The table below shows the height of a ball x seconds after being kicked.

A 2-column table with 5 rows. The first column is labeled time (seconds) with entries 0, 0.5, 1, 1.5, 2, 2.5, 3. The second column is labeled height (feet) with entries 0, 35, 65, 85, 95, 100, 95.
What values, rounded to the nearest whole number, complete the quadratic regression equation that models the data?

f(x) =
x2 +
x + 0
Based on the regression equation and rounded to the nearest whole number, what is the estimated height after 0.25 seconds?

feet

Answers

Answer 1

Answer:

-16

81

19

Explanation:

Answer 2

The quadratic regression equation is [tex]y = -16x^2 +81x[/tex], and the estimated height in 0.25 seconds is 19 feet

What are regression equations?

Regression equations are used to determine the relationship between sets of data

The dataset is given as:

Time (seconds)       Height (feet)

       0                               0

      0.5                            35

         1                              65

        1.5                             85

         2                              95

        2.5                           100

           3                             95

To determine the quadratic regression equation, we make use of a graphing calculator.

From the graphing calculator, we have the following calculation summary:

a = -16.429

b = 81.071

c = -0.357

A quadratic regression equation, is represented as:

[tex]y = ax^2 + bx + c[/tex]

So, we have:

[tex]y = -16.429x^2 +81.071x -0.357[/tex]

Approximate

[tex]y = -16x^2 +81x[/tex]

The estimated height in 0.25 seconds is:

[tex]y = -16* 0.25^2 +81 * 0.25[/tex]

[tex]y = 19.25\\[/tex]

Approximate

[tex]y = 19[/tex]

Hence, the estimated height in 0.25 seconds is 19 feet

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Answers

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Answers

Answer:

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Answers

Answer:

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Answers

Answer:

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Answers

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Answers

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Answers

Answer:

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Answers

Answer:

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Blank O2 + blank Mg arrow blank MgO
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Answers

The answer would be letter d

Answer:

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Answers

Answer: The new volume is 95.45 mL.

Explanation:

Given: [tex]V_{1}[/tex] = 45.0 mL,         [tex]P_{1}[/tex] = 0.98 atm

[tex]P_{2}[/tex] = 2.1 atm,               [tex]V_{2}[/tex] = ?

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Substitute the values into above formula as follows.

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