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24.03.2020 •
Mathematics
A concession stand at a sports stadium sells 600 bags of peanuts (each game) if the price of each bag is $4.00. For each $0.25 increase in price, 30 fewer bags are sold. What should the price of each bag of peanuts be so that the stand maximizes its revenue? What is the maximum revenue?
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Ответ:
Price = $4.50
Maximum revenue = $2,430
Step-by-step explanation:
The concession stand sells 600 bags of peanut at $4.00 (600, $4.00).
If there is a $1 increase in price, the number of bags old will decrease by 120, which mens that 480 bags of peanuts will be sold at $5.00 (480, $5.00).
Tracing a linear relationship between price and quantity sold with the two given points:
The revenue function is given by the price multiplied by the quantity sold:
The value of 'Q' for which the derivate of the revenue function is zero, is the output level for which revenue is maximum:
The total revenue of 540 units at $4.50 per unit is:
Ответ:
The weight that separate the top 8% by 5.2605 and the weight that separate bottom 8% by 5.1195.
Step-by-step explanation:
We are given that
Mean,
Standard deviation,
We have to find the two weights that separate the top 8% and the bottom 8%.
Let x1 and x2 the two weights that separate the top 8% and the bottom 8%.
Z-value for p-value =0.08 =
For 8% bottom
For 8% top
p-Value=1-0.08=0.92
Z- value=1.41
Now,
(x1,x2)=(5.1195,5.2605)