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tia_tia [17]
3 years ago
13

You might say that this experiment was an attempt to build a scale, and then calibrate it against a scale that we trust (the ele

ctronic scale). So as not to give away the answer to the preceding question, let's assume that we're talking about a different scale with the same design. Several trials are conducted, and the average and 95% confidence interval are calculated. The mass determined using this other balance scale does NOT agree with the mass measured on a digital scale. Is it possible that the balance scale and the electric scale are accurate, and the lack of agreement was just random?
Physics
2 answers:
hammer [34]3 years ago
8 0

Answer:

yes

Explanation:

Allushta [10]3 years ago
3 0

No.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a random error has been minimized or even eliminated.

<h3>What is a random error?</h3>

Random error is defined as the deviation of the total error from its mean value due to chance.

Random errors can result from the instrument not being precise or from mistakes by the researcher.

Random errors can be minimized by taking multiple readings and averaging the results.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a ransom error has been minimized.

Learn more about random errors at: brainly.com/question/22041172

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2 years ago
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus
babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

I = m\cdot (\vec{v}_{2} - \vec{v_{1}}) (1)

Where:

I - Impulse, in kilogram-meters per second.

m - Mass, in kilograms.

\vec{v_{1}} - Initial velocity of the hockey park, in meters per second.

\vec{v_{2}} - Final velocity of the hockey park, in meters per second.

If we know that m = 0.2\,kg, \vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right] and \vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right], then the impulse applied by the stick to the park is approximately:

I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]

I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

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3 years ago
What is the oxidation state of a hydrogen atom bound to an iron atom.?
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the answer is rust so the answer is rust
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4 years ago
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Suppose an endothermic reaction has a positive change in entropy greater than the heat absorbed. What can be said about the spon
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