A person places $661 in an investment account earning an annual rate of 7.1%, compounded continuously. Using the formula V = Pe^{rt}V=Pe rt , where V is the value of the account in t years, P is the principal initially invested, e is the base of a natural logarithm, and r is the rate of interest, determine the amount of money, to the nearest cent, in the account after 16 years.

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.