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Sedaia [141]
3 years ago
11

In a circle with a radius of 2.8 cm, an arc is intercepted by a central angle of π 5 radians. What is the arc length? Use 3.14 f

or π and round your final answer to the nearest hundredth. Enter your answer as a decimal in the box.  
Mathematics
1 answer:
Art [367]3 years ago
6 0
The answer was 1.76cm
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The concentration of hexane (a common solvent) was measured in units of micrograms per liter for a simple random sample of sixte
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Answer:

Yes, it can be concluded that the mean hexane concentration is less in treated water than in unsaturated water

Step-by-step explanation:

The number of of specimen in the samples of untreated water, n₁ = 16

The sample mean, \overline x_1 = 228.0

The sample standard deviation, s₁ = 4.3

The number of of specimen in the samples of treated water, n₂ = 20

The sample mean, \overline x_2 = 224.6

The sample standard deviation, s₂ = 5.0

The level of significance = 0.10

The null hypothesis, H₀; \overline x_1 ≥ \overline x_2

The alternative hypothesis, Hₐ;  \overline x_1 < \overline x_2

The degrees of freedom = 16 - 1 = 15

The test statistic, t_{\alpha} = 1.341

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}

Plugging in the values, we get;

t=\dfrac{(224.6- 228.0)}{\sqrt{\dfrac{5.0^{2}}{20} -\dfrac{4.3^{2} }{16}}} \approx -11.0675

Given that the t-value is large, the corresponding p-value is low, therefore, we fail to reject the null hypothesis and there is considerable statistical evidence to suggest that the mean hexane concentration is less in treated than in untreated water, therefore, we have; \overline x_1 ≥ \overline x_2

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How do i simplify each fraction 5/10 4/12 3/9 how do i solve by<br> doing what steps
horrorfan [7]
5/10 divide each number by 5, 5 divided by 5 is 1 and 10 divided by 5 is two so it would be 1/2

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3/9 divide each number by 3, 3 divided by 3 is 1, and 9 divided by 3 is 3 or it would also be 1/3

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