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Oduvanchick [21]
2 years ago
9

Digital cellphones use __________ to reduce the size of the channels that are required by scattering the digital fragments of co

nversations over many channels.
Computers and Technology
1 answer:
Yuri [45]2 years ago
7 0

Digital cellphones use Digital Compression to reduce the size of the channels that are required, by scattering the digital fragments of conversations over many channels

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Which organization supports professionals who assist families in making informed choices about products and services?
Naily [24]
The organisation that would best aid families and choose the right products to buy would be the American Association of Family and Consumer Sciences. In addition, the institution is primarily composed of home economists wherein these professionals promote efficient consumer consumption of goods.
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Theo is inspecting his document before he shares his PowerPoint. Which aspects can be inspected by the document inspector? Check
ryzh [129]

comments and annotations


document properties and personal information


invisible on-slide content


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2 years ago
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In a certain game, a player may have the opportunity to attempt a bonus round to earn extra points. In a typical game, a player
Assoli18 [71]

Answer:

С The player had 3 bonus round attempts and 7 of them earned extra points.

Explanation:

Given

See attachment for correct code segment

Required

Which of the options is not possible?

From the question, we understand that:

attempts \to [1,4] --- attempt can only assume values 1, 2, 3 and 4

The following "if statement" is repeated three times

<em>IF RANDOM[1,10] <= 7:</em>

<em>    success = success + 1</em>

This implies that the maximum value of the success variable is 3

The first printed value is the attempt variable

The second printed value is the success variable.

From the list of given options, (a), (b) and (d) are within the correct range of the two variable.

While (c) is out of range because the value printed for variable success is 7 (and 7 is greater than the expected maximum of 3)

8 0
1 year ago
Algorithmic Complexity: what is the asymptotic complexity (Big-O) of each code section? Identify the critical section of each.\
AleksandrR [38]

Answer:

Check the explanation

Explanation:

1) f(n) = O( 1 ), since the loops runs a constant number of times independent of any input size

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

2) f(n) = O( log n! ), the outer loop runs for n times, and the inner loop runs log k times when i = k,ie the total number of print will be – log 1 + log2 +log3 +log4+…...+ log n = log (1 . 2 . 3 . 4 . ……. . n ) =log n!

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

Note : Log (m *n) = Log m + Log n : this is property of logarithm

3) f(n) = O( n^2 ), since both outer and inner loop runs n times hence , the total iterations of print statement will be : n +n+n+…+n

for n times, this makes the complexity – n * n = n2

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

4 0
2 years ago
Write a recursive method called isReverse that accepts two strings as parameters and returns true if the two strings contain the
Virty [35]

Answer:

public static boolean isReverse(String s1, String s2)

{

if (s1.length() == 0 && s2.length() == 0)

{

return true;

}

else if (s1.length() == 0 || s2.length() == 0)

{

return false;

} else

{

String s1first = s1.substring(0, 1);

String s2last = s2.substring(s2.length() - 1);

return s1first.equalsIgnoreCase(s2last) &&

isReverse(s1.substring(1), s2.substring(0, s2.length() - 1));

}

}

public static boolean isReverse(String s1, String s2)

{

if (s1.length() != s2.length()) {

return false; // not same length

} else if (s1.length() == 0 && s2.length() == 0) {

return true;

} else

{

s1 = s1.toLowerCase();

s2 = s2.toLowerCase();

return s1.charAt(0) == s2.charAt(s2.length() - 1) &&

isReverse(s1.substring(1, s1.length()),

s2.substring(0, s2.length() - 1));

}

}

public static boolean isReverse(String s1, String s2)

{

if (s1.length() == s2.length())

{

return isReverse(s1.toLowerCase(), 0, s2.toLowerCase(), s2.length() - 1);

} else {

return false;

// not same length

}

}

private static boolean isReverse(String s1, int i1, String s2, int i2)

{

if (i1 >= s1.length() && i2 < 0) {

return true;

} else {

return s1.charAt(i1) == s2.charAt(i2) &&

isReverse(s1, i1 + 1, s2, i2 - 1);

}

}

public static boolean isReverse(String s1, String s2)

{

return reverse(s1.toLowerCase()).equals(s2.toLowerCase());

}

private static String reverse(String s)

{

if (s.length() == 0) {

return s;

} else {

return reverse(s.substring(1)) + s.charAt(0);

}

}

Explanation:

3 0
2 years ago
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