A stadium has 45,000 seats. Seats sell for $28 in Section A, $24 in Section B, and $20 in Section C. The number of seats in Section A equals the total number of seats in Sections B and C. Suppose the stadium takes in $1,139,600 from each sold-out event. How many seats does each section hold?

Step-by-step explanation:

given:

stadium has 45,000 seats.

section A = $28 per ticket

section B = $24 per ticket

section C = $20 per ticket

$1,139,600 from each sold out event

Seats on A = Seats B + Seats C

find:

How many seats does each section hold?

solution:

since A = B + C, substitute A into eq. 1

                       2

since A = 22,500

         B = 22,500 - C

substitute A and B into eq. 2 to get C

                 4

now plugin the value of C=15,200 into eq. 3

therefore, the number of seats on each sections are:

seats A = 22,500

seats B = 14,900

seats C = 7,600

Proof:

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

the answer is 0.3175 hope that you ^w^

also you might actually want to put this picture up if you want people to solve it luckily i remember this question