CPhill

You are correct....!!!....good job  !!!

Whoops...now I made an error...!!!!

[Trying to do math in my head....never a good idea  !!! ]

-72A  = - 216   divide both sides by -72

A  = 3    =   [$3  for adult tickets]

Putting this back into  23 - 6A    =  23 - 6(3) = 23  - 18  =  $5  for child's ticket 

But...you had the right idea...that was my stupid error...!!!

Not registred / Total   = 

18  / 120  =

0.15  = 

15%

OK...just simplify this

12A+14(23-6A)=106

12A  + 14*23  - 14*6A  = 106

12A  + 322 - 84A   = 106

-72A + 322  = 106   subtract 322 from each side

-72A  = - 216   divide both sides by -72

A  = 4    =   [4$ for an adult ticket ]

Now...put  4  bacK into   23  - 6A  to find the cost of a child's ticket  since  B = 23 - 6A 

Does that make sense  ???

Let's get the total number of dresses  = 14+ 37+22+56+33   = 162

And there are 14 red

So %  red  =  Red  /Total    = 14  / 162   ≈  .086  ≈ 8.6%   [round to 9%]  = "A"

OK...good job...look at your first equation...can't we rearrange it so that we have 

B  = 23  - 6A       ???

Knowing this...can you sub it into the second for  "B"   ????

No image is shown....

The Germans can be seated in 6! ways

The French can be seated in 5! ways

And the Italians can be seated in 3! ways

And...the groups themselves can  be  ordered in 3! ways on the 14 chairs

So....the total arrangements are   6! * 5! * 3! * 3!  =    720 * 120 * 6 * 6   =  3,110,400  arrangements

Can you set up the equations, RP???

[ Hint ...let the number of adult tickets be "A"  and the number of student tickets  be  "B" ]

EDIT...sorry  A should be the cost of an adult ticket and  B the cost of a student ticket

a) We will use  1000 - 2(180)  = 1000 - 360  = 640 ft for the other two sides

So...each of the longer sides  =  640 / 2  = 320  ft

b) The first figure must be correct....the other unidentified sides in the other three figures are all unequal

Just draw the first figure  and  put 340 ft on the top and bottom sides