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Vikki [24]
3 years ago
10

Which of the following is not an advantage of concurrent design?

Computers and Technology
2 answers:
andreyandreev [35.5K]3 years ago
8 0

the answer is E. i don't give damm

viva [34]3 years ago
7 0

The answer I believe is B

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malfutka [58]

When adding a header, you need to double tap the top of the document.

4 0
3 years ago
Read 2 more answers
Write a java program called allDigitsOdd that returns whether every digit of a positive integer is odd. Return true if the numbe
Vlada [557]

Answer:

public class Digits

{

   public static boolean allDigitsOdd(int num)

   {

       boolean flag=true;

       int rem;

       while(num>0)

       {

           rem=num%10;

           num=num/10;

           if(rem%2==0)    // if a even digit found immediately breaks out of loop

           {

               flag=false;

               break;

           }

       }

       return flag;     //returns result

   }

   public static void main(String args[])

   {

       System.out.println(allDigitsOdd(1375));    //returns true as all are odd digits

   }

}

OUTPUT :

true

Explanation:

Above program has 2 static methods inside a class Digits. Logic behind above function is that a number is divided by 10 until it is less than 0. Each time its remainder by 0 is checked if even immediately breaks out of the loop.

4 0
4 years ago
If I have an Animal superclass with a Mammal subclass, both concrete and both having a method called eat() with identical signat
Katarina [22]

Answer:

(c) the dynamic type of reference will determine which of the methods to call.

Explanation:

Polymorphism in Object Oriented Programming typically means the same method name can cause different actions depending on which object it is invoked on. Polymorphism allows for dynamic binding in that method invocation is not bound to the method definition until the program executes.

So in the case of Animal superclass and Mammal subclass, both having a method called eat() with identical signatures and return types, depending on which reference, the correct method eat() will be called dynamically upon execution.

For example, if we have the following;

================================

<em>Mammal mammal = new Animal();</em>

<em>mammal.eat()</em>

================================

The eat() method that will be called is the one in the Mammal subclass.

However, if we have;

================================

<em>Animal animal = new Animal();</em>

<em>animal.eat()</em>

================================

The eat() method of the Animal superclass will be called.

5 0
3 years ago
Which of the following jobs is in the agriculture career cluster
Elan Coil [88]
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3 years ago
(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-
I am Lyosha [343]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit  memory  to add in 2 location:

\to \frac{0}{1}  =2^1  \ (\frac{m}{m}  \ location)

The address of 2-bit  memory to add in 4 location:

\to \frac{\frac{00}{01}}{\frac{10}{11}}  =2^2  \ (\frac{m}{m}  \ location)

similarly,

Complete 'n'-bit memory address' location number is = 2^n.Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

= 2^n \\\\ = 2^{12} \\\\ = 4096

In point b:

\to Let \  Mega= 10^6

              =10^3\times 10^3\\\\= 2^{10} \times 2^{10}

So,

\to 2 \ Mega =2 \times 2^{20}

                 = 2^1 \times 2^{20}\\\\= 2^{21}

The memory position for '2^n' could be 'n' m bits'  

It can use 2^{21} bits to address the memory location of 21.  

That is to say, the 2-mega-location memory needs '21' bits.  

Memory Length = 21 bit Address

In point c:

i^{th} element array addresses are given by:

\to address [i] = B+w \times (i-LB)

_{where}, \\\\B = \text {Base  address}\\w= \text{size of the element}\\L B = \text{lower array bound}

\to B=\$ 52\\\to w= 4 byte\\ \to L B= 0\\\to address  = 10

\to address  [10] = \$ 52 + 4 \times (10-0)\\

                       =   \$ 52   + 40 \ bytes\\

1 term is 4 bytes in 'MIPS,' that is:

= \$ 52  + 10 \ words\\\\ = \$ 512

In point d:

\to  base \ address = \$ t 5

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are  < 4 byte

Calculating address:

\to address  [5] = \$ t5 + 4 \times (5-0)\\

                     = \$ t5 + 4 \times 5\\ \\ = \$ t5 + 20 \\\\= \$ t5 + 20  \ bytes  \\\\= \$ t5 + 5 \ words  \\\\= \$ t 10  \ words  \\\\

3 0
4 years ago
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