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djverab [1.8K]
2 years ago
9

To test the hypothesis that the population mean mu=3. 6, a sample size n=14 yields a sample mean 4. 007 and sample standard devi

ation 0. 607. Calculate the p-value and choose the correct conclusion
Physics
1 answer:
PolarNik [594]2 years ago
5 0

The P value for the given data set is 25127. For finding P value, we have to must find the Z value.

<h3>How to get the z scores?</h3>

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

The Z value is calculated as;

Z = \dfrac{X - \mu}{\sigma})

Z = (X - μ) / σ

Z = (4.007 - 3.6) / 0.607

Z = 0.67051

The P value for the given data set is 25127.

Learn more about z-score here:

brainly.com/question/21262765

#SPJ1

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Considerando que los coeficientes de dilatación de los siguientes metales son: hierro 11.7 x 10-6; plomo 27.3 x 10-6; cobre 16.7
Rasek [7]

Answer:

el plomo será el más largo

Explanation:

Dado que;

longitud inicial (l1) = 4m

Longitud final l2

aumento de temperatura (θ) = 10 ° C

Coeficiente de expansión lineal α

Ahora para el hierro;

α = 11,7 x 10-6

Desde;

l2-l / l1θ = α

l2 = α l1θ + l1

l2 = l1 (αθ + 1)

l2 = 4 ((11,7 x 10-6 * 10) + 1)

l2 = 4.00044 m

Para el plomo

l2 = 4 ((27,3 x 10-6 * 10) + 1)

l2 = 4,00109 m

Para cobre

l2 = 4 ((16,7 x 10-6 * 10) + 1)

l2 = 4.000668 m

Por lo tanto, el plomo será el más largo

7 0
2 years ago
You decide to impress Grandpa by showing him how fast sound travels. You have a piece of plastic pipe with an adjustable closed
fredd [130]

Answer:

336.96m/s

Explanation:

answer is in photo above

3 0
2 years ago
Which of the following is an example of a physical change but not a chemical change?
vazorg [7]

A water pipe freezes and cracks on a cold night. <em> (D)</em>

There are three physical changes going on in this scenario:

1).  The air gets cold and dark at night.

2).  The water in the pipe freezes and expands.

3).  The pipe cracks.

There are no chemical changes in the description.

7 0
3 years ago
Read 2 more answers
Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at
dybincka [34]

Answer:

The intensity at a spot 71 m away is 4.02*10^{-5}  Wm^{-2}

Explanation:

Given:

Initial intensity,I_{1}=3.0*10^{-4} Wm^{-2} at a distance, d_{1} = 26 m

Required:

New intensity, I_{2} =? at a distance, d_{2} = 71 m

Using the inverse square law,

I ∝ \frac{1}{d^{2} }

⇒I_{1}I_{1}d_{1}^{2}  =I_{2}d_{2}^{2}

I_{2} =\frac{I_{1} d_{1}^{2}  }{d_{2}^{2}} =\frac{3.0*10^{-4}*26^{2}  }{71^{2} } \\×

I_{2}=4.02*10^{-5}  Wm^{-2}

Thus, the intensity at a spot that is 71 m away is 4.02*10^{-5}  Wm^{-2}

5 0
3 years ago
The answer to this digram, the question is in the photo.
Alina [70]

there will be no noise with the ball

5 0
3 years ago
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