What doe decimals have that fractions don't have?

All vertices must be used once in Hamilton circuits. <br><br> true<br> false

kolbaska11 [484]

Answer:

false I believe

Step-by-step explanation:

hope its right

8 0

3 years ago

What are the solutions to the equation x² = 256?

tester [92]

The correct answer would be 128.

7 0

3 years ago

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1. Use a piece of wool, thread or chord to measure the circumference of each of the circles found on the last page. To do this j

spayn [35]

Answer:

just doing this for my points!!! lol

Step-by-step explanation:

6 0

3 years ago

What is the area of the following triangle in square meters? Do not round your answer. [Brainliest + 25 PTS. (EACH) + A shoutout

QveST [7]

57.4 because u add all the side up and u get a answer

4 0

3 years ago

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In a study of the accuracy of fast food drive-through orders, McDonald’s had 33 orders that were not accurate among 362 orders o

melomori [17]

Answer:

A. We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B. Null hypothesis: $p=0.1$

Alternative hypothesis: $p \neq 0.1$

C. $z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558$

D. $z_{\alpha/2}=-1.96$ $z_{1-\alpha/2}=1.96$

E. Fail to the reject the null hypothesis

F. So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.

Step-by-step explanation:

Data given and notation

n=362 represent the random sample taken

X=33 represent the number of orders not accurate

$\hat p=\frac{33}{363}=0.0912$ estimated proportion of orders not accurate

$p_o=0.10$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

A: Write the claim as a mathematical statement involving the population proportion p

We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B: State the null (H0) and alternative (H1) hypotheses

Null hypothesis: $p=0.1$

Alternative hypothesis: $p \neq 0.1$

When we conduct a proportion test we need to use the z statistic, 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 $p_o$ .

C: Find the test statistic

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

$z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558$

D: Find the critical value(s)

Since is a bilateral test we have two critical values. We need to look on the normal standard distribution a quantile that accumulates 0.025 of the area on each tail. And for this case we have:

$z_{\alpha/2}=-1.96$ $z_{1-\alpha/2}=1.96$

P value

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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

E: Would you Reject or Fail to Reject the null (H0) hypothesis.

Fail to the reject the null hypothesis

F: Write the conclusion of the test.

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.

6 0

3 years ago