The correct answer is A. true.
It was written in 1948 by the bureau of Labor statistics .
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Answer:
The answer is design.
Explanation:
I know this because based off the notes it states that "the designer can use a model or mock-up to illustrate the look and feel, to help gain a better understanding of the necessary elements and structures.
Answer:
b. will be lower if consumers perceive mobile phones to be a necessity.
Explanation:
The price elasticity of demand is described as the percentage variation in the demanded quantity of service or goods divided by the change in the percentage of the price. And henceforth it describes the responsiveness of the demanded quantity to a price change. And now if the mobile phones are thought of as being the necessity then the price will increase as demand will increase, and hence the price elasticity of demand will be lower. And if there is an improvement in the production technology then the price will be lowered, and hence price elasticity of demand will be less as the change in the percentage of the price will be negative. And the exact definition of it as we have described above. Hence, b is correct options.
A.) Those that work in the Printing Technology pathway are typically "<span>self-employed, and work indoors"
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Answer:
public class MagicSquare {
public static void main(String[] args) {
int[][] square = {
{ 8, 11, 14, 1},
{13, 2, 7,12},
{ 3, 16, 9, 6},
{10, 5, 4, 15}
};
System.out.printf("The square %s a magic square. %n",
(isMagicSquare(square) ? "is" : "is not"));
}
public static boolean isMagicSquare(int[][] square) {
if(square.length != square[0].length) {
return false;
}
int sum = 0;
for(int i = 0; i < square[0].length; ++i) {
sum += square[0][i];
}
int d1 = 0, d2 = 0;
for(int i = 0; i < square.length; ++i) {
int row_sum = 0;
int col_sum = 0;
for(int j = 0; j < square[0].length; ++j) {
if(i == j) {
d1 += square[i][j];
}
if(j == square.length-i-1) {
d2 += square[i][j];
}
row_sum += square[i][j];
col_sum += square[j][i];
}
if(row_sum != sum || col_sum != sum) {
return false;
}
}
return d1 == sum && d2 == sum;
}
}
