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miss Akunina [59]
3 years ago
12

A town official claims that the average vehicle in their area sells for more than the 40th percentile of your data set. Using th

e data, you obtained in week 1, as well as the summary statistics you found for the original data set (excluding the super car outlier), run a hypothesis test to determine if the claim can be supported. Make sure you state all the important values, so your fellow classmates can use them to run a hypothesis test as well. Use alpha = .05 to test your claim.(Note: You will want to use the function =PERCENTILE.INC in Excel to find the 40th percentile of your data set)First determine if you are using a z or t-test and explain why. Then conduct a four-step hypothesis test including a sentence at the end justifying the support or lack of support for the claim and why you made that choice.$29,925$46,561$46,711$29,235$42,342$27,249$60,428$59,512$49,196$82,135With Original 10Mean 47329.4Median 46636Standard Deviation 16978.06749Count 10
Mathematics
1 answer:
Hitman42 [59]3 years ago
7 0

Answer:

I'm no sure

Step-by-step explanation:

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1. Which would be a more helpful for solving the system: adding the two equations or subtracting
ira [324]

Answer:

1) Subtracting one from the other

2) x = 0.375 and y = 0.125

Step-by-step explanation:

Equation 1 :

\frac{2x+1}{2y}=7\\\\

Equation 2:

\frac{6x-1}{2y}=5

To solve these equations by the Elimination method we multiply equation 1 with 6 and multiply equation 2 with 2 so now,

\frac{2x+1}{2y} =7\\\\2x+1=14y\\\\Multiplying\ Equation\ 1\  with\ 6\\\\12x+6=84y

Now for the second equation:

\frac{6x-1}{2y}=5\\\\6x-1=10y\\\\Multiplying\ equation\ 2\ with\ 2 \\\\12x-2=20y

Now subtracting equation 2 from equation 1

12x+6=84y\\\\12x-2=20y\\\\Subtracting\ leads\ to\\12x-12x+6-(-2)=84y-20y\\0+8=64y\\8=64y\\8/64=y\\0.125=y

now insert this value of y into any equation

lets insert it into equation 1

\frac{2x+1}{2y} =7\\\\2x+1=14y\\2x+1=14(0.125)\\2x+1=1.75\\2x=1.75-1\\2x=0.75\\x=0.75/2\\x-0.375

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Answer:

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Given that :

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In which situation is the mean larger than the median?<br> a.the distribution is equal?
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