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Vladimir79 [104]
3 years ago
7

Kellens bill at a restaurant before tax and tip is $22 if tax is 5.25% and we wants to leave 15% of the bill including the tax o

n for a tip how much will he spend in total
Mathematics
2 answers:
maxonik [38]3 years ago
4 0
26.62   you would have to multiply 22 by 0.0525  and then add that to 22 and  u would get 23.15, then you would have to multiply that by 0.15 and u would get 3.47 and then you would add 3.47 and 23.15 and you get 26.62 witch is the answer 

hope this helped 
Musya8 [376]3 years ago
3 0

22x0.0525 = 1.155

22 x 0.15 = 3.3 (tip)

22 + tax + tip = 26.46 (unless the tax is actually 11.55)

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The length of time that an auditor spends reviewing an invoice is approximately normally distributed with a mean of 600 seconds
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Answer: 11.5% 

Explanation:


Since 1 minute = 60 seconds, we multiply 12 minutes by 60 so that 12 minutes = 720 seconds. Thus, we're looking for a probability that the auditor will spend more than 720 seconds. 

Now, we get the z-score for 720 seconds by the following formula:

\text{z-score} =  \frac{x - \mu}{\sigma}

where 

t = \text{time for the auditor to finish his work } = 720 \text{ seconds}
\\ \mu = \text{average time for the auditor to finish his work } = 600 \text{ seconds}
\\ \sigma = \text{standard deviation } = 100 \text{ seconds}

So, the z-score of 720 seconds is given by:

\text{z-score} = \frac{x - \mu}{\sigma}
\\
\\ \text{z-score} = \frac{720 - 600}{100}
\\
\\ \boxed{\text{z-score} = 1.2}

Let

t = time for the auditor to finish his work
z = z-score of time t

Since the time is normally distributed, the probability for t > 720 is the same as the probability for z > 1.2. In terms of equation:

P(t \ \textgreater \  720) 
\\ = P(z \ \textgreater \  1.2)
\\ = 1 - P(z \leq 1.2)
\\ = 1 - 0.885
\\  \boxed{P(t \ \textgreater \  720)  = 0.115}

Hence, there is 11.5% chance that the auditor will spend more than 12 minutes in an invoice. 
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