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Name [tex]$\qquad$[/tex] Per [tex]$\qquad$[/tex]

11. The table shows the heights [tex]$h$[/tex] (in feet) of a sponge [tex]$t$[/tex] seconds after it was dropped by a window cleaner on top of a skyscraper.

\begin{tabular}{|l|l|l|l|l|l|}
\hline Time, t & 0 & 1 & 1.5 & 2.5 & 3 \\
\hline Height, h & 280 & 264 & 244 & 180 & 136 \\
\hline
\end{tabular}

a. Use a graphing calculator to create a scatter plot. Which better represents the data: a line or a parabola? Explain.

b. Use the regression feature on your calculator to find the model that best fits the data. Round to the nearest hundredth if needed.

c. Use the model in part (b) to predict when the sponge will hit the ground. Round to the nearest hundredth if needed.

d. Identify and interpret the domain and range in this situation.

e. Describe the intervals where the graph is increasing or decreasing.



Answer :

Let's work through the given problem step by step:

### a. Scatter Plot and Model Representation
Using a graphing calculator, you would plot the points [tex]\((0, 280)\)[/tex], [tex]\((1, 264)\)[/tex], [tex]\((1.5, 244)\)[/tex], [tex]\((2.5, 180)\)[/tex], and [tex]\((3, 136)\)[/tex].

When you create a scatter plot with these points, you'll notice that the points appear to follow a curved path rather than a straight line. Therefore, a parabola (a quadratic function) would better represent the data compared to a linear model.

### b. Quadratic Regression Model
Using the regression feature on your graphing calculator (or statistical software), you perform a quadratic regression. The resulting quadratic equation that best fits the data is:

[tex]\[ h(t) = -16.00t^2 + 0.00t + 280.00 \][/tex]

Here, the coefficients are rounded to the nearest hundredth.

### c. Predicting When the Sponge Hits the Ground
To predict when the sponge hits the ground, you set the height [tex]\( h(t) \)[/tex] to 0 and solve for [tex]\( t \)[/tex]:

[tex]\[ -16.00t^2 + 0.00t + 280.00 = 0 \][/tex]

This simplifies to:

[tex]\[ -16.00t^2 + 280.00 = 0 \][/tex]

Solving for [tex]\( t \)[/tex]:

[tex]\[ t^2 = \frac{280.00}{16.00} \][/tex]
[tex]\[ t^2 = 17.50 \][/tex]
[tex]\[ t = \sqrt{17.50} \][/tex]
[tex]\[ t \approx 4.18 \][/tex]

So, the sponge will hit the ground approximately at [tex]\( t = 4.18 \)[/tex] seconds.

### d. Domain and Range
The domain of the situation refers to the time from when the sponge is dropped until it hits the ground. Thus, the domain is:

[tex]\[ \text{Domain: } (0, 4.18) \][/tex]

The range corresponds to the height of the sponge from the initial drop until it hits the ground. Since the maximum height is 280 feet (when [tex]\( t = 0 \)[/tex]) and the minimum height is 0 feet (when the sponge hits the ground), the range is:

[tex]\[ \text{Range: } (0, 280) \][/tex]

### e. Increasing or Decreasing Intervals
Examining the direction of the graph as time increases, we find that the height of the sponge decreases continuously over time. This means the graph is always decreasing in the interval from the initial drop to when it hits the ground. Therefore, the graph is decreasing on the interval:

[tex]\[ \text{Decreasing Interval: } (0, 4.18) \][/tex]

In summary:

- a. A parabola better represents the data.
- b. The quadratic model is [tex]\( h(t) = -16.00t^2 + 280.00 \)[/tex].
- c. The sponge will hit the ground approximately at [tex]\( t = 4.18 \)[/tex] seconds.
- d. Domain: [tex]\( (0, 4.18) \)[/tex], Range: [tex]\( (0, 280) \)[/tex].
- e. The graph is decreasing on the interval [tex]\( (0, 4.18) \)[/tex].