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Kipish [7]
4 years ago
12

Based on the data which company most likely has the longest average commute time per employee?

Mathematics
1 answer:
Natalija [7]4 years ago
3 0
Company 1 average= 24.4 minutes 
company 2 average= 19.8 minutes
company 3 average= 15 minutes 
company 4 average= 19 minutes

The answer is 24.4 or company one
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A quality control officer is randomly checking the weights of pumpkin seed bags being filled by an automatic filling machine. Ea
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If a bag of pumpkin seeds must weigh within 2.1 grams in order to be accepted, each bags must weigh between (400-2.1) grams and (400+2.1) grams. This means that if a bag weighs less than 397.9 grams (x<397.9) or more than 402.1 grams (x>402.1) it will be rejected. Therefore, 397.9>x>402.1 is the range of rejected bags.
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Let A = {1, 2, 3, 4, 5} and B = {2, 4}. What is A ∩ B? (1 point)
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Answer:

{2, 4}.

Step-by-step explanation:

A = {1, 2, 3, 4, 5} and B = {2, 4}.

A ∩ B?

This means intersection or what is in common

A ∩ B =  {2, 4}.

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3 years ago
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Answer:

The answer should be C

Step-by-step explanation:

I just got done calculating it and according to my calculations it should be C

5 0
3 years ago
I need help with the question pictured.​
Anna35 [415]
The correct answer would be option number 2
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The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales
andrezito [222]

Answer:

t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657    

p_v =P(t_{(39)}>6.657)=3.17x10^{-8}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 42 at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=44 represent the sample mean

s=1.9 represent the sample standard deviation

n=40 sample size  

\mu_o =42 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 42, the system of hypothesis would be:  

Null hypothesis:\mu \leq 42  

Alternative hypothesis:\mu > 42  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=40-1=39  

Since is a one side test the p value would be:  

p_v =P(t_{(39)}>6.657)=3.17x10^{-8}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 42 at 1% of signficance.  

3 0
3 years ago
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