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muminat
3 years ago
12

In Process Costing, the journal entry to record the cost of goods transferred out from the WIP shaping department to the WIP pac

kaging department would be a debit to WIP - shaping department and a credit to WIP - packaging department. True False
Engineering
1 answer:
ddd [48]3 years ago
4 0

Answer:

Journal entry :

Date Account and explanation Debit credit

                WIP-Packing department XXX  

                WIP-Shaping department            XXX

(To record cost of goods transferred out from the WIP shaping department to the WIP packaging department )  

So above statement is False

Explanation:

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D

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Masonry uses stone work, making a stone wall requires perfect masonry.

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What does Clay say will happen if the system is rejected?
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the nation will suffer terrible consequences

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I did that and got it right

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A blown fuse or tripped circuit breaker shows
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Your house is in need of a service upgrade, or it may indicate that your house has too few circuits.

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It's a sign that you are making excessive demands on the circuit and need to move some appliances and devices to other circuits.

6 0
3 years ago
A 75 ! coaxial transmission line has a length of 2.0 cm and is terminated with a load impedance of 37.5 j75 !. If the relative p
Cerrena [4.2K]

Answer:

4.26

Explanation:

The wavelength λ is given by:

\lambda=v/f=c/nf\\c=speed\ of\ light=3*10^8m/s,f=frequency=3*10^9Hz,n=permittivity=2.56\\\\\lambda=3*10^8/(2.56*3*10^9)=0.0625\ m\\

Phase constant (β) = 2π/λ

βl = 2π/λ × l

l = 2 cm = 0.02 m

βl = 2π/0.0625 × 0.02=2.01 rad = 115.3°

1 rad = 180/π degrees

Z_L=load\ impedance=37.5+j75\\\\Z_o=characteristic impedance = 75\ ohm\\\\\tilde {Z_L}=Z_L/Z_o=37.5+j75/75=0.5+j

\tilde {Z_{in}}=\frac{\tilde {Z_{L}}+jtan\beta l}{1+j\tilde {Z_{L}}tan\beta l}=\frac{0.5+j+jtan(115.2)}{1+j(0.5+j)tan(115.2)}=0.253-j0.274\\  \\Z_{in}=Z_o\tilde {Z_{in}}=75(0.253-j0.274)=19-j20.5\\\\\Gamma_L=\frac{Z_L-Z_0}{Z_L+Z_o}=\frac{37.5+j75-75}{37.5+j75+75}=0.62\angle 83^o\\\\\Gamma_{in}=\frac{Z_{in}-Z_0}{Z_{in}+Z_o}=\frac{19-j20.5-75}{19-j20.5+75}=0.62\angle -147^o\\\\VSWR=\frac{1+\rho}{1-\rho} =\frac{1+0.62}{1-0.62}=4.26

6 0
3 years ago
Past evidence shows that when a customer complains of an out-of-orderphone there is an 8% chance that the problem is with the in
Alexxandr [17]

Answer:

a. The expected number of failures due to a problem with inside wiring is 8 failures

b. The probability that at least 10 failures are due to inside wiring is approximately 0.176

c. It will be not unusual

Explanation:

The probability that the problem of an out of order is the inside wiring, P(x) = 8%

The number of complaints in a month period, x = 100

a. The expected number of failures due to a problem with inside wiring, E(x) = x·P(x)

∴ E(x) = 100 × 8% = 8

The expected number of failures due to a problem with inside wiring, E(x) = 8 failures

b. The probability that at least 10, P₁₀, failures are due to inside wiring is given as follows;

The probability of success, P = 0.08, therefore, the probability of failure, q = 1 - 0.08 = 0.92

P = _nC_r·P^r·q^{n-r}

P₀ = ₁₀₀C₀·(0.08)⁰·(0.92)¹⁰⁰ = 0.000239211874657

P₁ = ₁₀₀C₁·(0.08)¹·(0.92)⁹⁹ = 0.00208010325

P₂ = ₁₀₀C₂·(0.08)²·(0.92)⁹⁸ = 0.00895348793

P₃ = ₁₀₀C₃·(0.08)³·(0.92)⁹⁷ = 0.02543309616

P₄ = ₁₀₀C₄·(0.08)⁴·(0.92)⁹⁶ = 0.0536306593

P₅ = ₁₀₀C₅·(0.08)⁵·(0.92)⁹⁵ = 0.0895398833653

P₆ = ₁₀₀C₆·(0.08)⁶·(0.92)⁹⁴ = 0.123279549561

P₇ = ₁₀₀C₃·(0.08)⁷·(0.92)⁹³ = 0.143953759735

P₈ = ₁₀₀C₈·(0.08)⁸·(0.92)⁹² = 0.145518474516

P₉ = ₁₀₀C₉·(0.08)⁹·(0.92)⁹¹ = 0.129349755125

P₁₀ = ₁₀₀C₁₀·(0.08)¹⁰·(0.92)⁹⁰ = 0.10235502362

∴ P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀ = 0.82433004464

The probability of at least 10 failures are due problem with the inside wiring = 1 - (P₀ + P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + P₈ + P₉ + P₁₀) = 1 - 0.82433004464 = 0.175666995536

The probability of at least 10 failures are due problem with the inside wiring = 0.175666995536 ≈ 0.176

c. The probability of at most 5 failures are due problem with the inside wiring = P₀ + P₁ + P₂ + P₃ + P₄ + P₅ = 0.179876441908

Therefore, given that probability of at most 5 failures > The probability of 8 failures it will be not unusual since the cause of failure is more (92%) due to other causes which are more likely and therefore increase in the probability that there are fewer failures due inside wiring

6 0
3 years ago
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