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kondor19780726 [428]
1 year ago
13

Because you are working a late shift, you will get an increase in your wage. You will get and additional 10% per hour from 6:00p

m to 12:00 am and additional 15% from 12:00 am to 1:00 am
Mathematics
1 answer:
siniylev [52]1 year ago
7 0

CHECK THE COMPLETE QUESTION BELOW:

Supervisor: "Because you are working a late shift, you will get an increase in your wage. You will get an additional 10% per hour from 6:00 PM to 12:00 AM and additional 15% from 12:00 AM to 1:00 AM." Employee: "I am working from 2:00 PM to 12:00 AM. Between 2:00 PM and 6:00 PM, I will make $12.00 per hour. Between 6:00 PM and 12:00 AM, I will make __________ per hour

Answer:

I will make $13.2 per hour

Explanation:

It can be deducted from the question that that the money Employee gets money for Working from 2.00 PM to 6.00 PM = $12per hour

Also the Employee gets additional 10% of money for working from 6.00 PM to 12.00 AM which implies that

10% = (10/100)*$12per hour

=$1.2per hour

Again, Employee gets additional 15% of the money for working from 12.00 AM to 01.00 AM which can be calculated as:

15% = (15/100)*$12per hour

=$1.8 per hour

which means that Employee gets additional $1.8 per hour

of the money for working from 12.00 AM to 01.00 AM

Therefore, if Employee works from 2.00 PM to 12.00 AM,

Employee will gets money for Working from 2.00 PM to 6.00 PM as well as the 10\% extra wage which

= $12 + $1.2 = $13.2

= \$ 12 + \$ 1.2 = \$ 13.2

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Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

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            \mu = population mean

<em>Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

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So, 95% confidence interval for the population mean, \mu is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

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P(-1.645 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.645) = 0.90

P( -1.645 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

P( \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

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The range of the function is the set of all possible values that the function might take, that is {y|y=6x-4}. Recall that every real number y could be written of the form y=6x-4 for a particular x. So the range of the function is the set {y|y is a real number}.

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