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AleksandrR [38]
2 years ago
9

State if the triangles are similar. If so, how do you know they are

Mathematics
1 answer:
DIA [1.3K]2 years ago
6 0

Step-by-step explanation:

they're similar since 36 % 9 = 0 and 24 % 6 = 0 (they leave no remainder)

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A nationwide study of American homeowners revealed that 65% have one or more lawn mowers. A lawnequipment manufacturer, located
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Answer:

z=\frac{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589  

p_v =P(z>1.589)=0.056  

If we compare the p value obtained with the significance level assumed \alpha=0.05 we see that 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 homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

Step-by-step explanation:

Data given and notation

n=497 represent the random sample taken

X=340 represent the homes in Omaha with one or more lawn mowers

\hat p=\frac{340}{497}=0.684 estimated proportion of homes in Omaha with one or more lawn mowers

p_o=0.65 is the value that we want to test

\alpha represent the significance level

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 true proportion of homes in Omaha with one or more lawn mowers is higher than 0.65.:  

Null hypothesis:p\leq 0.65  

Alternative hypothesis:p > 0.65  

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{0.684 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.589  

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 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>1.589)=0.056  

If we compare the p value obtained with the significance level assumed \alpha=0.05 we see that 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 homes in Omaha with one or more lawn mowers is not ignificantly higher than 0.65

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