Question

an unknown metal has been found and the following experimental results have been tabulated in the table below. the table contains the grams of the unknown metal and the volume in milliliters of water displacement. find a linear model that expresses mass as a function of the volume.

a. write the linear regression equation for the data in the chart where x is the grams of the unknown metal. first the slope, then the y - intercept. round each answers to 3 decimal places:

volume = x+

b. if the mass of an unknown metal is 12, using your un - rounded regression equation find its predicted volume (in ml). round your answer to 1 decimal place. ml

Explanation:

Step1: Calculate the slope ($m$) and y - intercept ($b$)

We use the formula for the slope \(m=\frac{n\sum(xy)-\sum x\sum y}{n\sum(x^{2})-(\sum x)^{2}}\) and \(b = \frac{\sum y - m\sum x}{n}\), where \(n = 7\) (the number of data points).

Let \(x\) be the grams (mass) and \(y\) be the volume.

\(\sum x=14 + 16.5+19+21.5+24+26.5+29=140.5\)

\(\sum y=38 + 44.3+51.5+58.9+66.4+70.4+77.1 = 406.6\)

\(\sum(xy)=14\times38+16.5\times44.3+19\times51.5+21.5\times58.9+24\times66.4+26.5\times70.4+29\times77.1\)
\(=532+730.95+978.5+1266.35+1593.6+1865.6+2235.9\)
\(=9202.9\)

\(\sum(x^{2})=14^{2}+16.5^{2}+19^{2}+21.5^{2}+24^{2}+26.5^{2}+29^{2}\)
\(=196 + 272.25+361+462.25+576+702.25+841\)
\(=3410.75\)

\(m=\frac{7\times9202.9-140.5\times406.6}{7\times3410.75-(140.5)^{2}}\)
\(=\frac{64420.3-57127.3}{23875.25 - 19740.25}\)
\(=\frac{7293}{4135}\approx1.764\)

\(b=\frac{406.6-1.764\times140.5}{7}\)
\(=\frac{406.6 - 247.842}{7}=\frac{158.758}{7}\approx22.680\)

Step2: Predict the volume for \(x = 12\)

Using the un - rounded regression equation \(y=mx + b\) (where \(m=\frac{7293}{4135}\approx1.7637\) and \(b=\frac{158.758}{7}\approx22.6797\))

When \(x = 12\), \(y=1.7637\times12+22.6797\)
\(y = 21.1644+22.6797=43.8441\approx43.8\)

Answer:

A. Volume \(=1.764x + 22.680\)
B. \(43.8\)