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Wittaler [7]
3 years ago
7

It costs Charlie's Car Repair Shop $27 to change the oil. The shop charges their costumers $40 for an oil change. What is the pe

rcent of markup in the cost of an oil change?
Mathematics
1 answer:
frosja888 [35]3 years ago
3 0
Cost Price for oil change =$27 
Charge for the customers =$40 
Markup cost = Difference between selling price and cost price =$40-$27=$13  
To find the percent of markup, divide the markup cost by the cost price and multiply it by 100. 
Markup percent = (13/27)*100 
 =48.15%
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Shtirlitz [24]

Answer:

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Step-by-step explanation:

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The life expectancy (in hours) of an electric bulb is normally distributed with a mean of 5000 and a standard deviation of 1000.
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Answer:

P(z>1.3) = 0.9032

Step-by-step explanation:

We are given:

Mean = 5000

Standard deviation = 1000

x = 6300

P(x>6300)=?

z-score =?

z-score = x- mean/standard deviation

z-score = 6300 - 5000/1000

z- score = 1300/1000

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So, P(x>6300) = P(z>1.3)

Looking at the z-probability distribution table and finding value:

P(z>1.3) = 0.9032

So, P(z>1.3) = 0.9032

7 0
3 years ago
According to a Washington Post-ABC News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bother
k0ka [10]

Answer:

z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124  

p_v =P(z>7.124)=5.24x10^{-13}  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v 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 interest is significantly higher than 0.5.  

Step-by-step explanation:

Data given and notation

n=502 represent the random sample taken

X=331 represent the adults that said they would not be bothered if the NAtional security agency

\hat p=\frac{331}{502}=0.659 estimated proportion of people who would not be bothered if the NAtional security agency

p_o=0.5 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v 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 the true proportion is higher than the majority of 0.5:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124  

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 \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>7.124)=5.24x10^{-13}  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v 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 interest is significantly higher than 0.5.  

4 0
3 years ago
A multiple choice exam has 20 questions. Each question has 3 possible answers; there is no partial credit. Only 1 answer out of
amid [387]

Answer:To find out the theoretical probability of the case given, we need to make certain assumptions.

First, we'll assume that he'll attempt all of the questions, i.e he'll attempt all 10 questions.

Next assumption is that each option in each question is equally likely to be marked by the student.

This pretty much leads us to a binomial probability distribution.

Conditions are:

   Answers 10 questions.

   Each question has 4 options with only one correct answer and all other incorrect answers.

   Student is equally likely to pick any outcome in any given question.

   Hence, probability of choosing correct answer is 1/4 = 0.25. Probability of choosing incorrect answer is 1–1/4 = 3/4 = 0.75.

   The number of trials is 10.

   Total number of success is exactly 8 and failure is 2 amongst the 10 questions in any particular order.

Now, calculation is fairly simple.

Binomial probability distribution is such that…

P(8 correct ; 2 wrong)

= 10C8 × (0.25)^8 × (0.75)²

= 405/1048576 ≈ 3.862380981 × 10^-4 ≈ 0.000386

Step-by-step explanation:

6 0
3 years ago
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