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Vikki [24]
4 years ago
11

A Sampling of 409 motor Shafts from a very large production Batch shows a sample standard deviation in Diameter of 0.021 mm with

a sample mean of 9.251 mm Estimate the mean diameter of the entire batch at a %95 level of confidences?
Engineering
1 answer:
andrezito [222]4 years ago
7 0

Answer:

9.248 < \mu < 9.253 mm

Explanation:

Given data:

standard deviation  = 0.021 mm

sample mean = 9.251 mm

total sample = 409 m

confidence interval level = 95%

mean diameter of entire batch  \mu  = \bar X \pm z \frac{\sigma}{\sqrt{n}}

where

\bar X - sample mean = 9.251 mm

z = critical value

plugginf all value in the above relation to get the mean diameter

\mu = 9.251 \pm 1.96 \times \frac{0.021}{\sqrt{409}}

9.248 < \mu < 9.253 mm

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Dima020 [189]

Answer:

  248.756 mV

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Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

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The Thevenin equivalent source at the other terminal of the bridge is ...

  voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V

  impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω

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The open-circuit voltage is the difference between these terminal voltages:

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The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

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8 0
3 years ago
Resolver em c as equações:<br><br> Z elevado à 2 igual a 3+4i
Marrrta [24]

Answer:

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Explanation:

Um lado da equação será a potência e outro, o número inteiro. De outro modo, transforme a equação deixando-a isolada em um dos lados. Reescreva a equação. Prepare-a a fim de extrair o logaritmo de ambos os lados, que é o inverso da potência. Você pode calcular o logaritmo de base.

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Explanation:

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σadm = 900 / 5 = 180 MPa

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σ = P / A

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Rearranging:

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d = \sqrt{4*P / (\pi*\sigma)}

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elena-s [515]

Answer:

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Explanation:

given data

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solution

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so 1 revolution take = \frac{60}{15}

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and

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so 1 pulse rotate is = \frac{360}{40}

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