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katovenus [111]
3 years ago
8

the hight, the feet, of a javelin t seconds afther it is moldeled by the function (t) =4t^2 + 24t + 6. Afther how many seconds d

oes the javelin reah its maximum height of 42 feet?
Mathematics
1 answer:
ruslelena [56]3 years ago
8 0
The max height will be reached after 3 seconds.

You have a typo in your equation. For this type of quadratic equation, it always starts out with a negative coefficient of the t squared term.

To find the x-coordinate of the vertex, we just need to use the expression: -b/2a.

Then, we have:

-24 / (2(-4))

-24 / -8

3
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Answer:

Here's what I get.

Step-by-step explanation:

1. Representation of data

I used Excel to create a scatterplot of the data, draw the line of best fit, and print the regression equation.

2. Line of best fit

(a) Variables

I chose arm span as the dependent variable (y-axis) and height as the independent variable (x-axis).

It seems to me that arm span depends on your height rather than the other way around.

(b) Regression equation

The calculation is easy but tedious, so I asked Excel to do it.

For the equation y = ax + b, the formulas are

a = \dfrac{\sum y \sum x^{2} - \sum x \sumxy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\b = \dfrac{n\sumx y  - \sum x \sumxy}{n\sum x^{2}- \left (\sum x\right )^{2}}

This gave the regression equation:

y = 1.0595x - 4.1524

(c) Interpretation

The line shows how arm span depends on height.

The slope of the line says that arm span increases about 6 % faster than height.

The y-intercept is -4. If your height is zero, your arm length is -4 in (both are impossible).

(d) Residuals

\begin{array}{cccr}&\textbf{Arm Span} & \textbf{Arm Span}&\\\textbf{Height/in} &\textbf{Actual} & \textbf{Predicted}&\textbf{Residual}\\25 & 19 & 22.3 & -3.3\\40 & 41 & 38.2 & 2.8\\55 & 51 & 54.1 & -3.1\\65 & 67 & 62.6 & 4.4\\ \end{array}

The residuals appear to be evenly distributed above and below the predicted values.

A graph of all the residuals confirms this observation.  

The equation usually predicts arm span to within 4 in.

(e) Predictions

(i) Height of person with 66 in arm span

\begin{array}{rcl}y& = & 1.0595x - 4.1524\\66 & = & 1.0595x - 4.1524\\70.1524 & = & 1.0595x\\x & = & \dfrac{70.1524}{1.0595}\\\\& = & \textbf{66 in}\\\end{array}\\\text{A person with an arm span of 66 in  should have a height of about $\large \boxed{\textbf{66 in}}$}

(ii) Arm span of 74 in tall person

\begin{array}{rcl}y& = & 1.0595x - 4.1524\\& = & 1.0595\times74 - 4.1524\\& = & 78.4030 - 4.1524\\& = & \textbf{74 in}\\\end{array}\\\text{ A person who is 74 in tall should have an arm span of $\large \boxed{\textbf{74 in}}$}

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3 years ago
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