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Zarrin [17]
3 years ago
11

Distinguish between nonsampling error and sampling error. Choose the correct answer below.

Physics
1 answer:
kiruha [24]3 years ago
5 0

Answer:

The correct answer is D.

Non-sampling error is the error that results from​ under-coverage, non-response​ bias, response​ bias, or​ data-entry errors. Sampling error is the error that results because a sample is being used to estimate information about a population.

Explanation:

Sampling error is related to the variation between the true values of the sample and the population. If occurred, it is always random depending upon the sample chosen.

Non-sampling error can be random as well as non-random. Non-sampling error can occur irrespective of the sample chosen. It is related to the inappropriate analysis of the data.

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Write the chemical formula for the following diagrams.
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Answer:

hydrogen chloride...........

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Calculate the speed of an 8.0x10^4 kg airliner with a kinetic energy of 1.1x10^9 j ...?
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Kinetic energy is the energy possessed by an object on motion. it is expressed as follows:

KE = 0.5mv^2

where m is the mass and v is the velocity of the object. We calculate as follows:


KE = 0.5mv^2
1.1x10^9 J = 0.5(8.0x10^4 kg) v^2
v = 165.83 m/s
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How many significant figures does the number 91010.0 have?
kirza4 [7]

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6 (six)

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3 years ago
Read 2 more answers
A cylinder with rotational inertia I1 = 3.0 kg · m2 rotates clockwise about a vertical axis through its center with angular spee
erastova [34]

Answer:

ω = 1.83 rad/s clockwise

Explanation:

We are given:

I1 = 3.0kg.m2

ω1 = -5.4rad/s (clockwise being negative)

I2 = 1.3kg.m2

ω2 = 6.4rad/s  (counterclockwise being positive)

By conservation of the momentum:

I1 * ω1 + I2 * ω2 = (I1 + I2) * ω

Solving for ω:

\omega = \frac{I1 * \omega1 + I2*\omega2}{I1+I2}=-1.83rad/s

Since it is negative, the direction is clockwise.

8 0
2 years ago
A motorboat traveling on a straight course slows
never [62]

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

  • 65 km/h → 18.0556 m/s
  • 35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

  • v² = v₀² + 2aΔx

Substitute the known values into the equation.

  • (9.72222)² = (18.0556)² + 2a(45)
  • 94.52156173 = 326.0046914 + 90a
  • -231.4831296 = 90a
  • a = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

3 0
3 years ago
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