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Gnoma [55]
2 years ago
12

Create a class named BaseballGame that contains data fields for two team names and scores for each team in each of nine innings.

names should be an array of two strings and scores should be a two-dimensional array of type int; the first dimension indexes the team (0 or 1) and the second dimension indexes the inning. Create get and set methods for each field. The get and set methods for the scores should require a parameter that indicates which inning’s score is being assigned or retrieved. Do not allow an inning score to be set if all the previous innings have not already been set. If a user attempts to set an inning that is not yet available, issue an error message. Also include a method named display in DemoBaseballGame.java that determines the winner of the game after scores for the last inning have been entered. (For this exercise, assume that a game might end in a tie.) Create two subclasses from BaseballGame: HighSchoolBaseballGame and LittleLeagueBaseballGame. High school baseball games have seven innings, and Little League games have six innings. Ensure that scores for later innings cannot be accessed for objects of these subtypes.
Computers and Technology
1 answer:
m_a_m_a [10]2 years ago
6 0

Answer:

Check the explanation

Explanation:

BaseballGame:

public class BaseballGame {

  protected String[] names = new String[2];

  protected int[][] scores;

  protected int innings;

  public BaseballGame() {

      innings = 9;

      scores = new int[2][9];

      for(int i = 0; i < 9; i++)

          scores[1][i] = scores[0][i] = -1;

  }

 

  public String getName(int team) {

      return names[team];

  }

  public void setNames(int team, String name) {

      names[team] = name;

  }

 

  public int getScore(int team, int inning) throws Exception {

      if(team < 0 || team >= 2)

          throw new Exception("Team is ut of bounds.");

      if(inning < 0 || inning >= innings)

          throw new Exception("Inning is ut of bounds.");

     

      return scores[team][inning];

  }

  public void setScores(int team, int inning, int score) throws Exception {

      if(team < 0 || team >= 2)

          throw new Exception("Team is ut of bounds.");

      if(inning < 0 || inning >= innings)

          throw new Exception("Inning is ut of bounds.");

      if(score < 0)

          throw new Exception("Score is ut of bounds.");

      for(int i = 0; i < inning; i++)

          if(scores[team][i] == -1)

              throw new Exception("Previous scores are not set.");

     

      scores[team][inning] = score;

  }

 

}

HighSchoolBaseballGame:

public class HighSchoolBaseballGame extends BaseballGame {

  public HighSchoolBaseballGame() {

      innings = 7;

      scores = new int[2][7];

      for(int i = 0; i < 7; i++)

          scores[1][i] = scores[0][i] = -1;

  }

}

LittleLeagueBaseballGame:

public class LittleLeagueBaseballGame extends BaseballGame {

  public LittleLeagueBaseballGame() {

      innings = 6;

      scores = new int[2][6];

      for(int i = 0; i < 6; i++)

          scores[1][i] = scores[0][i] = -1;

  }

}

You might be interested in
Add the following 2's complement binary numbers. Also express the answer in decimal. a. 01+ 1011b. 11+ 01010101c. 0101+ 110d. 01
kenny6666 [7]

Answer:

(a) 01 + 1011

Convert the two numbers into their 4-bit representation.

=> 01 = 0001

=> 1011

Addition of the two numbers (in their 2's complement form) gives the following:

0 0 0 1       ------------------> + 1 (in decimal)

<u>1  0 1  1 </u>      ------------------> - 5 (in decimal)

<u>1  1  0 0 </u>    -------------------> - 4 (in decimal)

Explanation for (a)

<em>01 or 0001</em>

Since the most significant bit is 0 (leftmost bit), it shows that it is a positive number. Therefore, it can be directly converted to its decimal representation as follows:

0001 = 0 x 2^{3} + 0 x 2^{2} + 0 x 2^{1} + 1 x 2^{0} = 1

0001 is + 1 in decimal.

<em>1011</em>

Its most significant bit(MSB) is 1 showing that it is negative.

Flipping all its bits and adding 1 to the result, will convert it to it's positive counterpart.

i.e 1011 => 0100 + 1 = 0101.

Converting the result (0101) to decimal gives

0101 = 0 x 2^{3} + 1 x 2^{2} + 0 x 2^{1} + 1 x 2^{0} = 5

1011 is -5 in decimal

<em>01 + 1011 or 0001+ 1011 = 1100</em>

The result (1100), of the addition of the two numbers, is a negative number as the MSB is 1.

Convert it to it's positive counterpart.

i.e 1100 => 0011 + 1 = 0100.

Converting the result (0100) to decimal gives

0100 = 0 x 2^{3} + 1 x 2^{2} + 0 x 2^{1} + 0 x 2^{0} = 4

Therefore, the result, 1100 is -4 in decimal.

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

(b) 11 + 01010101

Convert the two numbers into their 8-bit representation since one of them is in 8-bit.

PS: Taking 11, its MSB is 1 showing that it is a negative number. Also, it's 8-bit representation will mean pre-padding it with ones(1s) rather than zeros as follows. In other words, pre-padding a 2's complement  number is done using its sign bit.

