Answer:
Number of fragments is 3
Explanation:
The maximum size of data field in each fragment = 4468 - 20(IP Header)
= 4448 bytes
Hence, the number of required fragment = (10000 - 20)/4448
= 3
Fragment 1
Id = 218
offset = 0
total length = 4468 bytes
flag = 1
Fragment 2
Id = 218
offset = 556
total length = 4468 bytes
flag = 1
Fragment 3
Id = 218
offset = 1112
total length = 1144 bytes
flag = 0
Answer:
<em>communications</em><em>,</em><em>risk</em><em>,</em><em>systems</em><em>,</em><em>test</em><em>ins</em><em>,</em><em>procedures</em><em>,</em><em>interviews</em><em>,</em><em>documents</em><em>.</em>
Explanation:
The auditor considers many factors in determining the nature, timing, and extent of auditing procedures to be performed in an audit of an entity's financial statements. One of the factors is the existence of an internal audit function. fn 1 This section provides the auditor with guidance on considering the work of internal auditors and on using internal auditors to provide direct assistance to the auditor in an audit performed in accordance with generally accepted auditing standards.Note: When performing an integrated audit of financial statements and internal control over financial reporting, refer to paragraphs 16-19 of PCAOB Auditing Standard No. 5, An Audit of Internal Control Over Financial Reporting That Is Integrated with An Audit of Financial Statements, for discussion on using the work of others to alter the nature, timing, and extent of the work that otherwise would have been performed to test controls.
Roles of the Auditor and the Internal Auditors
<em>02 </em>
<h2>
<em>I</em><em> </em><em>HOPE</em><em> </em><em>THIS</em><em> </em><em>HELPS</em><em> </em><em>ALOT</em><em>!</em><em> </em><em>:</em><em>3</em></h2>
C) 180 − (140 − 7a) = (70 − 3a)
Answer:
C) 180 − (140 − 7a) = (70 − 3a)
Explanation:
i got it wrong by clicking D on usatestprep