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lana [24]
3 years ago
6

You’re using Disk Manager to view primary and extended partitions on a suspect’s drive. The program reports the extended partiti

on’s total size as larger than the sum of the sizes of logical partitions in this extended partition. What might you infer from this information?
a. The disk is corrupted.
b. There’s a hidden partition.
c. Nothing; this is what you’d expect to see.
d. The drive is formatted incorrectly
Computers and Technology
1 answer:
Sever21 [200]3 years ago
6 0

Answer:

<em>b. There’s a hidden partition. </em>

Explanation:

The hidden partition <em>is a separate section set aside on OEM computer hard drives, often alluded to as the recovery partition and restore partition.</em>

This portion of memory is used by the manufacturer to preserve the data used to restore your computer to its default settings.

This function is particularly useful as it does not involve the CD or DVD of the operating system.

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You are given 6 training examples for a binary classification problem as follows:
slavikrds [6]

Answer:To simplify the discussion, we will only consider two-class classifiers in this section and define a linear classifier as a two-class classifier that decides class membership by comparing a linear combination of the features to a threshold.

Figure 14.8: There are an infinite number of hyperplanes that separate two linearly separable classes.

\includegraphics[width=6cm]{vclassline.eps}

In two dimensions, a linear classifier is a line. Five examples are shown in Figure 14.8 . These lines have the functional form $w_1x_1+w_2x_2=b$. The classification rule of a linear classifier is to assign a document to $c$ if $w_1x_1+w_2x_2>b$ and to $\overline{c}$ if $w_1x_1+w_2x_2\leq b$. Here, $(x_1, x_2)^{T}$ is the two-dimensional vector representation of the document and $(w_1, w_2)^{T}$ is the parameter vector that defines (together with $b$) the decision boundary. An alternative geometric interpretation of a linear classifier is provided in Figure 15.7 (page [*]).

We can generalize this 2D linear classifier to higher dimensions by defining a hyperplane as we did in Equation 140, repeated here as Equation 144:

\begin{displaymath}

\vec{w}^{T}\vec{x} = b

\end{displaymath} (144)

The assignment criterion then is: assign to $c$ if $\vec{w}^{T}\vec{x} > b$ and to $\overline{c}$ if $\vec{w}^{T}\vec{x} \leq b$. We call a hyperplane that we use as a linear classifier a decision hyperplane .

Figure 14.9: Linear classification algorithm.

\begin{figure}\begin{algorithm}{ApplyLinearClassifier}{\vec{w},b,\vec{x}}

score ...

...in{IF}{score>b}

\RETURN{1}

\ELSE

\RETURN{0}

\end{IF}\end{algorithm}

\end{figure}

The corresponding algorithm for linear classification in $M$ dimensions is shown in Figure 14.9 . Linear classification at first seems trivial given the simplicity of this algorithm. However, the difficulty is in training the linear classifier, that is, in determining the parameters $\vec{w}$ and $b$ based on the training set.

Explanation:

3 0
3 years ago
Name the months that have 30 days​
faust18 [17]

<em>Answer:</em>

<em>September, April, June, and November 29th in each leap year. Hope this helps! Have a blessed day!</em>

6 0
3 years ago
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taking a guess here : a lone.

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The Polish mathematician Wacław Sierpiński described the pattern in 1915, but it has appeared in Italian art since the 13th cent
soldi70 [24.7K]

Answer:

/ Sierpinski.java

public class Sierpinski {

     // method to find the height of an equilateral triangle with side length =

     // length

     public static double height(double length) {

           // formula= length*sqrt(3)/2

           double h = (length * Math.sqrt(3)) / 2.0;

           return h;

     }

     // method to draw a filled triangle (pointed downwards) with bottom vertex

     // x,y

     public static void filledTriangle(double x, double y, double length) {

           // finding height

           double h = height(length);

           // filling triangle as a polygon

           // passing x, x-length/2, x+length/2 as x coordinates

           // and y, y+h, y+h as y coordinates

           StdDraw.filledPolygon(new double[] { x, x - (length / 2.0),

                       x + (length / 2.0) }, new double[] { y, y + h, y + h });

     }

     // method to draw an n level sierpinski triangle

     public static void sierpinski(int n, double x, double y, double length) {

           // checking if n is above 0 (base condition)

           if (n > 0) {

                 // drawing a filled triangle with x, y as bottom coordinate, length

                 // as side length

                 filledTriangle(x, y, length);

                 // drawing the triangle(s) on the top

                 sierpinski(n - 1, x, y + height(length), length / 2);

                 // drawing the triangle(s) on the left side

                 sierpinski(n - 1, x - (length / 2.0), y, length / 2);

                 // drawing the triangle(s) on the right side

                 sierpinski(n - 1, x + (length / 2.0), y, length / 2);

           }

     }

     public static void main(String[] args) {

           // parsing first command line argument as integer, if you dont provide

           // the value while running the program, this program will cause

           // exception.

           int n = Integer.parseInt(args[0]);

           // length of the outline triangle

           double length = 1;

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           double h = height(length);

           // drawing a triangle (pointed upwards) to represent the outline.

           StdDraw.polygon(new double[] { 0, length / 2, length }, new double[] {

                       0, h, 0 });

           // drawing n level sierpinski triangle(s) with length / 2, 0 as x,y

           // coordinates and length / 2 as initial side length

           sierpinski(n, length / 2, 0, length / 2);

     }

}

7 0
3 years ago
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