lObjective 6-6) The following are various management assertions (a. through m.)
related to sales and accounts receivable.
Management Assertion
a. Recorded sales transactions have occurred.
b. There are no liens or other restrictions on accounts receivable.
c. All sales transactions have been recorded.
d. Receivables are appropriately classified as to trade and other receivables in the financial statements and are clearly described.
e. Sales transactions have been recorded in the proper period.
f. Accounts receivable are recorded at the correct amounts.
g. Sales transactions have been recorded in the appropriate accounts.
h. All required disclosures about sales and receivables have been made.
i. All accounts receivable have been recorded.
j. Disclosures related to receivables are at the correct amounts.
k. Sales transactions have been recorded at the correct amounts.
1. Recorded accounts receivable exist.
m. Disclosures related to sales and receivables relate to the entity.
a. Explain the differences among management assertions about classes of transactions and events, management assertions about account balances, and management assertions about presentation and disclosure.
b. For each assertion, indicate whether it is an assertion about classes of transactions and events, an assertion about account balances, or an assertion about presentation and disclosure.
c. Indicate the name of the assertion made by management. (Hint: See Table 6-2

Answer:

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Explanation:

Note that when communicating with parents for a meeting, the most effective way to do it is to make it an in-person communication.

In-person communication is precisely what it sounds like, and it is one of the most successful methods of forming parent connections. In-person communication is used whenever you interact with a parent in person.

Communication in our daily lives helps us develop connections by allowing us to share our experiences and needs, as well as connect with others. It is the vitality of life, allowing us to express our emotions, convey information, and share our opinions. We must all communicate.

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Survey responses to the question. The regions are:

1) East: Qualitative.

2) North: Neither.

3) South: Nominal.

4) West: continuous.

Variables with no numerical value are labeled using nominal data. Male/female (albeit slightly archaic), hair color, ethnicities, names of people, etc. are typical instances.

Data is information that has been transformed into a format that is useful for transfer or processing in computing. Data is information that has been transformed into binary digital form for use with modern computers and communication mediums.

Therefore, the correct option is 1) East: Qualitative. 2) North: Neither. 3) South: Nominal, and 4) West: continuous.

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:

The probability that fewer than 3 of the 15 smartphone users use their phones in meetings or classes is 7.46%.

The binomial probability formula.

P(X = k) = [tex]^nC_r[/tex] [tex]P^k[/tex] [tex](1 - P)^{n - k}[/tex]

where n is the sample size, k is the number of successes, P is the probability of success, (P - 1) is the failure.

We have,

The required probability.

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

Where X is the number of smartphone users out of 15 who use their phones in meetings or classes.

Now,

Using the binomial probability.

P(X = k) = [tex]^nC_r[/tex] [tex]P^k[/tex] [tex](1 - P)^{n - k}[/tex]

where n is the sample size, k is the number of successes (using their smartphones in meetings or classes), p is the probability of success (49%), and (n choose k) is the number of ways to choose k successes out of n trials.

We get,

P(X = 0) = [tex]^{15}C_0[/tex] x [tex]0.49^0[/tex] x [tex]0.51^{15}[/tex] = 0.0018

P(X = 1) = [tex]^{15}C_1[/tex] x [tex]0.49^1[/tex] x [tex]0.15^{14}[/tex] = 0.0149

P(X = 2) = [tex]^{15}C_2 \times 0.49^2 \times 0.49^{13}[/tex] = 0.0579

The probability of fewer than 3 smartphone users using their phones in meetings or classes.

P(X < 3)

= P(X = 0) + P(X = 1) + P(X = 2)

= 0.0018 + 0.0149 + 0.0579

= 0.0746

= 0.0746 x 100

= 7.46%

Thus,

The probability that fewer than 3 of the 15 smartphone users use their phones in meetings or classes is 7.46%.

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a) The probability of the panel consisting of at least 3 senior auditors is of: 0.3485 = 34.85%.

b) The probability of the panel consisting of at most 2 senior auditors is of: 0.6515 = 65.15%.

The mass probability formula is presented as follows:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

The values of these parameters for this problem are given as follows:

[tex]N = 11, k = 6, n = 4[/tex]

The probability of the panel consisting of at least 3 senior auditors is given as follows:

P(X ≥ 3) = P(X = 3) + P(X = 4)

In which:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 3) = h(3,11,4,6) = \frac{C_{6,3}C_{5,1}}{C_{11,4}} = 0.303[/tex]

[tex]P(X = 4) = h(4,11,4,6) = \frac{C_{6,4}C_{5,0}}{C_{11,4}} = 0.0455[/tex]

Then the probability is of:

P(X ≥ 3) = P(X = 3) + P(X = 4) = 0.303 + 0.0455 = 0.3485 = 34.85%.

The probability of at most 2 senior auditors is of:

P(X ≥ 2) = 1 - P(X ≥ 3) = 1 - 0.3485 = 0.6515 = 65.15%.

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Answer: The average age of a vehicle registered in the United States is 8 years or 96 months. Assume the standard deviation is 16 months. If a random sample of 36 vehicles is selected, find the probability that the mean of their age is between 90 and 100 months.

---------------------------

Find the t-scores of 90 and 100.

---------------------------

t(90) = (90-96)/[16/sqrt(36)]

= -6/(16/6) = - 2.25

---------------------------

t(100) = (100-96)/[16/36)] = 4*6/16 = 1.5

----------------------------

P(90 < t-bar < 100) = P(-2.25< t < 1.5) = tcdf(-2.25,1.5,35) = 0.9133

========================

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Explanation:

A frequency polygon displays class frequencies while an ogive displays cumulative frequencies.

Procedure to find 5 number summary -

Step 1: First we need to arrange the date set in ascending order.

Step 2: Find minimum and maximum number in the data set.

Step 3: Find Median.

Median = Median is the center element (Ascending order)

Step 4: Find Q1 and Q2

Q1 is the median in the lower half of the data, and Q3 is the median in the upper half of data.

Now, consider the given numbers -

224, 203, 207, 201, 212, 225, 196, 221, 210, 195, 206, 206, 221, 191, 217

Here,

Ascending order : 191, 195, 196, 201, 203, 206, 206, 207, 210, 212, 217, 221, 221, 224, 225

Minimum : 191

Maximum : 225

First Quartile(Q1) : 201

Third Quartile(Q2) : 221

Median : 207

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Answer:

The answer is 3mm.

Explanation: