Believe it or not, :), the width of the ribbon needs to be known to get the right answer. If the ribbon had an infinitely small width the ribbon needed would just be the perimeter.
P=2(L+W)
P=2(12+15)
P=54 in
However an infinitely small width is not something that has any practical application. For example, if the width of the ribbon were 1/2 in.
P=2((L+1)+W)
P=2(16+12)
P=56 in So the perimeter increased by 2 in when the ribbon was only 1/2 in wide.
The "infinitely small" width and the perimeter of 54 in is only a decent approximation if the ribbon were a very thin string.
The first digit can be any one of the four. For each of those . . .
The second digit can be any one of the remaining 3 .
The total number of possible arrangements is (4 x 3) = 12 .
Answer:
Step-by-step explanation:
Did you mean.






No option Solution.
Answer:
$88,000
Step-by-step explanation:
40,000*.12= 4,800
4,800*10= 48,000
40,000+48,000= 88,000
Step-by-step explanation:
