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3 years ago

A customer got debited twice, how would you investigate and resolve this conflict on the customer side and on the company side?

Engineering

1 answer:

eimsori [14]3 years ago

8 0

Answer:

First, the question seems to assume that the debit is erroneous. Because, if it's intentional, then there is little need for an investigation.

So if there was an unintentional duplication of transaction, this would totally depend on what kind of debit it is. So to answer the question, we'll explore as many scenarios as possible.

Debit on a Bank Account (Business to Bank)

If it is a debit on the bank account, then the statement of account from the company side and the account activity(ies) leading up to the debits will reveal why there is a double debit. It is possible the transaction was entered twice. This can happen when there is a bad connection that creates a failed transaction message on the customer's end but still carries through with the payment instruction. In this case, the customer will think the debit or payment request didn't go through and proceed to repeat the same transaction.

Banks do execute debits on accounts for charges such as Bank Account Maintenance Charges, Credit card charges etc. Errors could also arise internally that led to the duplication of a debit instruction either due to a bug in the system.

So in this case, we'll need to access the customer's account internally whilst externally, the client will need to provide the message containing the alerts, date and time of debits etc. to help the banks further the investigation.

Business to Business

This could arise out of a "failed POS" transaction. Similar to the condition stated above, when there is a glitch with the network and the POS fails, it should print out a receipt that reads failed or declined.

This receipt will contain references to the transaction. This will aid the investigation of the debits to the account.

Explanation:

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A sandy soil has a total unit weight of 120 pcf, a specific gravity of solids of 2.64, and a water content of 16 percent. Comput

olchik [2.2K]

Answer:

A). Dry unit weight = 1657.08Kg/m3

B). Porosity = 0.37

C). Void ratio = 0.593

D). 0.712

Explanation:

Total unit weight, Y = 120pcf =1922.2 Kg/m3

Specific gravity of solids, Gs = 2.64

Water content, w = 16%

A). Dry unit weight

Yd = Y/(1+w)

= 1922.2/(1+0.16) = 1657.08Kg/m3

B). Porosity

However void ratio, e = Gs×Yw/Yd, where Yw = 1000Kg/m3

Void ratio = 2.64×1000/1657.08 = 0.593

And porosity = e/(1+e) =0.593/(1+0.593) = 0.37

C). void ratio, e = 0.593

D). Degree of saturation, S = m×Gs/e where m =water content

S = 0.16×2.64/0.593 = 0.712

5 0

3 years ago

Consider steady heat transfer through the wall of a room in winter. The convection heat transfer coefficient at the outer surfac

AveGali [126]

The heat transfer which is in steady state, the heat transfer rate to the wall is equal to the wall.

<u>Explanation:</u>

The convection transfer of heat to the wall is

$Q=h A\left(T_{s}-T_{f}\right)$

Here, $T_{S}$ is the temperature of solid surface, $T_{f}$ is the temperature of moving fluid stream which is adjacent of solid surface, h is the heat transfer coefficient.
The coefficient of convection heat transfers outer surface contains 3 times to the inner surface which experience smaller drop of temperature for 3 times that compares to inner surface.
Hence, the temperatures outer surface get close to the surroundings of air temperature.

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3 years ago

I can't log in even though it says all ages are accepted.Instead when I TRY logging in it says this. "We're sorry, but we are no

uysha [10]

Answer:

Explanation:

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8 0

3 years ago

Can anyone answer this question

Solnce55 [7]

Answer:

a) 2.18 m/s^2

b) 9.83 m/s

Explanation:

The flywheel has a moment of inertia

J = m * k^2

Where

J: moment of inertia

k: radius of gyration

In this case:

J = 144 * 0.45^2 = 29.2 kg*m^2

The block is attached through a wire that is wrapped around the wheel. The weight of the block causes a torque.

T = p * r

r is the radius of the wheel.

T = m1 * g * r

T = 18 * 9.81 * 0.6 = 106 N*m

The torque will cause an acceleration on the flywheel:

T = J * γ

γ = T/J

γ = 106/29.2 = 3.63 rad/s^2

SInce the block is attached to the wheel the acceleration of the block is the same as the tangential acceleration at the eddge of the wheel:

at = γ * r

at = 3.63 * 0.6 = 2.81 m/s^2

Now that we know the acceleration of the block we can forget about the flywheel.

The equation for uniformly accelerated movement is:

X(t) = X0 + V0*t + 1/2*a*t^2

We can set a frame of reference that has X0 = 0, V0 = 0 and the X axis points in the direction the block will move. Then:

X(t) = 1/2*a*t^2

Rearranging

t^2 = 2*X(t)/a

$t = \sqrt{\frac{2*X(t)}{a}}$

$t = \sqrt{\frac{2*18}{2.81}} = 3.6 s$

It will reach the 1.8 m in 3.6 s.

Now we use the equation for speed under constant acceleration:

V(t) = V0 + a*t

V(3.6) = 2.81 * 3.6 = 9.83 m/s

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4 years ago

The technique of smoothing out joint compound on either side of a joint is known as which of the following

Black_prince [1.1K]

Answer:

a is the answer to your questions

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3 years ago