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Mathematics, 20.04.2020 18:01 JamesLachoneus
On their birthdays, employees at a large company are permitted to take a 60-minute lunch break instead of the usual 30 minutes. Data were obtained from 10 randomly selected company employees on the amount of time that each actually took for lunch on his or her birthday. The company wishes to investigate whether these data provide convincing evidence that the mean time is greater than 60 minutes. Of the following, which information would NOT be expected to be a part of the process of correctly conducting a hypothesis test to investigate the question, at the 0.05 level of significance?
Being willing to assume that the distribution of actual birthday lunch times for all employees at the company is approximately normal
Knowing that there are no outliers in the data as indicated by the normal probability plot and boxplot
Using a t-statistic to carry out the test
Using 9 for the number of degrees of freedom
Given that the p-value is greater than 0.05, rejecting the null hypothesis and concluding that the mean time was not greater than 60 minutes.
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Mathematics, 22.06.2019 01:10, hellicuh
Evaluate 8x2 + 9x − 1 2x3 + 3x2 − 2x dx. solution since the degree of the numerator is less than the degree of the denominator, we don't need to divide. we factor the denominator as 2x3 + 3x2 − 2x = x(2x2 + 3x − 2) = x(2x − 1)(x + 2). since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand has the form† 8x2 + 9x − 1 x(2x − 1)(x + 2) = correct: your answer is correct. to determine the values of a, b, and c, we multiply both sides of this equation by the product of the denominators, x(2x − 1)(x + 2), obtaining 8x2 + 9x − 1 = a correct: your answer is correct. (x + 2) + bx(x + 2) + cx(2x − 1).
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