Question

Hailey is going to use a computer at a Internet cafe. The cafe charges $1.50 for every minute using a computer on top of an initial charge of $4. Make a table of values and then write an equation for C, in terms of t, representing the total cost of using a computer for t minutes at the Internet cafe.

Answer 1

Answer:

The equation is C(t) = 1.50t + 4.00 and the data table is below.

Step-by-step explanation:

We are given that a cafe charges $1.50 for every minute that someone uses the internet. The cafe also charges an initial $4 fee to use the internet.

We need to find a function, in terms of t, that will find these values.

We can treat this as a linear equation. If we initially pay $4, this is acquired before we pay the fee every minute. Therefore, this would be our y-intercept at x = 0. Our first coordinate is (0, 4).

Secondly, if we are charged every minute, this would be a recurring payment and therefore we would spend $1.50 after each minute elapses. This gives us our slope: 1.50 or $\frac{3}{2}$.

Therefore, we can set up our function.

C is the total cost and t is the time, in minutes, that internet is used at the cafe. Our equation can be set up as $C(t) = 1.50t + 4.00$.

Now, we can create a table of values for t = 0 to t = 5 to determine what C(t) is at each of these prime intervals.

$\begin{array}{|c|c|} \cline{1-2} \textbf{C} & \textbf{t} \\ \cline{1-2} a & 0 \\ \cline{1-2} b & 1 \\ \cline{1-2} c & 2 \\ \cline{1-2} d & 3 \\ \cline{1-2} e & 4 \\ \cline{1-2} f & 5 \\ \cline{1-2} \end{array}$

This is what our table looks like, but we need to fill the C column. Therefore, we can use substitution of our t-values into the equation and solve for C.

$\bullet \ \text{For t = 0,}$

$C(0) = 1.50(0) + 4\\\\C(0) = 0 + 4\\\\C(0) = 4.00$

$\bullet \ \text{For t = 1,}$

$C(1) = 1.50(1) +4\\\\C(1) = 1.50 + 4\\\\C(1) = 5.50$

$\bullet \ \text{For t = 2,}$

$C(2) = 1.50(2) + 4\\\\C(2) = 3.00 + 4\\\\C(2) = 7.00$

$\bullet \ \text{For t = 3,}$

$C(3) = 1.50(3) + 4\\\\C(3) = 4.50 + 4\\\\C(3) = 8.50$

$\bullet \ \text{For t = 4,}$

$C(4) = 1.50(4) + 4\\\\C(4) = 6.00 + 4\\\\C(4) = 10.00$

$\bullet \ \text{For t = 5,}$

$C(5) = 1.50(5) + 4\\\\C(5) = 7.50 + 4\\\\C(5) = 11.50$

Now, we can insert these values into our table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{C} & \textbf{t} \\ \cline{1-2} 4.00 & 0 \\ \cline{1-2} 5.50 & 1 \\ \cline{1-2} 7.00 & 2 \\ \cline{1-2} 8.50 & 3 \\ \cline{1-2} 10.00 & 4 \\ \cline{1-2} 11.50 & 5 \\ \cline{1-2} \end{array}$

Therefore, our equation is C(t) = 1.50t + 4.00 and our data table is above.