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stiks02 [169]
3 years ago
7

Wright Company deposits all cash receipts on the day when they are received and it makes all cash payments by check. At the clos

e of business on May 31, Its Cash account shows a $27,900 debit balance. The company's May 31 bank statement shows $26,200 on deposit in the bank. a. The May 31 bank statement lists $120 in bank service charges; the company has not yet recorded the cost of these services b. Outstanding checks as of May 31 total $5,800. c. May 31 cash receipts of $6,400 were placed in the bank's night depository after banking hours and were not recorded on the May 31 bank statement. d. In reviewing the bank statement, a $420 check written by Smith Company was mistakenly drawn against Wright's account. e. The bank statement shows a $560 NSF check from a customer; the company has not yet recorded this NSF check. Prepare a bank reconciliation for the company using the above information.
Engineering
1 answer:
alina1380 [7]3 years ago
8 0

Answer:

                                              Wright Company

                                             Bank Reconciliation

                                                 May 31, 2013

Credit side                                                                                   Debit side

Bank statement $26200                 |                          Book balance $27900

<em>Add;                                                    </em>

Deposit on May 31 $6400

Bank error $420

Sub-total=$33020

Deductions;                                        |                       Deduct

ions

Outstanding checks $5800              |                 Bank service charge $120

Adjusted bank balance $27220       |                  NSF check $560

                                                                             Total deduction $680

                                                      Adjusted book balance $27220

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Air (ideal gas) is contained in a cylinder/piston assembly at a pressure of 150 kPa and a temperature of 127°C. Assume that the
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Answer:

The process is not possible.

Explanation:

We know for ideal condition, the work done for isothermal process is

W_{ideal} = P_{1}.V_{1} ln\frac{V_{2}}{V_{1}}

and for ideal gas, we know  PV = mRT

Therefore, W_{ideal} = mRTln\frac{V_{2}}{V_{1}}

                                                  = mRTln\frac{P_{1}}{P_{2}}

                                                  =  0.287 x 400ln\frac{150}{450}

                                                  = -126.12 kJ/kg (negative sign indicates that the process is compressive, so work input to the compressor is 126.12 kJ/kg )

Now we know for adiabatic compression process

                    PV^{\gamma } = C

We know \frac{T_{2}}{T_{1}}=(\frac{P_{2}}{P_{1}})^{\frac{\gamma -1}{\gamma }}

T_{2} = 556 K

For adiabatic work done, W_{adiabatic} = \frac{P_{1}\times V_{1}-P_{2}\times V_{2}}{\gamma -1}

                                                                       = \frac{mR(T_{1}-T_{2})}{\gamma -1}

                                                                       = \frac{0.287(400-556)}{1.45 -1}

                                                                       = -99.49 kJ/kg (negative sign indicates that the process is compressive, so work input to the compressor is 99.49 kJ/kg )

We know that in isothermal process, work input to the compressor is minimum. But in the above adiabatic polytropic process, work input to the compressor is less than the work done in the isothermal process.

Thus the process is not possible.

                                                             

7 0
4 years ago
Technician A says that wheel lug nuts can be loosened when the vehicle is off the ground with an impact
Pavlova-9 [17]

Answer:

technician a is correct, if you were to torque down a wheel with an impact then you could risk damaging the wheel and over tightening the lugs

Explanation:

6 0
2 years ago
1. Two aluminium strips and a steel strip are to be bonded together to form a composite bar. The modulus of elasticity of steel
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Answer:

1.933 KN-M

Explanation:

<u>Determine the largest permissible bending moment when the composite bar is bent  horizontally </u>

Given data :

modulus of elasticity of steel = 200 GPa

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Allowable stress for steel = 220 MPa

Allowable stress for Aluminum = 100 MPa

a = 10 mm

<em>First step </em>

determine moment of resistance when steel reaches its max permissible stress

<em>next </em>: determine moment of resistance when Aluminum reaches its max permissible stress

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<em>attached below is a detailed solution </em>

5 0
3 years ago
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Angelina_Jolie [31]

Answer:

Oil is graded by the Society of Automotive Engineers - B.

5 0
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1. Implement the k-means clustering algorithm either in Java or Python. • The program should be executable with at least 3 param
givi [52]

Answer:

The code for this Question in Python is as follows:

matplotlib inline

from copy import deepcopy

import numpy as np

import pandas as pd

from matplotlib import pyplot as plt

plt.rcParams['figure.figsize'] = (16, 9)

plt.style.use('ggplot')

# Importing the dataset

data = pd.read_csv('xclara.csv')

print(data.shape)

data.head()

# Getting the values and plotting it

f1 = data['V1'].values

f2 = data['V2'].values

X = np.array(list(zip(f1, f2)))

plt.scatter(f1, f2, c='black', s=7)

# Number of clusters

k = 3

# X coordinates of random centroids

C_x = np.random.randint(0, np.max(X)-20, size=k)

# Y coordinates of random centroids

C_y = np.random.randint(0, np.max(X)-20, size=k)

C = np.array(list(zip(C_x, C_y)), dtype=np.float32)

print(C)

# To store the value of centroids when it updates

C_old = np.zeros(C.shape)

# Cluster Lables(0, 1, 2)

clusters = np.zeros(len(X))

# Error func. - Distance between new centroids and old centroids

error = dist(C, C_old, None)

# Loop will run till the error becomes zero

while error != 0:

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   for i in range(len(X)):

       distances = dist(X[i], C)

       cluster = np.argmin(distances)

       clusters[i] = cluster

   # Storing the old centroid values

   C_old = deepcopy(C)

   # Finding the new centroids by taking the average value

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       points = [X[j] for j in range(len(X)) if clusters[j] == i]

       C[i] = np.mean(points, axis=0)

   error = dist(C, C_old, None)

# Initializing KMeans

kmeans = KMeans(n_clusters=4)

# Fitting with inputs

kmeans = kmeans.fit(X)

# Predicting the clusters

labels = kmeans.predict(X)

# Getting the cluster centers

C = kmeans.cluster_centers_

fig = plt.figure()

ax = Axes3D(fig)

ax.scatter(X[:, 0], X[:, 1], X[:, 2], c=y)

ax.scatter(C[:, 0], C[:, 1], C[:, 2], marker='*', c='#050505', s=1000)

4 0
4 years ago
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