Consider the following code segments for a Date class:

public class Date
{
private int month;
private int day;
private int year;

public Date( int m, int d, int y ) { ... }
public Date( Date dt ) { / ... missing code ... / }
public String getDate() { ... } // returns date in m/d/y format
public int getMonth() { ... } // returns month
public int getDay() { ... } // returns day
public int getYear() { ... } // returns year
public void setDate( int m, int d, int y ) { ... } // sets date

}

Which of the following code segments correctly implements the missing code for the second constructor?

I. setDate( dt. getMonth(), dt. getDay(), dt. getYear() );

II. month = dt. getMonth();
day = dt. getDay();
year = dt. getYear();

III. month = dt. month;
day = dt. day;
year = dt. year; Select one:

A. I only
B. II only
C. III only
D. I and II only
E. I and III only
F. I, II, and III

3. (6 pts) internally in the computer, with few exceptions, all numerical computation is done using binary numbers. output, however, often uses ascii, which is formed by appending 011 to the left of a bcd code. thus, an algorithm that directly converts a binary integer to a bcd integer is very useful. here is one such algorithm 1) draw lines to the left of the binary number to bound the expected bcd decades. (each decade is a group of 4 bits.) move the binary number one bit to the left. add 0011 to each bcd decade containing a binary value> 0100 repeat steps 2-3 until the last bit in the binary number has been moved into the least significant decade position. (note that when the last bit has been shifted into bcd decade, step 3 is not repeated.) read the bcd result. 2) 3) 4) 5) a) execute the algorithm for the binary number 1101101 b) execute the algorithm for the binary number 01110101110 4. (4 pts) represent the decimal number 3568 in bcd; excess-3 code; ascil; and hex.

You have a large, late-model pick-up truck with a rear seat. the pick-up truck weighs 6,500 pounds. the florida seat belt law

Consider a direct-mapped cache with 216 words in main memory. the cache has 16 blocks of 8 words each. it is a word-addressable computer (rather than a byte-addressable computer which we normally discuss). (a) how many blocks of main memory are there? (b) what is the format of a memory address as seen by the cache, that is, what are the sizes of the tag, cache block, and block offset fields (if they apply)? (c) to which cache block will the memory reference db6316 map?