# The scatter plot isSince the points are in linear pattern and decr.pdf

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The scatter plot is Since the points are in linear pattern and decreasing porosity with increasing pcf, the relation is strong negative Least square regression output is: Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1.7754 -0.5727 -0.1325 0.6034 1.6818 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 118.90992 4.49912 26.43 1.10e-12 *** x -0.90473 0.04109 -22.02 1.12e-11 *** --- Signif. codes: 0 Solution The scatter plot is Since the points are in linear pattern and decreasing porosity with increasing pcf, the relation is strong negative Least square regression output is: Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1.7754 -0.5727 -0.1325 0.6034 1.6818 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 118.90992 4.49912 26.43 1.10e-12 *** x -0.90473 0.04109 -22.02 1.12e-11 *** --- Signif. codes: 0.

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### The scatter plot isSince the points are in linear pattern and decr.pdf

• 1. The scatter plot is Since the points are in linear pattern and decreasing porosity with increasing pcf, the relation is strong negative Least square regression output is: Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1.7754 -0.5727 -0.1325 0.6034 1.6818 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 118.90992 4.49912 26.43 1.10e-12 *** x -0.90473 0.04109 -22.02 1.12e-11 *** --- Signif. codes: 0 Solution The scatter plot is Since the points are in linear pattern and decreasing porosity with increasing pcf, the relation is strong negative Least square regression output is: Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1.7754 -0.5727 -0.1325 0.6034 1.6818 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 118.90992 4.49912 26.43 1.10e-12 *** x -0.90473 0.04109 -22.02 1.12e-11 *** --- Signif. codes: 0
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