n an orchid show, seven orchids are to be placed along one side of the greenhouse. of the orchids, we see that the number of lin
1 answer:
Answer:
35
Step-by-step explanation:
7 orchids can be lined as 7!. This means that for the first orchid of the line, you can select 7 options. When you place the first orchid, for the second option you can select among 6 since 1 orchid has already been placed. Similarly, for the 3rd orchid of the line, you have left 5 options. The sequence goes in this fashion and for 7 orchids, you have 7*6*5*4*3*2*1 possibilities. However, there is a restriction here. 3 of the orchids are white and 4 are levender. This means that it does not make a difference if we line 3 white orchids in an arbitrary order since it will seem the same from the outside. As a result, the options for lining the 7 orchids diminish. The reduction should eliminate the number of different lining within the same colors. Similar to 7! explanation above, 3 white orchids can be lined as 3! and 4 levender orchids can be lined as 4!. To eliminate these options, we divide all options by the restrictions. The result is:
= 35. [(7*6*5*4*3*2*1/(4*3*2*1*3*2*1)]
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Answer:
So the p value obtained was a very low value and using the significance level given we have
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 more than one-third of employers have sent an employee home to change clothes is higher than 1/3 or 0.333.
Step-by-step explanation:
Data given and notation
n=2759 represent the random sample taken
X=974 represent the people saying that they had sent an employee home for inappropriate attire
estimated proportion of people saying that they had sent an employee home for inappropriate attire
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
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 more than one-third of employers have sent an employee home to change clothes.:
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statistic, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
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 . The next step would be calculate the p value for this test.
Since is a right tailed the p value would be:
So the p value obtained was a very low value and using the significance level given we have
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 more than one-third of employers have sent an employee home to change clothes is higher than 1/3 or 0.333.
Answer:
See Explanation
Step-by-step explanation:
Your question is poorly formatted. However, I'm able to deduce that your question involves at least 2 algebraic fractions that needs to be simplified.
So, I'll make use of the following expression:
When you follow the steps I'm about to provide, you'll arrive at the right answer
Required
Determine an equivalent expression
Step 1: Take L.C.M of the denominators
Step 2: Evaluate the fractions
Step 3: Open the brackets at the numerator
Step 4: Collect and evaluate like terms
<em>Follow the above steps, and you'll be able to solve your question.</em>