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2 years ago

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Fossils are remains of organisms found in sedimentary rock. These remains are found in thousands of rock layers around the world

and provide evidence for Earth's long history of changing lifeforms. Rock layers that formed more recently A. contain fossils of animals only. B. contain fossils of plants only. C. contain fossils that resemble existing species less closely than do fossils found in older rock layers. D. contain fossils that resemble existing species more closely than do fossils found in older rock layers.

1 answer:

8_murik_8 [283]2 years ago

5 0

C- they contain fossils that resemble existing species less closely than do fossils found in older rock layers

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Quadrilateral QRST is similar to quadrilateral EDGF. What's the scale factor from EDGF to QRST

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Answer:

C. $\frac{6}{5}$

Step-by-step explanation:

The ratio of the corresponding sides of a polygon is said to be equal. The ratio gives us the scale factor.

Therefore, assuming that quadrilateral QRST was the original polygon that was scaled up to give a similar polygon, EDGF, the scale factor would be expressed as $\frac{42}{35}$

The scale factor = $\frac{42}{35} = \frac{6*7}{5*7} = \frac{6}{5}$

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2 years ago

Tessa had $90 in her checking account. She paid her cable/internet bill for $60. She deposited $50 from her part-time job before

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90-60+50-65=15

$15 in her account

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2 years ago

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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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2 years ago

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Dominik [7]

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$2906.17 am average is calculated by dividing the sum of all monthly expenditures by the number of months reported.

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