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belka [17]
3 years ago
6

It has been observed that electrical connectors manufactured by Jolt Electrical Supply Company last an average of 18.2 months an

d follow a normal distribution with a standard deviation of 1.7 months. Jolt agrees to replace any connector that fails within 19 months. Out of 500 connectors sold, how many does Jolt expect to replace, on average?
Physics
1 answer:
Viktor [21]3 years ago
7 0

Answer:

Jolt can expect to replace 160 units.

Explanation:

given data:

standard deviation 1.7

average lfe of connectors = 18.2 months

number of connectors = 5000

z(19) = \frac{(19-18.2)}{1.7}

z(19)  = 0.4706

P(x< 19) = P(z< 0.4706)

P(x< 19) = 0.6810   from z table

therefore 68.1% fails within 19 months.

hence 0.3190   {100-68.1} will fail during that period.

number of unit of Jolt that need to be replace, on average is  0.319*500 = 160 units.

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Explanation:

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Using parallel axis theorem, the moment of inertia of each rod is:

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The total moment of inertia is:

I = 2Idisk + 5Irod

I = 2 (1/2 MR²) + 5 [1/2 mr² + m (R − r)²]

I = MR² + 5/2 mr² + 5m (R − r)²

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I = 23,750 g cm²

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Explanation:

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Solve this system of equations without graphing and show your reasoning <br><br> 5x+y=7<br> 20x+2=y
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Answer:

Explanation:

5x + y = 7

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Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stoc
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Answer:

ΔR_{e} = 84   Ω,     R_{e} = (40 ± 8) 10¹   Ω

Explanation:

The formula for parallel equivalent resistance is

          1 / R_{e} = ∑ 1 / Ri

In our case we use a resistance of each

           R₁ = 500 ± 50  Ω

          R₂ = 2000 ± 5%

This percentage equals

        0.05 = ΔR₂ / R₂

        ΔR₂ = 0.05 R₂

        ΔR₂ = 0.05 2000 = 100   Ω

We write the resistance

        R₂ = 2000 ± 100    Ω

We apply the initial formula

        1 / R_{e} = 1 / R₁ + 1 / R₂

        1 / R_{e} = 1/500 + 1/2000 = 0.0025

        R_{e}  = 400    Ω

Let's look for the error  (uncertainly) of Re

      R_{e} = R₁R₂ / (R₁ + R₂)

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       R_{e} = R₁R₂ / R’

Let's look for the uncertainty of this equation

      ΔR_{e} / R_{e} = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

      ΔR’= ΔR₁ + ΔR₂

We substitute the values

     ΔR_{e} / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

     ΔR_{e} / 400 = 0.1 + 0.05 + 0.06

     ΔR_{e} = 0.21 400

     ΔR_{e} = 84   Ω

Let's write the resistance value with the correct significant figures

    R_{e} = (40 ± 8) 10¹   Ω

6 0
3 years ago
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