The booster club voted on
where they would go for their
annual trip. a majority of the
club voted to go to a baseball
game. they bought 29 tickets.
some of the tickets cost $21
each and some cost $27 each.
the total cost of all the tickets
was $675. how many tickets
of each price did they buy?

The number of tickets purchased costing $21 each =   18

The number of tickets purchased costing $ 27 each =   11

Step-by-step explanation:

The total number of tickets purchased = 29

Here, let us assume that:

The number of tickets  purchased costing $21   = m

The number of tickets purchased costing $ 27  = 29 - m

So, now the cost of m tickets costing $21 each = m x ( $21)  = 21 m

Also, the cost of purchasing ( 29-m) tickets costing $27 each

= (29-m)x   $27   = 783  - 27 m

Also, the total cost of purchasing  29 tickets  = $ 675

⇒ The total cost of m tickets  +  (29- m ) tickets  = $ 675

or,  21 m + 783  - 27 m  =  675

⇒  - 6 m  = 675  - 783  =  -108

or, m  108/6 =  18

⇒ m  = 18

Hence the number of tickets purchased costing $21 each = m = 18

The number of tickets purchased costing $ 27 each  = 29 - m

= 29 - 18  =  11

answer: it's a

step-by-step explanation:

homozygous. one is homozygous, while two is heterozygous.