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Kendra is considering two different long-distance phone plans. Phone plan A charges a $100 sign-up fee and 3 cents per minute. Phone plan B does not charge a sign-up fee, but it charges 5 cents per minute.

Respuesta :

calculista

Answer:

Part 1) The number of minutes in a month must be greater than 50 in order for the plan A to be preferable

Part 2) The number of minutes in a month must be equal to 50 minutes

Step-by-step explanation:

The question is

Part 1) How many minutes would Kendra have to use in a month in order for the plan A to be preferable? Round your answer to the nearest minute

Part 2) Enter the number of minutes where Kendra will pay the same amount for each long distance phone plan

Part 1)

Let

x ---> the number of minutes

we have

Cost Plan A

[tex]3x+100[/tex]

Cost Plan B

[tex]5x[/tex]

we know that

In order for plan A to be cheaper than plan B, the following inequality must hold true.

cost of plan A < cost of plan B

substitute

[tex]3x+100 < 5x[/tex]

solve for x

subtract 3x both sides

[tex]100< 5x-3x\\100<2x[/tex]

divide by 2 both sides

[tex]50 < x[/tex]

Rewrite

[tex]x> 50\ min[/tex]

therefore

The number of minutes in a month must be greater than 50 in order for the plan A to be preferable

Part 2)

Let

x ---> the number of minutes

we have

Cost Plan A

[tex]3x+100[/tex]

Cost Plan B

[tex]5x[/tex]

we know that

In order for plan A cost the same than plan B, the following equation must hold true.

cost of plan A = cost of plan B

substitute

[tex]3x+100= 5x[/tex]

solve for x

[tex]5x-3x=100\\2x=100\\x=50\ min[/tex]

therefore

The number of minutes in a month must be equal to 50 minutes

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