Question

corinne has a cell phone plan that includes 200 minutes for phone calls and unlimited texting. an additional fee is charged for using more than 200 minutes for phone calls. the figure below is the graph of ( c = f(m) ), where ( c ) is the monthly cost after ( m ) minutes used.
cell phone plan
part a
what is the minimum monthly cost for corinnes cell phone plan?
part b
for what ( m ) is ( f(m)=55? )

Explanation:

Part A

Step1: Analyze the graph for minimum cost

The minimum cost occurs when the number of minutes \(m\) is within the included 200 - minute limit. Looking at the graph, when \(m\leq200\), the cost \(C\) is constant. The value of \(C\) for \(m\leq200\) is 30.

Part B

Step1: Determine the equation of the line for \(m > 200\)

First, find two points on the line for \(m>200\). Let's take \((200,30)\) and \((250,60)\).
The slope \(k=\frac{60 - 30}{250 - 200}=\frac{30}{50}=0.6\).
Using the point - slope form \(y - y_1=k(x - x_1)\) with \((x_1,y_1)=(200,30)\), the equation is \(C-30 = 0.6(m - 200)\), which simplifies to \(C=0.6m-120 + 30=0.6m - 90\).

Step2: Solve for \(m\) when \(C = 55\)

Set \(C = 55\) in the equation \(C=0.6m-90\).

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But we can also use the graph's grid. Since the slope is \(0.6=\frac{3}{5}\) (rise over run). From the point \((200,30)\), if \(C\) increases by 3 (from 30 to 55, the increase in \(C\) is \(55 - 30=25\)), and since the slope \(k = 0.6=\frac{\Delta C}{\Delta m}\), then \(\Delta m=\frac{\Delta C}{k}\). \(\Delta C = 25\), so \(\Delta m=\frac{25}{0.6}=\frac{250}{6}\approx41.67\). So \(m=200+\frac{250}{6}=\frac{1200 + 250}{6}=\frac{1450}{6}\approx241.67\). Looking at the graph's grid (assuming each small square has a certain value, if we consider the linear part after \(m = 200\) and count the units based on the slope - related rise and run), we can also see that when \(C = 55\), \(m = 250\) (by observing the graph's grid structure where for the linear part \(C\) increases by 6 for every 10 units of \(m\) increase. From \(C = 30\) (\(m = 200\)) to \(C = 55\) (an increase of \(25\)), and if we assume a more approximate grid - based counting (since \(C\) increases by 6 when \(m\) increases by 10, \(25\div6\approx4.17\) times of 10 - unit increase in \(m\)), but a more accurate graph - reading (assuming the graph is on a grid where for the linear part \(C\) and \(m\) have a relationship \(C-30=\frac{3}{5}(m - 200)\)) gives \(m = 250\)

Answer:

Part A: 30
Part B: 250