19.1k views
3 votes
Caroline is considering two video game rental plans. Plan A can be modeled with the equation C = 2n, and Plan B can be modeled with the equation C = n + 6, where C represents the cost in dollars and n represents the number of games rented each month. Which statement would justify selecting Plan A instead of selecting Plan B?

User Forien
by
8.0k points

2 Answers

4 votes

Final answer:

To justify selecting Plan A instead of selecting Plan B, we need to compare their cost equations. Plan A has a lower cost for each game rented compared to Plan B.

Step-by-step explanation:

To justify selecting Plan A instead of selecting Plan B, we need to compare their cost equations. Plan A is modeled by the equation C = 2n, where C represents the cost and n represents the number of games rented. Plan B is modeled by the equation C = n + 6. Comparing the two equations, we can see that Plan A has a lower cost for each game rented compared to Plan B. For example, if you rent 5 games, Plan A would cost $10 (2 * 5) while Plan B would cost $11 (5 + 6).

Therefore, Plan A would be the better choice if you want to save money on each game rented.

User Durgesh Kumar
by
8.3k points
4 votes

Options

A. Caroline rents exactly 7 games each month.

B. Caroline rents exactly 6 games each month.

C. Caroline rents 6 or more games each month.

D. Caroline rents from 1 to 5 games each month.

Answer:

D. Caroline rents from 1 to 5 games each month.

Step-by-step explanation:

Given

Plan A:


C = 2n

Plan B:


C = n + 6

Required

Which options justifies A over B

The solution to this question is option (d).

In option d, n = 1,2,3,4,5

When any of the values of n is substituted in plan A and B, respectively; the cost of plan A is cheaper than plan B.

This is not so, for other options (A - C)

To show:

Substitute 1 for n in A and B

Plan A:


C = 2n
= 2 * 1 = 2

Plan B:


C = n + 6
= 1 + 6 = 7

Substitute 5 for n in A and B

Plan A:


C = 2n
= 2 * 5 = 10

Plan B:


C = n + 6
= 5 + 6 = 11

See that A < B

User Mnestorov
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories