Mathematics, 03.12.2019 18:31, Ayalij
K=u n _19 ans: d exhibit 13-5 part of an anova table is shown below. ss . df ms source of sum of degrees of mean variation squares freedom square k-1 between treatment 180 ntk within treatment 300 is (error) total 480 18 ni 80 27. refer to exhibit 13-5. the mean square between treatments (mstr) is 3k- 19-4 us < 20 b. 60 c. 300 d. 15 ans: b 28. refer to exhibit 13-5. the mean square within treatments (mse) is a. 60 b. 15 c. 300 d. 20 ans: d 29. refer to exhibit 13-5. the test statistic is a. 2.25 b. 6 c. 2.67 d. 3 ez ans: d 30. refer to exhibit 13-5. if at 95% confidence, we want to determine whether or not the means of the populations are equal, the p-value is a. between 0.01 to 0.025 b. between 0.025 to 0.05 c. between 0.05 to 0.10 d. greater than 0.1 0.05 dftr aff 60 ans: c
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To solve the system of equations below, pedro isolated the variable y in the first equation and then substituted it into the second equation. what was the resulting equation? { 5y=10x {x^2+y^2=36
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K=u n _19 ans: d exhibit 13-5 part of an anova table is shown below. ss . df ms source of sum of de...
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