A swimmer trains both in the water and on land. Each day the time spent in the water accounts for 60% of the total time she spends training.
On Thursday the swimmer trains for 126 minutes in the water. How many minutes did she train for in total on Thursday?

Factor the following expression. 27y3 – 343 a. (3y + 7)(9y2 + 2ly + 49) b. (3y – 7)(9y2 + 2ly + 49) c. (3y – 7)(932 – 217 + 49) d. (3y + 7)(92 – 2ly + 49)

Solve each quadratic equation by factoring and using the zero product property. [tex]x^2+6x+8=0[/tex]

On a certain portion of an experiment, a statistical test result yielded a p-value of 0.21. what can you conclude? 2(0.21) = 0.42 < 0.5; the test is not statistically significant. if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 21% of the time, so the test is not statistically significant. 0.21 > 0.05; the test is statistically significant. if the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 79% of the time, so the test is not statistically significant. p = 1 - 0.21 = 0.79 > 0.05; the test is statistically significant.