A frequency polygon displays class frequencies while an ogive displays cumulative frequencies.
Is an ogive and a frequency polygon the same?A frequency polygon is a line graph while an orgive is a histogram . A frequency polygon displays class frequencies while an orgive displays cumulative frequencies there is no difference between a frequency polygon and ogive.The area under the frequency polygon is the same as the area under the histogram and is, therefore, equal to the frequency values that would be displayed in a distribution table. The frequency polygon also shows the shape of the distribution of the data, and in this case, it resembles a bell curve.The frequency polygon is a curve that is drawn on the x-axis and the y-axis. The x-axis represents the values in the dataset, while the y-axis shows the number of occurrences of each distinct category.An Ogive Chart is a curve of the cumulative frequency distribution or cumulative relative frequency distribution. For drawing such a curve, the frequencies must be expressed as a percentage of the total frequency.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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When communicating with parents for a meeting which is the most effective way to do it
Note that when communicating with parents for a meeting, the most effective way to do it is to make it an in-person communication.
What is in-person communication and why is it important?
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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Assume that when adults with smartphones are randomly selected, 49% use them in meetings or classes. If 15 adult smartphone users are randomly selected find the probability that fewer than 3 of them use their smartphones in meetings or classes.
The probability that fewer than 3 of the 15 smartphone users use their phones in meetings or classes is 7.46%.
What is the binomial probability theorem?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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6. The training will help me ask my supervisor when I am unsure of how to do a task safely.
(1 point)
A. Strongly Agree
B. Agree
C. Neutral
D. Disagree
E. Strongly Disagree
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:
m. Disclosures related to sales and receivables relate to the entity. a. Explain the differences among management assertions about classes of transactions and ...
Explanation:
Write your answer in simplest form.
Be sure to include the correct unit in your answer.
Answer:
The answer is 3mm.
Explanation:
how to develop an e portfolio
Answer:
1. Decide on the purpose of your e portfolio. Ask yourself why you are creating the e portfolio. Consider what you want to communicate and the audience you are targeting.
2. Gather the content you want to include in your e portfolio. This could include resumes, work samples, awards, certifications, and other relevant documents.
3. Choose a platform for your e portfolio. There are many online tools that can help you create an e portfolio, such as WordPress, Weebly, Wix, and Squarespace.
4. Design your e portfolio. Consider the overall look and feel that you want to achieve. Think about the colors, fonts, and images you want to use.
5. Upload your content to your e portfolio. Make sure to include descriptions and captions for each item.
6. Test your e portfolio. Check for any errors or typos and make sure that all links are working properly.
7. Publish your e portfolio. Once you are satisfied with the final product, you can share it with your intended audience.
Explanation:
Classify the following data. Indicate whether the data is qualitative or quantitative, indicate whether the data is discrete, continuous, or neither, and indicate the level of measurement for the data.
Survey responses to the question "In which of the following regions of the country do you live?"
1) East 2) North 3) South 4) West
Survey responses to the question. The regions are:
1) East: Qualitative.
2) North: Neither.
3) South: Nominal.
4) West: continuous.
What is nominal data?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 average age of a vehicle registerd in the united state in 8 year or 96 month . assume the standard deviation is 16 month .if random sample 36 month is selected .find the probabilty that the mean of their age a. betwen 90 and 100
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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Muhammad Zayaan Yasir
Explanation:
A government audit panel of 4 professionals is to be constituted from 6 senior auditors and 5 chartered accountants. Selection is random. (a) Determine the probability of the panel consisting of at least 3 senior auditors; (b) What is the probability of the panel consisting of at most 2 senior auditors.?
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%.
What is the hypergeometric distribution formula?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:
x is the number of successes.N is the size of the population.n is the size of the sample.k is the total number of desired outcomes.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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