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Juliette [100K]
3 years ago
11

D=7.9cm. what is the circumference and area of the circle?

Mathematics
2 answers:
In-s [12.5K]3 years ago
8 0
Circumfrence: pi times diameter
7.9*\pi
circumfrence is approximatley 24.806

area: \pi r^2
a=\pi*(1/2(d))^2
a= 48.99
Lera25 [3.4K]3 years ago
6 0
You multiply 7.9 by pi to get an answer of 24.8 cm for the curcumference, and then square the raduis, 3.95, to get 15.6025, and then multiply that by pi to get 48.9 cm squared for the area. (I used 3.14 for pi and rounded to the first decimal place)
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Answer:

\frac{2}{\sqrt{5} }

Step-by-step explanation:

To solve this, all we need to do is draw a triangle.  

From the arctan(2) we can deduce that tan(x)=\frac{2}{1}

From that, we can draw our triangle as we know that tan(x) is the opposite side over the adjacent side. Attached is that triangle. Through the Pythagorean Theorem, we can find that the hypotenuse of this right triangle is  \sqrt{5}

Now all we need to do is take the sin of this triangle, which is the opposite side over the hypotenuse

This gives us the value of \frac{2}{\sqrt{5} } which is our answer.

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In a recent poll, 778 adults were asked to identify their favorite seat when they fly, and 492 of them chose a window seat. Use
seropon [69]

Answer:

Null hypothesis:p \leq 0.5  

Alternative hypothesis:p > 0.5  

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

p_v =P(Z>7.36)=9.19x10^{-14}  

Since the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .  

Step-by-step explanation:

1) Data given and notation

n=778 represent the random sample taken

X=492 represent the people that chose a window seat.

\hat p=\frac{492}{778}=0.632 estimated proportion of people that chose a window seat.

p_o=0.5 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the majority of adults prefer window seats when they fly:  

Null hypothesis:p \leq 0.5  

Alternative hypothesis:p > 0.5  

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.

3) Calculate the statistic  

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

z=\frac{0.632 -0.5}{\sqrt{\frac{0.5(1-0.5)}{778}}}=7.36  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.01. 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>7.36)=9.19x10^{-14}  

5) Conclusion

Since the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of adults prefer window seats when they fly is significantly higher than 0.5 .  

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