Solve the equation m2 + 6m = –4 using the quadratic formula. A. m = –3±√ 5 B. m = 5±√ 3 C. m = 3±√ 5 D. m = –5±√ 3

Mathematics

1 answer:

USPshnik [31]3 years ago

7 0

Answer: m = –3±√ 5

−b±√b2−4ac

−(6)±√(6)2−4(1)(4)

2(1)

−6±√20

m=−3+√5 or m=−3−√5

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a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

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Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

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$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05