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steposvetlana [31]
3 years ago
14

museum passes cost $5 for adults and $2 for children. one day the museum sold 1820 passes for $6100. how many of each type were

sold
Mathematics
1 answer:
mr Goodwill [35]3 years ago
3 0
1820 divide by 5 =364
1820 divide by 2=910

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A newspaper poll asked respondents if they trusted "eco friendly" labels on cleaning products. Out of 1000 adults surveyed, 498
Paul [167]

Answer:

So the p value obtained was a very high value and using the significance level assumed \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 respondents that trust these labels is not significanlty less than 0.5 or 50%.  

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=498 represent the adults that trust these labels

\hat p=\frac{498}{1000}=0.498 estimated proportion of respondents that trust these labels

p_o=0.5 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of respondents that trust these labels is at least 50%:  

Null hypothesis:p\geq 0.5  

Alternative hypothesis:p < 0.5  

When we conduct a proportion test we need to use the z statisitc, 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.

3) Calculate the statistic  

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

z=\frac{0.498 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=-0.126  

4) 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 is not provided but we can assume it as \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

So the p value obtained was a very high value and using the significance level assumed \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 respondents that trust these labels is not significanlty less than 0.5 or 50%.  

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A local gym offers a trial membership for 3 months. It discounts the regular monthly fee x by $25. If the total cost of the tria
adoni [48]
X-25 < 100

x minus the $25 discount has to be less than 100
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The first one is 8 and here's a graph. the second one is 5 again here's a graph

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natita [175]

Answer:

Yes

Step-by-step explanation:

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An artist makes a cube-shaped sculpture.
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All the sides of a cube are always the same.
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