Answer:
for(j = n; j > 0; j--)
System.out.print(\"*\");
Step-by-step explanation:
-A for loop is a repetition control structure.
-It allows the efficiency to write a loop that would otherwise be written a couple of times.
-The output is a single line comprising n asterisks,
-The number of asterisks printed will be equivalent to the n-value declared.
F(x) is a quadratic equation with the x-side squared and a is positive which means that the graph of the function is a parabola facing up. The range of f(x) is given by {y|y ≥ k}, where k is the y-coordinate of the vertex.

, written in vertex form is

, where (h, k) = (-1, -11)
Therefore, range ={y|y ≥ -11}
Answer:
Step-by-step explanation:
This is one minus the probability that all the girls audition for different roles. The total number of ways of assigning roles to the girls is 12^7, because to each of the 7 girls, you have a choice of 12 roles.
Then if each girl is to receive a different role, then there are 12!/5! possibilities for that. If you start assigning roles to the girls, then for the first girl, there are 12 choices, but for the next you have to choose one of the 11 different ones, so 11 for the next, and then one of the 10 remaining for the next etc. etc., and this is 12*11*10*...*6 = 12!/(12-7)! =12!/5!
The probability that a random assignment of one of the 12^7 roles would happen to be one of the 12!/5! roles where each girl has a different role, is
(12!/5!)/12^7 = 12!/(12^7 5!)
Then the probability that two or more girls addition for the same part is the probability that not all the girls are assigned different roles, this is thus:
1 - 12!/(12^7 5!)
Answer:
option D
Step-by-step explanation:
x-2y≥-12
-2y≥-x-12
y≤0.5x+6
has to be a solid line since y is less than or equal to 0.5x+6
so, options 1 and 3 are not applicable
also, in y≤0.5x+6 , y is less than or equal to this line, thus it only exists on the line and to the right of it (below it)
so, option 2 is wrong also, leaving us with option 4 or D, which in this case is the answer
I attached a file of this inequality's graph to prove my answer once more