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o-na [289]
4 years ago
10

Natalie walked 35 mile in 12 hour. How fast did she walk, in miles per hour? A. 310 mile per hour B. 56 mile per hour C. 115 mil

es per hour D. 2 miles per hour
Mathematics
2 answers:
vichka [17]4 years ago
6 0
2 miles per hour
Hope this helps
Paha777 [63]4 years ago
5 0
12 hours = 35 miles
1 hour = 35 / 12
= 2.91666667 miles
= 2 miles per hour.

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Step-by-Step Explanation:

Parenthesis first...

39.64-5*(2)

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According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100:114 (46.7%
mojhsa [17]

Answer:

z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154  

p_v =2*P(z  

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 proportion of girls born is not significantly different from 0.467

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=63 represent the number of girls born

\hat p=\frac{63}{150}=0.42 estimated proportion of girls born

p_o=0.467 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion if girls is 0.467.:  

Null hypothesis:p=0.467  

Alternative hypothesis:p \neq 0.467  

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.

Calculate the statistic  

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

z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154  

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=0.05. 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  

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 proportion of girls born is not significantly different from 0.467

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