Y=-6/8X-3/8 is your answer
If the price of one ticket of bus is $180 and the bus has 56 seats then the maximum revenue that it can earn is $5107.6
Given that the price of a bus ticket to Saskatoon is $180 and the bus has 56 seats.
We are required to find the maximum revenue that the bus company can earn.
Suppose x represents the number of seats, y represents the total amount.
Price=$156
Seats=56
When the bus is of half capacity the bus seats will be 28.
As price decreases th rider gains 2 more.
Revenue equation.
y=(156.5x)(28+2x)-------------1
Expanding the equation.
y=4368-140x+312-10![x^{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D)
Differentiating with respect to x.
dy/dx=0-140+312-20x
=172-20x
Put dy/dx=0
172-20x=0
x=8.6
Substitute the value of variable x in the equation 1.
y=(156-5x)(28+2x)
=$5107.6
Hence the maximum revenue that the bus company can earn is $5107.6.
Learn more about differentiation at brainly.com/question/954654
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2x-4
If x is the amount the other team scored then the 2x is double. This gives that 4 less than double the amount of the other team would be minus 4
(-oo, 42)
Mark brainliest
Hope this helps you
Graph that then it won’t let me send pic
Answer:
D. 0.098
Step-by-step explanation:
The probabilities of all choices must total 1, so we have ...
p(never) + p(married) + p(widowed) + p(divorced) = 1
p(divorced) = 1 - (p(never) + p(married) + p(widowed))
= 1 - (0.239 + 0.595 + 0.068) = 1 - 0.902
p(divorced) = 0.098