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show both decimal places (5.06).
A phone company set the following rate schedule for an m-minute call from any of its pay phones.
0.70
when m 6
0.70+ 0.24(m - 6)
when m> 6 and m is an integer
0.70+ 0.24([m-6] + 1) when m > 6 and m is not an integer
c(m) =
a. What is the cost of a call that is under six minutes?
b. What is the cost of a 14-minute call?
c. What is the cost of a 9 ½2-minute call?

Respuesta :

a. the cost of a call for six minutes is 0.70.

b.the cost of a 14-minute call is 2.62.

c.the cost of a 9 ½2-minute call is 1.66.

How to find the cost of each call ?

given that

rate schedule for an m-minute call from any of its pay phones 0.70.

c(m) = 0.70 when m<=6

c(m) = 0.70+ 0.24(m-6) when m>6 & m is an integer

c(m) = 0.70+0.24((m-6)+1) when m>6 & m is not an integer.

a. to find the cost of a call for six minutes.

already given that c(m) =0.70 when m=6.

Therefore, the cost of a call for six minutes is 0.70.

b. to find the cost of a 14-minute call.

according to given information

[tex]c(14) = 0.70 + 0.24(m - 6) \\ c(14) = 0.70 + 0.24(14 \: - 6) \\c(14) = 0.70 + (0.24 \times 14) - (0.24 \times 6) \\c(14) = 0.70 + (3.36 - 1.44) \\ c(14) = 0.70 + 1.92 \\ c(14) = 2.62[/tex]

Therefore,the cost of a 14-minute call is 2.62.

c.to find the cost of a 9 ½2-minute call

according to given information

[tex]c(9 \frac{1}{2} 2) = 0.70 + 0.24((9 \frac{1}{2} 2 - 6) + 1) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.24((9 - 6) + 1) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.24(3+ 1) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.24(4) \\ c(9 \frac{1}{2} 2) = 0.70 + 0.96 \\ c(9 \frac{1}{2} 2) = 1.66[/tex]

Therefore,the cost of a 9 ½2-minute call is 1.66.

Learn more about problems on cost of call, refer:

https://brainly.com/question/16584226

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