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Sergio [31]
3 years ago
10

Ron is interested at working at Gas’N’Go, but whenever he looks on their company Web site he doesn’t see any openings listed. Wh

at should Ron do?
a.
Ron should wait for six months before checking the company site again.
b.
Ron should check the company Web site once a week to look for openings.
c.
Ron should send a prospecting letter to see if there are any unadvertised openings
d.
Ron should give up and look at different companies
Business
2 answers:
Brrunno [24]3 years ago
4 0

Answer: c). Ron should send a prospecting letter to see if there are any unadvertised openings

Explanation: Since Ron is interested at working at Gas’N’Go, but is unable to find an opening on the company Web site he should try and send a prospecting letter to see if there are any unadvertised openings. It might happen at times that companies do not post online for certain positions. but when Ron send a letter he might have a chance to get a call from them.

Whitepunk [10]3 years ago
4 0
Questions like this usually want an immediate result so sending the email or C. is most likely the correct answer
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Answer:

t=\frac{5000-5600}{\frac{800}{\sqrt{16}}}=-3      

Explanation:

Data given and notation      

\bar X=5000 represent the sample mean      

s=800 represent the standard deviation for the sample      

n=16 sample size      

\mu_o =5600 represent the value that we want to test    

\alpha 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 determine if the mean is lower than 5600, the system of hypothesis would be:      

Null hypothesis:\mu \geq 5600      

Alternative hypothesis:\mu < 5600      

We don't know the population deviation, so for this case 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{5000-5600}{\frac{800}{\sqrt{16}}}=-3      

4 0
3 years ago
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