Two groups are going on a trip to the theater. The first group has 30 students in for adult chaperones. The second group has 25 students in for adult chaperones. The cost, in dollars, for each student ticket, S, and each adult ticket, A, can be determined using the system of equations below

$20

Step-by-step explanation:

Two groups are going on a trip to a theater. The first group has 30 students and 4 adult chaperones. The second group has 25 students and 4 adult chaperones. The cost, in dollars, for each student ticket, s, and each adult ticket, a, can be determined using the system of equations below.

30s+4a=720

25s+4a=620

What is the cost for each student ticket?

Solution:

30s+4a=720

(1)

25s+4a=620 (2)

Subtract (2) from (1)

30s - 25s = 720 - 620

5s = 100

Divide both sides by 5

s = 100/5

= 20

s = $20

Substitute the value of s into (1)

30s + 4a = 720

30(20) + 4a = 720

600 + 4a = 720

4a = 720 - 600

4a = 120

a = 120 / 4

= 30

a = $30

Each student tickets cost $20

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

there is no question content here, but you for the free points! : d