A stadium has 45,000 seats. Seats sell for $28 in Section A, $24 in Section B, and $20 in Section C. If section C held 300 fewer seats, then it would have half the number of seats as section B. Suppose the stadium takes in $1,139200 from each sold-out event. How many seats does each section hold?

Let's define the variables:

A = number of seats in section A.

B = number of seats in section B.

C = number of seats in section C.

We have the equations:

A + B + C = 45,000.

C - 300 = B/2

A*$28 + B*$24 + C*$20 = $1,139,200

This is a system of equations, the first step to solve this is to isolate one variable in one of the equations, and then replace it in the others.

I will isolate C in the second equation:

C = B/2 + 300.

Now let's replace this in the other two equations:

A + B + B/2 + 300 = 45,000

A*$28 + B*$24 + (B/2 + 300)*$20 = $1,139,200

Let's simplify these equations:

A + B*(3/2) = 44,700

A*$28 + B*$34 + $6,000 = $1,139,200

Now let's isolate A in the first equation:

A = 44,700 - B*(3/2)

Let's replace this in the other equation:

(44,700 - B*(3/2))*$28 + B*$34 + $6,000 = $1,139,200

Now let's solve this for B.

-B*$8 + $1,252,200 = $1,139,200

-B*$8 =  $1,139,200 - $1,252,200  = -$113,000

B = $113,000/8 = 14,125

Now we can replace that in the equations:

A =  44,700 - B*(3/2) =  44,700 - 14,125*(3/2) = 23,512.5, that we should round up to 23,513.

And C = B/2 + 300 = 7362.5

As we rounded the previous one up, we should round this one down to 7362.

idk

step-by-step explanation:

use that smart brain

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )