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3 years ago

8.) Simplify.

Mathematics

1 answer:

Aleks04 [339]3 years ago

7 0

Answer:

The answer to your question is letter E. cos²α

Step-by-step explanation:

cot²Ф - cot²Фcos²Ф

Remember that cot = $\frac{cos \alpha}{sin \alpha }$ ; substitute in the previous equation

$\frac{cos^{2}\alpha }{sin^{2} \alpha} - \frac{cos^{2}\alpha }{sin^{2}\alpha } cos^{2}\alpha$

$\frac{cos^{2}\alpha - cos^{4}\alpha }{sin^{2}\alpha }$ Factor cos²α

$\frac{cos^{2}\alpha(1 - cos^{2}\alpha }{sin^{2}\alpha }$

Remember that sin²α = 1 - cos²α

$\frac{cos^{2}\alpha sin^{2}\alpha}{sin^{2} \alpha}$

Simplify

cos²α

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Find the sumofthe geometrical progression of five terms, of which the first term is 7 and the multiplier is 7.Verify that the su

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Answer:

The sum of first five term of GP is 19607.

Step-by-step explanation:

We are given the following in the question:

A geometric progression with 7 as the first term and 7 as the common ration.

$a, ar, ar^2,...\\a = 7\\r = 7$

$7, 7^2, 7^3, 7^4...$

Sum of n terms in a geometric progression:

$S_n = \displaystyle\frac{a(r^n - 1)}{(r-1)}$

For sum of five terms, we put n= 5, a = 7, r = 7

$S_5 = \displaystyle\frac{7(7^5 - 1)}{(7-1)}\\\\S_5 = 19607$

The sum of first five term of GP is 19607.

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3 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

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

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

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

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

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

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

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

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