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Delicious77 [7]
3 years ago
14

The pyrotechnic team for Fist Bump’s tour want to know how high the arena is to 2 decimal places. The arena is 61.3786 metres hi

gh. How high is it to 2 d.p?
Mathematics
1 answer:
Aloiza [94]3 years ago
4 0
When you find any value to x amount of decimal places, you need to go to the term AFTER that to determine the correct value.  So, for this, take 61.378 metres.  From here, look if the last number is 4 or less.  If the number is 4 or less, take the leftmost number off, and you will have your answer (if we had 61.373, the correct answer would be 61.37).  If the number is 5 or more, take the leftmost number off, and add 1 to the new leftmost number.  Since the last number is 8, the correct answer here is 61.38 metres high.  If adding 1 makes the number 10, keep the 0 at the end.  So, if you had 61.399, you would make it 61.40
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Arm Span(x) Height(y)
natita [175]

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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