The first one is 10.5
The second is y10
And the third is 7
Answer:
Step 1
Step-by-step explanation:
Step one is incorrect because you need to add 3 but it subtracted.
Answer:
- r = 12.5p(32 -p)
- $16 per ticket
- $3200 maximum revenue
Step-by-step explanation:
The number of tickets sold (q) at some price p is apparently ...
q = 150 + 25(20 -p)/2 = 150 +250 -12.5p
q = 12.5(32 -p)
The revenue is the product of the price and the number of tickets sold:
r = pq
r = 12.5p(32 -p) . . . . revenue equation
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The maximum of revenue will be on the line of symmetry of this quadratic function, which is halfway between the zeros at p=0 and p=32. Revenue will be maximized when ...
p = (0 +32)/2 = 16
The theater should charge $16 per ticket.
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Maximum revenue will be found by using the above revenue function with p=16.
r = 12.5(16)(32 -16) = $3200 . . . . maximum revenue
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<em>Additional comment</em>
The number of tickets sold at $16 will be ...
q = 12.5(32 -16) = 200
It might also be noted that if there are variable costs involved, maximum revenue may not correspond to maximum profit.
Answer:
it would be 7/10
Step-by-step explanation:
7 over 10.
Hope this helps:)
9x - 27 < 4x + 40
-> 5x < 67
-> x < 67/5