Q45. - Algebra 2

Question

For exercise 43-45, use the following information.

The parking garage at Burrough’s Department store charges $$\mathbf{1}\mathbf{.}\mathbf{50}$ for each hour or fraction of an hour for parking.

45. Jada went shopping at Burrough’s Department store yesterday. She left her car in the parking garage for two hours and twenty-seven minutes. How much did Jada pay for parking?

Step-by-Step Solution

Answer

Jada paid $$\mathbf{4}\mathbf{.}\mathbf{50}$ for parking.

Step 1 ­- Description of step.

It is given that the parking garage at Burrough’s Department store charges $$1.50$ for each hour or fraction of an hour for parking.

That implies the charge for the time greater than 0 but less than or equal to 1 hour is $$1.50$.

Therefore, the charge for the time greater than 1 hour and less than or equal to 2 hours is:

$\$1.50\times 2=\$3.00$

Therefore, the charge for the time greater than 2 hour and less than or equal to 3 hours is:

$\$1.50\times 3=\$4.50$

Similarly, the pattern continues.

Step 2 ­- Description of step.

Therefore, it can be noticed that for the time $t$:

If $0<t\le 1$, charge is $$1.50$.

If $1<t\le 2$, charge is $$3.00$.

If $2<t\le 3$, charge is $$4.50$.

Step 3 &shy;- Description of step.

As, Jada parked her car for 2 hours and twent-seven minutes, which is greater than 2 hours and less than 3 hours.

Therefore, the charge for parking is $$4.50$.

Therefore, Jada paid $$4.50$ for parking.