Answer:
phones = {'John': '1234567', 'Julie' : '7777777'}
Explanation:
In the code given in the question phones dictionary contains contains two keys John and Julie and have the values '5555555' and '7777777' respectively. Now in the code the key John in the dictionary phones is assigned the value '1234567' .So the value corresponding to the key John becomes '1234567'.
Answer:
In Word, click Mailings > Start Mail Merge > Step-by-Step Mail Merge Wizard to start the mail merge wizard. Choose Labels, and then click Next: Starting document. Choose Label options, select your label vendor and product number, and then click OK. Click Next: Select recipients.
Explanation:
Paint,calculator,camera,snipping tool,file,store,and photis
An unfavorable opinion or feeling formed beforehand or without knowledge, thought, or reason.
2.
any preconceived opinion or feeling, either favorable or unfavorable.
3.
unreasonable feelings, opinions, or attitudes, especially of a hostile nature, regarding an ethnic, racial, social, or religious group.
4.
such attitudes considered collectively:
The war against prejudice is never-ending.
5.
damage or injury; detriment:
a law that operated to the prejudice of the majority.
JAVA programming was employed...
What we have so far:
* Two 2x3 (2 rows and 3 columns) arrays. x1[i][j] (first 2x3 array) and x2[i][j] (second 2x3 array) .
* Let i = row and j = coulumn.
* A boolean vaiable, x1rules
Solution:
for(int i=0; i<2; i++)
{
for(int j=0; j<3; j++)
{
x1[i][j] = num.nextInt();
}
}// End of Array 1, x1.
for(int i=0; i<2; i++)
{
for(int j=0; j<3; j++)
{
x2[i][j] = num.nextInt();
}
}//End of Array 2, x2
This should check if all the elements in x1 is greater than x2:
x1rules = false;
if(x1[0][0]>x2[0][0] && x1[0][1]>x2[0][1] && x1[0][2]>x2[0][2] && x1[1][0]>x2[1][0] && x1[1][1]>x2[1][1] && x1[1][2]>x2[1][2])
{
x1rules = true;
system.out.print(x1rules);
}
else
{
system.out.print(x1rules);
}//Conditional Statement
