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cupoosta [38]
3 years ago
12

10.0 ppm 10.4 ppm 10.2 ppm 10.8 ppm 10.1 ppm 10.5 ppm 10.1 ppm Using the new instrument on the first day the technicians got a v

alue of 9.0 ppm. Are the technicians right to exclude the measurement of the first day as an outlier at the 90% confidence level?
Physics
1 answer:
iren [92.7K]3 years ago
3 0

Answer:

The datapoint 9.0 ppm is outlier at the 90% confidence level.

Explanation:

The old data has following values

mean=10.5 mm

standard deviation 0.2 mm

Now the mean of new values is calculated as following

mean=\frac{10+10.4+10.2+10.8+10.1+10.5+10.1}{7}\\mean=10.3 ppm

So the value as 9.0 ppm can be considered easily as outlier in  this regard.

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Darryl finds a bottle of what looks like clear water, with dirt settled at the bottom. When he shakes the bottle, the water gets
Alexus [3.1K]
Was there any choice of answers so i can help or something u have to figure out
7 0
4 years ago
. Two identical vehicles traveling at the same speed are made to collide with barriers in an insurance company collision test. T
Serggg [28]

Answer:

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

Explanation:

For this exercise let's use the relationship between momentum and momentum.

         I = F t = Δp

in this case the final velocity is zero

        F t = 0 -m v₀

        F = m v₀ / t

in order to answer the question we must assume that the two vehicles have the same mass and speed

concrete barrier

        F₁ = -p₀ / 0.1

        F₁ = - 10 p₀

barrier collapses

         F₂ = -p₀ / 1

let's look for the relationship of the forces

        F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

5 0
3 years ago
In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le
expeople1 [14]

Answer:

The  value  is   m \approx   310

Explanation:

From the question we are told that

     The  focal length of the objective is  f_o =  1.0 \ cm

    The  focal length of the eyepiece is  f_e  =  2.0 \  cm

    The  tube length is  L  =  25 \  cm

Generally the magnitude of the overall magnification is mathematically represented as

            m =  m_o  *  m_e

Where  m_o is the objective magnification which is mathematically represented as

        m_o  =  \frac{L}{f_o }

=>      m_o  =  \frac{25}{1 }

=>      m_o  =  25

m_e is the eyepiece magnification which is mathematically evaluated as

     m_e  =  \frac{L }{f_e }

     m_e  =  \frac{25 }{ 2}

      m_e  =  12.5 \  cm

So

    m =  25 * 12.5

     m \approx   310

6 0
3 years ago
¿Que es un átomo???????
disa [49]
Un átomo es una porción material menor de un elemento químico que interviene en las reacciones químicas y posee las propiedades características de dicho elemento.
5 0
3 years ago
How much current is in a circuit that includes a 9-volt battery with a resistance of 3
Darya [45]
Just remember
Voltage = current times resistance

current = voltage over resistance

Current = 9/3 = 3
7 0
3 years ago
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