The sum of two prime numbers is always even if the number two is not included. Since the values on the list are all odd, one of the numbers has to be two. (D) fits the scenario since 29+2=31 and 29 is prime. The other numbers will be composite when two is subtracted from it.

We have: \(x-3, x-2, x-1, x, x+1, x+2, x+3\) . Adding all of these together, we get \(7x=63, x=9\) . The smallest number is \(6 \) and the largest number is \(12.\) Thus, the answer is \(6+12=\boxed{18}.\) (A).

You're right. I forgot the 0 at the end.

We have the equation: \(1.25x=10.50\), so \(x=\boxed{8.4}\) dollars/

In fact, this is an application os stars and bars, so \(\binom{8+4-1}{3}=\binom{11}{3}=\boxed{165}.\)

Just copy the link here.

We have two equations: x+y=55, x=9+y, so 9+y+y=55, 2y=46, y=23.

x=55-23=32, so 23 and 32.

We have: \(56x+569=54x+811\) . So, \(2x=242\), and \(x=121\) .

The total cost is \(56(121)+569=7345.\)

If the party had \(\boxed{121}\) attendees, the cost at each venue would be \(\boxed{\$7345}\)

Yeah, we need a picture for this.

We have 5a=6p and a=15+p

So, 5(15+p)=6p, 75+5p=6p, p=75 cents.

Now, 5a=6(75)

5a=450 cents

So, 450+450=900 cents or 9 dollars.