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Rzqust [24]
3 years ago
9

Highway transportation officials are trying to determine if the average speed on Archer road is above 45mph, the post speed limi

t. They took a sample of 50 cars. The sample mean was 44.9 and the standard deviation was 10. Find the test statistic.0.010.0707-0.01-0.0707
Physics
1 answer:
USPshnik [31]3 years ago
4 0

Answer:

Option D) -0.0707

Explanation:

We are given the following in the question:  

Population mean, μ = 45 mp

Sample mean, \bar{x} = 44.9

Sample size, n = 50

Sample standard deviation, s = 10

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{44.9 - 45}{\frac{10}{\sqrt{50}} } = -0.0707

Thus, the test statistic is

Option D) -0.0707

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Answer:

The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.

See attachment below for full solution

Explanation:

This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.

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3 years ago
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3 years ago
The temperature at 8:00
musickatia [10]
Change in temperature = final temperature - Initial temperature 
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Hope this helps!
6 0
2 years ago
A bicycle racer inflates their tires to 7.1 atm on a warm autumn afternoon when temperatures reached 27 °C. By morning the tempe
natulia [17]

Answer:

The required pressure is 6.4866 atm.

Explanation:

The given data : -

In the afternoon.

Initial pressure of tire ( p₁ ) = 7 atm = 7 * 101.325 Kpa =  709.275 Kpa

Initial temperature ( T₁ ) = 27°C = (27 + 273) K = 300 K

In the morning .

Final temperature ( T₂ ) = 5°C = ( 5 + 273 ) K = 278 K

Given that volume remains constant.

To find final pressure ( p₂ ).

Applying the ideal gas equation.

p * v = m * R * T

\frac{p}{T}  = constant

\frac{p_{1} }{T_{1} }  = \frac{p_{2} }{T_{2} }

p_{2}  = \frac{T_{2} }{T_{1} } *p_{1}  

p_{2}  = \frac{278}{300}  * 709.275  = 657.2615 Kpa = 6.486 atm

8 0
2 years ago
What are some things that doesn't move in a circular motion or are non examples of circular motion?
photoshop1234 [79]

Answer:In non-uniform circular motion, an object's motion is along a circle, but the object's speed is not constant. In particular, the following will be true. The object's velocity vector is always tangent to the circle. The speed and angular speed of the object are not constant

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7 0
3 years ago
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