Answer:
21 child tickets.
Step-by-step explanation:
Let x represent number of adult tickets and y represent number of child tickets.
We have been given that on Thursday, three times as many adult tickets as child tickets were sold. We can represent this information in an equation as:

We have been given that at a the movie theater, child admission is $6.10 and adult admission is $9.20.The all tickets were sold for a total of $707.70. We can represent this information in an equation as:
Upon substituting equation (1) in equation (2), we will get:




Therefore, 21 child tickets were sold on Thursday.
Answer:
0.15 for each message
Step-by-step explanation:
The charge per message can be found by dividing the charge by the number of messages. Using the first line of the table, we get ...
charge per message = (total charge)/(number of messages) = $38.40/256
charge per message = $0.15
Alyssa is charged 0.15 for each text message.
_____
We can check other lines of the table to see if charges are really proportional:
$26.55/177 = $0.15
$31.35/209 = $0.15
The given function is,

The graph can be drawn as,
The period will be distance between two consequetive maximas,

Thus, period is correct.
The amplitude can be determined as,

Thus, amplitue is incorrect.
The range is,

Thus, the range is correct.
The midline is,

Thus, the midline is correct.
Thus, option (b) is the solution.
Answer:
D
Step-by-step explanation:
Probability of passing history course = 0.59
Probability of passing math course = 0.60
Probability of passing both courses = 0.49
Probability of passing at least one = 0.59 + 0.60 - 0.49 = 0.7 ##