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jok3333 [9.3K]
2 years ago
4

You are reviewing the Client Overview tab for a new client to determine the scope of a bookkeeping clean-up engagement. You noti

ce that there is a large amount in the Opening Balance Equity account.

Computers and Technology
2 answers:
kicyunya [14]2 years ago
7 0

Answer:

The scope of a bookkeeping clean-up engagement include the following:

Quick books automatically records the following transactions to the opening equity account.

Opening balances entered when new customers or vendors are setup.

The inventory total balances entered in the new item dialogue.

The three reasons causing or affecting the balance includes the following:

An opening balance was entered by the client when creating a new other current asset account.

The balances to one or more product/service items were entered during the setup process.

The balance opening were included when importing customers, by using a tool called the import data tool.

Explanation:

Solution:

  • Quick books automatically records the following transactions to the opening equity account.
  • The bank statement ending balance transaction when a new bank account is created.
  • Opening balances to other balance sheet accounts created in the add new account dialogue box.
  • Opening balances entered when new customers or vendors are setup or processed.
  • The inventory total balances entered in the new item dialogue.
  • Bank reconciliation adjustment for quick book versions 2005 alike.

The Three reasons causing or resulting in this balance are:

<em>Options</em>

  • Opening balances to one or more product/service items were entered during the setup process.
  • An opening balance was entered by the client when creating a new other current  asset account.
  • Opening balance were included when importing customers, by using a tool called the import data tool.

Note: Kindly find an attached copy of the complete question to this solution below.

Guest1 year ago
0 0

first and last one

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Answer:

Explanation:

Let us first consider the procedure procA; the caller in the example given.

  Some results: $s0,$s1,$s2, $t0,$t1 and $t2 are being stored by procA. Out of these registers, few registers are accessing by procA after a call to procB. But, procB might over-write these registers.

       Thus, procA need to save some registers into stack first before calling procB, .

      only $s1,$t0 and $t1 are being used after return from procB in the given example,

       Caller saves and restores only values in $t0-$t9, according to MIPS guidelines for caller-saved and callee-saved registers, .

       Thus, procA needs to save only $t0 and $t1.

    jal instruction overwrites the register $ra by writing the address, to which the control should jump back, after completing the instructions of procB, when procB is called,.

       Therefore, procA also need to save $ra into stack.

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   Also, procA needs to save frame pointer, which points the start of the stack space for each procedure.

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Let us consider the procedure procB; the callee in the given example.

 The callee is responsible for saving values in $s0-$s7 and restoring them before returning to caller, this is according to MIPS guidelines for caller-saved and callee-saved registers,

   procB is expected to over-write the registers $s2 and $t0. Nonetheless, in the first two lines, procB might over-write the registers $s0 and $s1.

   Thus, procB is responsible for saving and restoring $s0,$s1 and $s2.

X:

We need to create space for 5 values on the stack since procA needs to save $a0,$ra,$t0,$t1 and $fp(frame pointer), . Each value(word) takes 4 bytes.

$sp = $sp – 20 # on the stack, create space for 5 values

sw $a0, 16($sp) # store the result in $a0 into the memory address

               # indicated by $sp+20

sw $ra, 12($sp) # save the second value on stack

sw $t0, 8($sp) # save the third value on stack

sw $t1, 4($sp) # save the fourth value on stack

sw $fp, 0($sp) #  To the stack pointer, save the frame pointer

$fp = $sp      #  To the stack pointer, set the frame pointer

Y:

lw $fp, 0($sp) #  from stack, start restoring values

lw $t1, 4($sp)

lw $t0, 8($sp)

lw $ra, 12($sp)

lw $a0, 16($sp)

$sp = $sp + 20 # decrease the size of the stack

P:

$sp = $sp – 12 #  for three values, create space on the stack

sw $s0, 0($sp) # save the value in $s0

sw $s1, 0($sp) # save the value in $s1

sw $s2, 0($sp) # save the value in $s2

Q:

lw $s0, 0($sp) #  from the stack, restore the value of $s0

lw $s1, 0($sp) #  from the stack, restore the value of $s1

lw $s2, 0($sp) #  from the stack, restore the value of $s2

$sp = $sp + 12 # decrease the stack size

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The population of town A is less than the population of town B. However, the population of town A is growing faster than the pop
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Answer:

#include<iostream>

using namespace std;

void main()

{

int townA_pop,townB_pop,count_years=1;

double rateA,rateB;

cout<<"please enter the population of town A"<<endl;

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cout<<"please enter the grothw rate of town B"<<endl;

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Explanation:

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2.32 LAB: Musical note frequencies On a piano, a key has a frequency, say f0. Each higher key (black or white) has a frequency o
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Answer:

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using namespace std;

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for(int n = 1; n<=5;n++)

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Explanation:

The programming language is not stated; However, one can easily deduce that the program is to be written using C++

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using namespace std;

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<em>for(int n = 1; n<=5;n++) </em>

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<em> }  </em>

return 0;

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