in a flow

Talja [164]

Answer:

$250\ \text{lbm/min}$

$625\ \text{ft/min}$

Explanation:

$A_1$ = Area of section 1 = $10\ \text{ft}^2$

$V_1$ = Velocity of water at section 1 = 100 ft/min

$v_1$ = Specific volume at section 1 = $4\ \text{ft}^3/\text{lbm}$

$\rho$ = Density of fluid = $0.2\ \text{lb/ft}^3$

$A_2$ = Area of section 2 = $2\ \text{ft}^2$

Mass flow rate is given by

$m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}$

The mass flow rate through the pipe is $250\ \text{lbm/min}$

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

$m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}$

The speed at section 2 is $625\ \text{ft/min}$ .

3 0

3 years ago

9. To forward bias an NPN transistor, the

murzikaleks [220]

Answer:

a. the base must be more positive than the emitter

Explanation:

A transistor can be defined as a semiconductor component that is used to control the flow of voltage or current and as a gate (switch) for electronic signals. Thus, a transistor allows for the amplification, control and generation of electronic signals in a circuit. The three (3) basic parts of a transistor are; base, emitter and collector.

Basically, there are two (2) main types of transistor and these are;

1. PNP transistor.

2. NPN transistor.

Biasing of a transistor can be defined as the process of providing the controlled amount of direct current (DC) voltage or current conditions so as to enable the transistor amplify the alternating current (AC) input signal correctly.

Hence, to forward bias an NPN transistor, the base must be more positive than the emitter because the majority carriers are electrons which are moved from the n-type region to the p-type region while the minority carriers are holes.

This ultimately implies that, for an NPN transistor to conduct current in milliamps, its base-emitter junction must be forward biased.

4 0

3 years ago

Truckco manufactures two types of trucks: 1 and 2. Each truck must go through the painting shop and assembly shop. If the painti

o-na [289]

He started work at 16 for the North West Company and then the Hudson's Bay Company, becoming a high-ranking officer. From 1851 to 1864, he was Governor of the Colony of Vancouver Island

5 0

3 years ago

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