A commuter railway has 800 passengers a day and charges each one 2 dollars. ?
For each 10 cents the fair is increased, 2 fewer people will ride the train. Express the income I from the train in terms of the ticket price p (in dollars). T/F

The income I from the train in terms of the ticket price p (in dollars) is;

I = $(-20p² + 840p)

Number of passengers a day = 800 passengers

Charge for each passenger = $2

We are told that for each $0.1 there is an increase in ticket price and the number of passengers will be 2 fewer people.

Thus;

Number of passengers is now (800 - 2x)

The ticket price is now; p = $(2 + 0.1x)

Thus;

Let us make x the subject of the formula from the price equation to get;

x = (p - 2)/0.1

x = 10p - 20

Putting 10p - 20 for x in (800 - 2x) gives;

Number of passengers = 800- 2(10p- 20)

Number of passengers = 800 - 20p + 40

Number of passengers = 840- 20p

I = p(840 - 20p)

I = -20p² + 840p

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I= -20p^2 + 840p

Step-by-step explanation:

When the ticket price is $2 there are 800 passengers daily, but every $0.1 increase in ticket price the number of passengers will be decreased by 2.  

You can put information into these equations of:

passenger- = (800-2x)

ticket price= p = $2 + 0.1x

Income is calculated by multiplying the number of the passenger with the ticket price. The answer will be expressed in terms of the ticket price, so we need to remove x from the passenger equation.

p= $2 +0.1x

p-$2 = 0.1x

x= 10p- $20

If  p= ticket price, the function for the number of passengers it will be:

passenger = (800-2x)

passenger = 800- 2(10p- $20)

passenger =800- 20p+40

passenger =840- 20p

The function of I will be:

I= passenger x ticket price

I=  840- 20p * p

I= -20p^2 + 840p