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maks197457 [2]
2 years ago
13

A researcher conducted an experiment to investigate the effects of working in crowded versus uncrowded conditions on the job sat

isfaction of employees in a variety of work settings. In this study, the dependent variable is
Mathematics
1 answer:
Artyom0805 [142]2 years ago
4 0

In a research of investigating the effects of working in crowded versus uncrowded conditions on the job satisfaction of employees in a variety of work setting , working is a depending variable and crowding is an independent variable.

Given aim of the research is to investigate the effects of working in crowded conditions and uncrowded conditions.

We know that depending variable is a variable whose value is dependent on the other variable. Independent variable is a variable whose value does not dependent on the other variables.

In the given question working may be affected negatively in crowded environment means there will be less work in crowded environment and more work in uncrowded environment.

Hence the depending variable in research is working.

Learn more about depending variables at brainly.com/question/25223322

#SPJ4

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The expression below represents the number of bacteria in a petri dish after t hours. Interpret the meaning of the expression.
konstantin123 [22]

The answer is B) The initial number of bacteria is 57, and the growth rate is 31% per hour.


The expression shown is an <em>exponential expression</em>.  An exponential expression usually takes the form ab^{x}, as this one does.  A represents the <em>initial value</em>, in this case-the initial number of bacteria.  B represents the <em>growth rate</em>.  When it is an increasing growth rate, it will begin with 1, because it will have everything it had before, plus the percent increase, the decimal portion.  In this case, a=57, and b=0.31.


Hope this makes sense!

3 0
3 years ago
suppose that a given country of 72 million people about 3,600 die from drowning each year what is the percent chance that a pers
Trava [24]
Your answer is 0.003%
4 0
3 years ago
Use this information to answer the questions. University personnel are concerned about the sleeping habits of students and the n
Oksanka [162]

Answer:

z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

p_v =P(Z>2.097)=0.018  

If we compare the p value obtained and 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% of significance the proportion of  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Step-by-step explanation:

1) Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

\hat p=\frac{209}{377}=0.554 estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

p_o=0.5 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

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.

3) Calculate the statistic  

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

z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

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 provided \alpha=0.05. 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>2.097)=0.018  

If we compare the p value obtained and 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% of significance the proportion of  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

6 0
3 years ago
Which set of graphs can be used to find the solution set to 3 log2(x + 4) &lt; 6?
alexgriva [62]

Answer:

It's graph D

Step-by-step explanation:

5 0
3 years ago
10% of Kindergarteners know how to read before school starts. If there are 150 kindergarteners at a
Tamiku [17]

Answer:

15

Step-by-step explanation:

To get 10 percent, you divide by 10.

150 divided by 10 equals 15.

3 0
4 years ago
Read 2 more answers
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