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sergij07 [2.7K]
3 years ago
15

Find the lateral surface area of the figure below. Round the the nearest tenths place.

Mathematics
1 answer:
nadya68 [22]3 years ago
4 0
Answer: lateral surface area = 119.4cm^2
r = 2 cm.
h = 9.5 cm.

Steps:
Lateral surface area of a cylinder = 2πrh
r = radius h = height π = pi

2π(2)(9.5)
2π(19)
38π = 119.38052
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The mayor of a town has proposed a plan for the construction of an adjoining bridge. A political study took a sample of 1700 vot
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z=\frac{0.66 -0.63}{\sqrt{\frac{0.63(1-0.63)}{1700}}}=2.562  

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Step-by-step explanation:

Data given and notation

n=1700 represent the random sample taken

\hat p=0.66 estimated proportion of voters that favored construction

p_o=0.63 is the value that we want to test

\alpha=0.02 represent the significance level

Confidence=98% or 0.98

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that percentage of residents who favor construction is more than 63%.:  

Null hypothesis:p \leq 0.63  

Alternative hypothesis:p > 0.63  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.66 -0.63}{\sqrt{\frac{0.63(1-0.63)}{1700}}}=2.562  

Statistical decision  

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The significance level provided \alpha=0.02. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>2.562)=0.0052  

So the p value obtained was a very low value and using the significance level given \alpha=0.02 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that at 2% of significance the proportion of voters that favored construction is higher than 0.63 or 63%.

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