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3 years ago

Which ics function is responsible for documentation for mutual aid agreements?

Business

2 answers:

Softa [21]3 years ago

6 0

Planning

The Incident Command System is a management construct which functions to help resource management during incidents. It involves Mutual-Aid Agreement which is the document between agencies in writing.

The Incident Command System- Planning Section is tasked to collect, evaluate, disseminate information and prepares and documents Incident Action Plans. Documentation is one of the main responsibilities of the Planning Section.

Nataly_w [17]3 years ago

3 0

Answer:

planning is responsible for documentation for mutual aid agreements?

Explanation:

Planning can be described as the process of thinking about such activities that are required to achieve a goal.

planning is the first step to achieve desired results.

planning involves the creation of a plan, such as psychological aspects that require conceptual skills.

planning is also responsible for documentation for mutual aid agreement as without planning we cannot make a successful or meaningful agreement

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laiz [17]

Answer:

40%

Explanation:

Contribution margin = Contribution ÷ Sales × 100

= 72,000 ÷ $180,000 × 100

= 0.4 × 100

= 40%

Please not that other information given in the question are not relevant in arriving at the contribution margin ratio hence will be ignored.

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4 years ago

On Mar 3, Lyons Company paid dividends of $1,000. Use your knowledge of what a correct journal entry should look like to identif

uysha [10]

Answer:

Debit : Dividend $1,000

Credit : Cash $1,000

Explanation:

The Journal entry to record dividend payment include a Debit to Dividend Account and a Credit to Cash Account to depict the outflow of cash.

8 0

3 years ago

Sosa Company reported net income of $190,000 for 2017. Sosa Company also reported depreciation expense of $35,000 and a loss of

mart [117]

Answer:

$228,000

Explanation:

Preparation of the operating activities section of the statement of cash flows for 2017 for Sosa Company

Sosa Company operating activities section of the statement of cash flows for 2017

Net income $190,000

Add:Depreciation expenses $35,000

Loss on disposal of plant assets $5,000

Increase in accounts payable $17,000

Less: Increase in accounts receivable($15,000)

Increase in prepaid expenses ($4,000)

Net cash flow of the operating activities $228,000

Therefore the operating activities section of the statement of cash flows for 2017 for Sosa Company will be $228,000

7 0

3 years ago

Xie Company identified the following activities, costs, and activity drivers for this year. The company manufactures two types o

san4es73 [151]

Answer:

Results are below.

Explanation:

First, we need to calculate the plantwide predetermine manufacturing overhead rate:

Predetermined manufacturing overhead rate= total estimated overhead costs for the period/ total amount of allocation base

total estimated overhead costs for the period= (625,000 + 900,000 + 105,000 + 175,000 + 300,000 + 75,000)

total estimated overhead costs for the period= $2,180,000

Predetermined manufacturing overhead rate= 2,180,000 / 125,000

Predetermined manufacturing overhead rate= $17.44 per direct labor hour

Now, we can allocate overhead to each product line:

Allocated MOH= Estimated manufacturing overhead rate* Actual amount of allocation base

Deluxe:

Allocated MOH= 17.44*2,500

Allocated MOH= $43,600

Basic:

Allocated MOH= 17.44*6,000

Allocated MOH= $104,640

6 0

3 years ago

Find the mean, median and mode for the above set of data

qwelly [4]

Answer:

a. Mean = 35.2 ≈ 35

b. Median = 35.6 ≈ 36

c. Mode = 36.6 ≈ 37

Step-by-step Explanation:

==>Given:

Class of ages in yrs

No. of cases of each class = f

Midpoint of each class = x

Product of midpoint and no. of cases of each class = fx

==>Required:

a. Mean

b. Median

c. Mode

==>SOLUTION:

a. Mean = (Σfx)/Σf

Σf = sum of no. of cases = 5+10+20+22+13+5 = 75

Σfx = 47.5+195+590+869+643.5+297.5 = 2,642.5

Mean = 2,642.5/75 = 35.2 ≈ 35

b. Median = Lm + [((Σf/2) - Cfb)/fm]Cw

Our median is between the 37th and the 38th term, which can be found in the class interval 35-44. This is our median class.

Lm = Lower class boundary of the median class = lower limit of the Medina class + upper limit of the class before the median class ÷ 2 = (35+34)/2 = 34.5

Σf/2 = 75/2 = 37.5

Cfb = Cumulative frequency of class before the median class = 5+10+20 = 35

fm = frequency of the Medina class = 22

Cw = Class width = upper class boundary - lower class boundary = 44.5-34.5 = 10

Median = 34.5 + [(37.5-35)/22] × 10

= 34.5 + [2.5/22] × 10

= 34.5 + [25/22]

= 34.5 + 1.1

= 35.6 ≈ 36

c. Mode = Lm + [∆¹/(∆¹+∆²)]Cw

Modal class = (35-44) [i.e. the class with the highest frequency, which is where our mode falls in]

Lm = lower class boundary of the modal class = lower limit of the modal class + upper limit of the class before the modal class ÷ 2 = (35+34)/2 = 34.5

∆¹ = difference between the frequency of the modal class & the frequency of the class before the modal class = 22 - 20 = 2

∆² = difference between the frequency of the modal class & the frequency of the class after the modal class = 22 - 13 = 9

Cw == Upper class boundary - Lower class boundary = 44.5 - 34.5 = 10

Mode = 34.5 + [2/(2+9)] × 10

= 34.5 + [2/11] × 10

= 34.5 + [20/11]

= 34.5 + 1.8

Mode = 36.6 ≈ 37

8 0

3 years ago