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