=> 11 = 11111111

=> 01010101

Addition of the two numbers (in their 2's complement form) gives the following:

  1  1  1  1  1  1  1  1       ------------------> - 1 (in decimal)

<u>   0 1  0 1  0  1 0 1 </u>      ------------------> + 85 (in decimal)

<u>1 0 1  0 1  0  1 0 0</u>    ------------------->  + 84 (in decimal)

Discard the carry-out bit (1) making the result 01010100

Explanation for (b)

<em>11 or 11111111</em>

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

11111111 => 00000000 + 1 = 00000001

Converting the result (00000001) to decimal gives 1

11 or 11111111 is  - 1 in decimal.

<em>01010101</em>

Its MSB is 0 showing that it is also positive. Now, convert to decimal

01010101 = 85

01010101 is + 85 in decimal.

<em>11111111 + 01010101 = 01010100</em>

The result (01010100), is also positive as the MSB is 0. Now, convert to decimal.

01010100 = 84

Therefore the result (01010100) is + 84 in decimal.

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

(c) 0101 + 110

Convert the two numbers into their 4-bit representations.

=> 0101 = 0101

=> 110 = 1110

Addition of the two numbers (in their 2's complement form) gives the following:

     0 1 0 1       ------------------> + 5 (in decimal)

<u>      1  1 1  0 </u>      -----------------> - 2 (in decimal)

<u>   1 0 0 1  1 </u>    -------------------> + 3 (in decimal)

Dicard the carry-out bit (leftmost bit) to get 0011 as result.

Explanation for (c)

<em>0101</em>

Since the MSB is 0 (leftmost bit), it shows that it is a positive number. Now, convert to decimal.

0101 =  5

0101 is +5 in decimal.

<em>110 or 1110</em>

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

i.e 1110 => 0001 + 1 = 0010.

Converting the result (0010) to decimal gives

0010 =  2

1110 is -2 in decimal

<em>0101 + 110 = 0101+ 1110 = 0011</em>

The result (0011), is a positive number as the MSB is 0.

Converting the result (0011) to decimal gives

0011 = 3

Therefore, the result, 0011 is 3 in decimal.

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

(d) 01 + 10

Convert the two numbers into their 4-bit representation.

=> 01 = 0001

=> 10 = 1110

Addition of the two numbers (in their 2's complement form) gives the following:

0 0 0 1       ------------------> + 1 (in decimal)

<u>1  1  1  0 </u>      ------------------> - 2 (in decimal)

<u>1  1  1  1 </u>    -------------------> - 1 (in decimal)

Explanation for (d)

<em>01 or 0001</em>

Since the MSB is 0 (leftmost bit), it shows that it is a positive number. Now convert to decimal.

0001 = 1

0001 is + 1 in decimal.

<em>1110</em>

Its MSB is 1 showing that it is negative.

Convert it to it's positive counterpart.

i.e 1110 => 0001 + 1 = 0010.

Converting the result (0010) to decimal gives

0010  = 2

1110 is -2 in decimal

<em>01 + 1011 or 0001+ 1110 = 1111</em>

The result (1111), is a negative number as the MSB is 1.

Convert it to it's positive counterpart.

i.e 1111 => 0000 + 1 = 0001.

Converting the result (0001) to decimal gives

0001 = 1

Therefore, the result, 1111 is -1 in decimal.

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Early Intel processors (e.g., the 8086) did not provide hardware support for dual-mode operation (i.e., support for a separate u
Firlakuza [10]

Answer:

Uneven use of resources

Explanation:

Potential problem associated with supporting multi - user operation without hardware support is:

Uneven use of resources: In a situation where we assign a set of resources to user 1 and if a new user comes, then it would be difficult to allocate new resources to him. The processor would get confused between the two users. And the tasks would not be completed. This can affect task processing.

8 0
2 years ago
3. Megan and her brother Marco have a side business where they shop at flea markets, garage sales, and estate
Neko [114]

Answer:

a

Explanation:

Megan doesn't have a registered business. She can't claim insurance

4 0
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IANA has primarily been responsible with assigning address blocks to five regional internet registries (RIR). A tech needs to re
VikaD [51]

Answer:

The American Registry for Internet Numbers ARIN

Explanation:

The American Registry for Internet Numbers (ARIN) is a not for profit organization that serves as the administrator and distributor of Internet numeric resources such as IP addresses (IPv4 and IPv6) ASN for the United States, Canada, as well as North Atlantic and Caribbean islands

There are four other Regional Internet Registry including APNIC, RIPE NCC, LACNIC and AFRINIC.

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A packet analyzer is a program that can enable a hacker to do all of the following EXCEPT ________. Select one: A. assume your i
bonufazy [111]

Answer:

Option (B) is the correct answer of this question.

Explanation:

Packet analyzer is a software application or set of infrastructure capable of unencrypted and recording communication that travels through a virtual system of a computer system.A packet analyzer used to detect network activity is recognized as a broadband monitoring system.

A packet analyzer is a code application that is used for monitoring, intercepting, and recording http requests with the help of a virtual interface.

Other options are incorrect because they are not related to the given scenario.

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