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1 year ago

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Mathematics

1 answer:

JulijaS [17]1 year ago

5 0

Answer:

67,296.75

Step-by-step explanation:

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Surveys were sent to a random sample of owners of all-wheel-drive (AWD) vehicles and to a random sample of owners of front-wheel

salantis [7]

Answer:

No, the confidence interval does not provide evidence that this belief is plausible as the confidence interval range contains the null value (0)

Step-by-step explanation:

Given that the confidence interval for the mean difference of AWD and FWD vehicles is :

(−0.01,0.12)

When defining the hypothesis for mean difference :

H0 : μ1 - μ2 = 0

H1 : μ1 - μ2 ≠ 0

Using the confidence interval result, reject the Null if the confidence interval value at the given α - level does not contain 0 ; otherwise fail to reject the Null ;

Since the (−0.01,0.12) contains 0 ; then we fail to reject the Null and conclude that there is no statistically significant difference between the two groups

8 0

1 year ago

Shaded area=40<br> Find x <br> Width=3x<br> Length=4x<br> Not shaded area=2x

Mashutka [201]

In the process of laying out the question, you have given us important facts about the shaded area and the unshaded area. Now step back for just a second. If they're that deeply involved in the question, do you honestly expect that anybody can answer the question without seeing the picture of them ? ! ?

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2 years ago

I need help with this

Aliun [14]

-55+13= -42

94i -12i = 82i

answer = -42+82i

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2 years ago

1) Two apples weight________ kilograms.

ratelena [41]

Answer:

12, 72, 180

Step-by-step explanation:

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2 years ago

Read 2 more answers

The following data set represents the age of

Julli [10]

$\huge\text{Hey there!}$

$\huge\textsf{Breaking the meaning down in simpler terms}$
$\huge\textbf{Median:}\\\large\text{is understood to be \bf the number in the center (also known as}\\\large\textbf{the middle number)}$

$\huge\textbf{Mean:}\\\large\text{is understood to be the \bf total.}$

$\huge\textsf{Formulas for each term}$

$\huge\text{Median:}\\\large\textsf{To find the median you have to put each of your numbers in your data}\\\large\textsf{set from LEAST (smallest) to GREATEST (biggest). }$

$\huge\text{Mean:}\\\mathsf{\dfrac{sum\ of\ all\ terms}{number\ of\ terms}= mean}$

$\huge\textbf{SOLVING FOR THE QUESTION}$

$\huge\textsf{Median:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textbf{Conversion: }\large\text{27, 33, 38, 41, 42, 52, 64, 68, 72}\\\\\large\textsf{Make sure it is even between both sides of the data plot.}\\\\\large\text{It seems to even on BOTH sides of \boxed{\textsf{14}} so it could possibly be your}\\\large\text{median.}$

$\huge\textsf{Mean:}\\\\\large\textbf{Original data set: }\large\text{27, 52, 64, 41, 33, 38, 42, 60, 72, 68}\\\\\large\textsf{Your equation: }\mathsf{\dfrac{27 + 52 + 64 + 41 + 33 + 38 + 42 + 60 +72 + 68}{10}}\\\\\mathsf{\dfrac{79 + 64 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{143 + 41 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{184 + 33 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{217 + 38 + 42 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{255 + 42 + 60 + 72 + 68}{10}}$

$\mathsf{\dfrac{297 + 60 + 72 + 68}{10}}\\\\\mathsf{\dfrac{357 + 72 + 68}{10}}\\\\\mathsf{\dfrac{429 + 68}{10}}\\\\\mathsf{\dfrac{497}{10}}\\\\\mathsf{\approx 49.70}\\\\\large\text{From the looks of it \boxed{\rm{\dfrac{497}{10}}}\ or \boxed{\text{49.70}} could possibly be your mean.}$

$\huge\text{Good luck on your assignment \& enjoy your day!}$

<h3>
~ $\frak{Amphitrite1040:)}$ </h3>

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1 year ago