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The swinging pendulum Mrs. Hanrahan's precalculus class collected data on the length (in centimeters) of a pendulum and the time (in seconds) the pendulum took to complete one back-and-forth swing (called it's period). The theoretical relationship between a pendulum's length and its period is

$\text{period}=\frac{2蟺}{\sqrt{g}}\sqrt{\text{length}}$

where $\mathrm{g}$is a constant representing the acceleration due to gravity (in this case, $\mathrm{g}=980\mathrm{cm}/\mathrm{s}2g=980\mathrm{cm}/{\mathrm{s}}^2$). Here is a graph of the period versus length, $\sqrt{\text{length}}$, along with output from a linear regression analysis using these variables.

a. Give the equation of the least-squares regression line. Define any variables you use.

b. Use the model from part (a) to predict the period of a pendulum with length $80$cm.

Step by step solution

01

Part (a) Step 1: Given Information

Given data:

02

Part (a) Step 2: Explanation

Least square regression line's general equation

$\hat{y}=b_0+b_{1}x$

In the row "constant" and the column "Coef" of the computer's output, the calculated constant $b_0$is mentioned.

$b_0=-0.08594$

The computed slope $b_1$ is listed in the row "sqrt(length)" and the column "Coef" of the computer output.

$b_1=0.209999$

03

Part (a) Step 3: Explanation

Putting the value of $b_0$and $b_1$:

$\hat{y}=b_0+b_{1}x$

$\hat{y}=-0.08594+0.209999x$

The square root of the length is $x$, and the period is $y$.

$\hat{y}=-0.08594+0.209999\sqrt{x}$

Where $x$ denotes length and $y$ denotes time.

04

Part (b) Step 1: Given Information

Given data:

05

Part (b) Step 2: Explanation

From the part (a)

$\hat{y}=-0.08594+0.209999\sqrt{x}$

Where $x$denotes length and $y$denotes time.

Using the pendulum length as a variable in the equation

$\hat{y}=-0.08594+0.209999\sqrt{80}$

$\hat{y}=1.7923$

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