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Anastasy [175]
2 years ago
14

Use the box plot to answer the questions.

Mathematics
2 answers:
mixas84 [53]2 years ago
6 0

Answer:

20 90 8

Step-by-step explanation:

Anna007 [38]2 years ago
3 0

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

Answer:

The answer is... 20, 90, 8

Step-by-step explanation:

What is the range of Charles’s scores?

<h3>20</h3>

What is the median of Charles’s scores?

<h3>90</h3>

What is the IQR of Charles’s scores?

<h3>8</h3>

If you move the graph to its correct spot, your answers will be clear.

I hope this helps!

- sincerelynini

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Person A buys 10 granola bars and 6 cups of yogurt for 518 Person B buys 5 granola bars and 4 cups of yogurt for $9.50. Find the
Alja [10]

Person A buys 10 granola bars and 6 cups of yogurt for <em>$18</em>

Person B buys 5 granola bars and 4 cups of yogurt for <em>$9.50</em>

Let <u>x</u> represent the granola bars

Let <u>y</u> represent the yogurt

10x + 6y = 18 << ( Divide both sides by 2 )

The second equation is 5x + 4y = 9.50. Now subtract the two equations.

5x + 4y = 9.50

-5x - 3y = -9

         y = $.50

5x + 3(.50) = 9

5x + 1.50 = 9

5x = 7.5

  x = $1.50

I hope this answered your question  

Please remember to rate my answer and comment bellow if you have any questions

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Last year, 46% of business owners gave a holiday gift to their employees. A survey of business owners indicated that 35% plan to
DiKsa [7]

Answer:

a) Number = 60 *0.35=21

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

p_v =P(Z  

c) If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

Step-by-step explanation:

Data given and notation  

n=60 represent the random sample taken

X represent the business owners plan to provide a holiday gift to their employees

\hat p=0.35 estimated proportion of business owners plan to provide a holiday gift to their employees

p_o=0.46 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)  

Part a

On this case w ejust need to multiply the value of th sample size by the proportion given like this:

Number = 60 *0.35=21

2) Part b

We need to conduct a hypothesis in order to test the claim that the proportion of business owners providing holiday gifts had decreased from the 2008 level:  

Null hypothesis:p\geq 0.46  

Alternative hypothesis:p < 0.46  

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.

Calculate the statistic  

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

z=\frac{0.35 -0.46}{\sqrt{\frac{0.46(1-0.46)}{60}}}=-1.710  

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 left tailed test the p value would be:  

p_v =P(Z  

Part c

If we compare the p value obtained and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% the proportion of business owners providing holiday gifts had decreased from the 2008 level.  

We can use as smallest significance level 0.044 and we got the same conclusion.  

8 0
2 years ago
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