Select the correct answer.

Jane has 6/7 of a yard of ribbon. She uses 3/4 of a yard to trim an ornament. how much ribbon is left?

Degger [83]

Answer:

2/28

Step-by-step explanation:

3 0

3 years ago

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The probability that the bus is late on Monday is 20%, and the probability that the bus is late on Tuesday is 10%. What is the p

Aleks [24]

Add 20% and 10% (30%). Then, because it is two days combined, it would be 30% of 200. Divide each by 2, and there is your answer.

<span>The probability that the bus will be late on both days is: 15%.</span>

7 0

3 years ago

Kara drew a picture of a monarch butterfly. A monarch butterfly has an average wingspan of 4 inches. In her picture, the butterf

Artyom0805 [142]

5.5x Scale
22/4=5.5
She drew the butterfly 5.5 times bigger than it is in real life.

8 0

3 years ago

Name three pairs of numbers that have 5 as their gfc (Greatest Common factor). Use each number once in your answer. I really nee

katen-ka-za [31]

20,25 and 35,50 and 10,15. Each pair would have a GCF of 5.

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3 years ago

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a national survey conducted among a simple random sample of 1,507 adults shows that 56% of americans think the civil war is stil

soldier1979 [14.2K]

a)

$z=4.658\\p_v=P(z > 4.65)=1-P(z < 4.65)=1.581x*10^{-6}$

b) By assuming a significance level of α = 0.05 and observing that $p_v$ > α, we can reject the null hypothesis at this level of significance. And in this instance, it makes sense that more than 50% of Americans believe that the Civil War still has an impact on American politics and political life.

c) The confidence interval at 90% would be shown (0.527;0.593).

About 54% to 59% of all Americans, in our opinion, believe that the Civil War is still significant today.

<h3>What is a survey?</h3>

A survey is a set of questions used in human subject research with the goal of gathering specific information from a certain population. Surveys can be carried out via the phone, by mail, online, at street corners, and even in shopping centers.

a)

1) The presented data and the notation "n=1507" refer to the random sample of Americans, "X" denoting the proportion of those Americans in the sample who believe that the Civil War is still significant to American politics and political life, and "p cap" denoting the estimated number of those Americans.

Since the problem specifies that majority α=0.05 represents the significance threshold, $p_0$ is the number that we wish to test (no given, but is assumed)

Z would stand for the statistic (interesting variable), and $p_v$ for the p value (variable of interest)

P is the percentage of the public that believes that the Civil War still has an impact on American politics and political life.

2) Useful concepts and formulas

The allegation that more Americans than 50% believe the Civil War is still significant to American politics and political life needs to be tested through a hypothesis (Majority). :

Alternative Hypothesis: p $\geq$ 0.5

Null Hypothesis: p $\leq$ 0.5

The proportion is thought to have a normal distribution.

This test for the percentage of union membership has one tail higher.

Verify whether the sample conforms to the presumptions required for the test to be applied.

a) The random sample must be representative; in this situation, there is no mention of this issue, but it is safe to presume that it exists.

b) The sample size must be sufficient.

$np_0=1507*0.5=753.5 > 10\\n(1-p_0)=1507*(1-0.5)=753.5 > 10$

3. Determine the statistic.

The formula used to calculate the statistic is as follows:

$z=\frac{p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n} } }= \frac{0.56-0.5}{\sqrt{\frac{0.5(1-0.5)}{1507} } }=4.658$

P value= $p_v=P(z > 4.65)=1-P(z < 4.65)=1.581*x*10^{-6}$

According to the alternative hypothesis, the following would be the p value:

P value= $p_v=P(z > 4.65)=1-P(z < 4.65)=1.581*x*10^{-6}$

Using this formula, the confidence interval would be determined.

p ± $z_{\frac{\alpha }{2}}\sqrt{\frac{p(1-p)}{n} }$

With the value of $\alpha$ and $\alpha _2$ , we can get the quantile necessary for the interval in the normal standard distribution for the 90% confidence interval.

$z_{\frac{\alpha }{2} }=2.58$

adding the following to the formula for the confidence interval:

$0.56-2.58\sqrt{\frac{0.56(1-0.56)}{1507} } = 0.527\\0.56+2.58\sqrt{\frac{0.56(1-0.56)}{1507} } = 0.593$

Also provided would be the 90% confidence interval (0.527;0.593).

About 54% to 59% of all Americans, in our opinion, believe that the Civil War is still significant today.

And because the interval does not contain the value of 0.5, we can conclude that, at a 90% level of confidence, the percentage of Americans who believe that the Civil War is still relevant to American politics and political life is higher than 0.5. This result is consistent with the result of part b.

To know more about survey, visit:

brainly.com/question/13192080

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6 0

2 years ago