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Juliette [100K]
4 years ago
11

Which is a secant of circle A?

Mathematics
1 answer:
sergey [27]4 years ago
7 0

Answer: line CD (fourth choice)

====================================================

Let's go through the choices one by one

1) Segment AB is a radius, which is not a secant line. We can rule this out.

2) line DE is a tangent line which only touches the circle at exactly one spot. We need something that cuts the circle in 2 spots for it to be a secant line. This can also be crossed off the list.

3) segment HG is a chord, which is fairly close to a secant line, but it must extend infinitely in both directions. In other words, it needs to be a line instead of a line segment. This is crossed off the list.

4) line CD is a secant line. It is a geometric line in that it goes on forever in both directions (it's not a segment, and not a ray either). Also, this line crosses the circle at 2 points, which contrasts it from a tangent line. This is the answer we want.

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Annette is stacking boxes in her closet. There are 15 boxes in all. If each box weighs 7.5 pounds, his much do the boxes weigh t
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4 years ago
The countries of Europe report that 48% of the labor force is female. The United Nations wonders if the percentage of females in
Ostrovityanka [42]

Answer:

a) 0.436 - 1.64\sqrt{\frac{0.436(1-0.436)}{525}}=0.400

0.436 + 1.64\sqrt{\frac{0.436(1-0.436)}{525}}=0.472

The 90% confidence interval would be given by (0.400;0.472)

b) We are confident at 90% that the true proportion of female in the Labor Force is between 0.400 and 0.472

c) Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

For this case since the upper limit of the confidence interval is lower than 0.48 (0.472<0.48) we can conclude that the true proportion of female is actually lower than 0.48 or 48% at 10% of signficance.

Step-by-step explanation:

Part a

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. And the critical value would be given by:

z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64

The estimated proportion for female is \hat p = \frac{229}{525}= 0.436

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.436 - 1.64\sqrt{\frac{0.436(1-0.436)}{525}}=0.400

0.436 + 1.64\sqrt{\frac{0.436(1-0.436)}{525}}=0.472

The 90% confidence interval would be given by (0.400;0.472)

Part b

We are confident at 90% that the true proportion of female in the Labor Force is between 0.400 and 0.472

Part c

We need to conduct a hypothesis in order to test the claim that the true proportion is lower than 0.48:  

Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

For this case since the upper limit of the confidence interval is lower than 0.48 (0.472<0.48) we can conclude that the true proportion of female is actually lower than 0.48 or 48% at 10% of signficance.

5 0
3 years ago
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