The ticket price that would maximize the total revenue would be $ 23.
Given that a football team charges $ 30 per ticket and averages 20,000 people per game, and each person spend an average of $ 8 on concessions, and for every drop of $ 1 in price, the attendance rises by 800 people, to determine what ticket price should the team charge to maximize total revenue, the following calculation must be performed:
- 20,000 x 30 + 20,000 x 8 = 760,000
- 24,000 x 25 + 24,000 x 8 = 792,000
- 28,000 x 20 + 28,000 x 8 = 784,000
- 26,000 x 22.5 + 26,000 x 8 = 793,000
- 27,200 x 21 + 27,200 x 8 = 788,000
- 26,400 x 22 + 26,400 x 8 = 792,000
- 25,600 x 23 + 25,600 x 8 = 793,600
- 24,800 x 24 + 24,600 x 8 = 792,000
Therefore, the ticket price that would maximize the total revenue would be $ 23.
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A) > since it is 50c per weekday and 75c each weekend assuming it allows for the 2 days each saturday/sunday.
50c * 5 = $2.50 since there are 5 days in weekdays
75c * 2 = $1.50 since there are 2 days in the weekend
Add $2.50 and $1.50 to get $4.00
b) For 3 school days we know it is a weekday on the school week.
So perform 50c * 3 which gives us <span>$1.50
</span>c) 12 days off from school is 10 weekdays and 1 weekend or 2 days of 75c
So now just do 50c * 10 which is $5.00 and 75c * 2 which is $1.50
Add $5.00 and $1.50 and we get $6.50
d) 4 weeks = 20 weekdays since 5 *4 = 20 and 8 days in each weekend since 2 * 4 = 8
Now that we have the amount of weekdays and weekend days we can multiply.
50c * 20 = $10.00
75c * 4 = $3.00
Add $10.00 and $3.00 to get $13.00 for 4 weeks.
e) We have 1 day of the weekend and 2 weekdays here.
50c * 2 = $1.00
75c * 1 = 75c
<span>$1.00 + 75c = $1.75 in those 3 days listed.
</span>
Add all these together to get your total value.
$4.00 + $1.50 + $6.50 + $13.00 + $1.75 = $26.75
Answer:
<u>Answer</u><u>:</u><u> </u><u>B</u>
If the -1 is not included, then your answer is 3.17.
2(3x-4)=11
3x2=6 and -4x2= -8
6x-8=11
+8 +8
6x=19
/6 /6
X= 3.17
