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Westkost
11 days ago
11

In a car lot, the ratio of the number of new cars to the number of pre owned cars is 6 to 5. The total number of new and preowne

d cars on the lot is 66. If 4 new cars and 2 preowned cars are sold and are removed from the lot, what fraction of the remaining cars on the lot are preowned
Mathematics
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Suppose that the last four months of sales were 8, 10, 15, and 9 units, respectively. Suppose further that the last four forecas
Leona [12618]

Response:

MAD value comes out to be 3.

Detailed Breakdown:

The given sales forecasts for the last four months are 5, 6, 11, and 12 units.

To calculate the Mean Absolute Deviation (MAD) for these forecasts:

The average of the forecasts across four months is \frac{5 + 6 + 11 + 12}{4} = 8.5.

Thus, the total of absolute differences between the forecast values and the average is = |5 - 8.5| + |6 - 8.5| + |11 - 8.5| + |12 - 8.5| = 3.5 + 2.5 + 2.5 + 3.5 = 12.

Hence, the MAD value will be = \frac{12}{4} = 3 (Final Answer)

4 0
2 months ago
To better understand how husbands and wives feel about their finances, Money Magazine conducted a national poll of 1010 married
Svet_ta [12734]

Answer:

  • a. Refer to the table below
  • b. Refer to the table below
  • c. 0.548
  • d. 0.576
  • e. 0.534
  • f) i) 0.201, ii) 0.208

Explanation:

To begin with, organize the data provided:

Table: "Who excels at obtaining deals?"

                       Who Excels?

Respondent      I Am        My Spouse     We are Equal

Husband           278             127                 102

Wife                   290            111                   102

a. Create a joint probability table and utilize it to respond to the ensuing inquiries.

The joint probability table presents identical details expressed as proportions. The values from the table need to be divided by the total number of responses involved.

1. Total responses: 278 + 127 + 102 + 290 + 111 + 102 = 1,010.

2. Determine each proportion:

  • 278/1,010 = 0.275
  • 127/1,010 = 0.126
  • 102/1,010 = 0.101
  • 290/1,010 = 0.287
  • 111/1,010 = 0.110
  • 102/1,010 = 0.101

3. Construct the table containing these values:

Joint probability table:

Respondent      I Am        My Spouse     We Are Equal

Husband           0.275           0.126                 0.101

Wife                   0.287           0.110                  0.101

This table illustrates that the joint probability of identifying as a husband while choosing 'I am' equals 0.275. Each cell conveys the joint probability associated with each gender's response.

Consequently, this delineates the purpose of a joint probability table.

b. Generate marginal probabilities for Who Excels (I Am, My Spouse, We Are Equal). Provide commentary.

Marginal probabilities are computed for each row and column of the table, indicated in the margins, which is their namesake.

For the column titled "I am," it amounts to: 0.275 + 0.287 = 0.562

Similarly, perform calculations for the other two columns.

For the row designated 'Husband,' it would thus be 0.275 + 0.126 + 0.101 = 0.502. Apply the same for the row labeled 'Wife.'

Table Marginal probabilities:

Respondent      I Am        My Spouse     We Are Equal     Total

Husband           0.275           0.126                 0.101             0.502

Wife                   0.287           0.110              0.101             0.498

Total                 0.562           0.236            0.202             1.000

Notably, when summing the marginal probabilities for both rows and columns, the results will always equate to 1. This is a consistent truth for marginal probabilities.

c. Given the respondent is a husband, what is the likelihood that he believes he is better at securing deals than his wife?

This requires the utilization of conditional probability.

The goal here is to ascertain the probability of the response being "I am" when the respondent identifies as a "Husband."

Using conditional probability:

  • P ( "I am" / "Husband") = P ("I am" ∩ "Husband) / P("Husband")

  • P ("I am" ∩ "Husband) = 0.275 (obtained from the intersection of columns "I am" and rows "Husband")

  • P("Husband") = 0.502 (derived from total of row "Husband")

  • P ("I am" ∩ "Husband) / P("Husband") = 0.275 / 0.502 = 0.548

d. In the instance that the respondent is a wife, what probability exists that she believes she is superior to her husband in acquiring deals?

We seek to identify the probability wherein the response claims "I am" while the respondent is labeled a "Wife," applying the conditional probability formula again:

  • P ("I am" / "Wife") = P ("I am" ∩ "Wife") / P ("Wife")

  • P ("I am" / "Wife") = 0.287 / 0.498

  • P ("I am" / "Wife") = 0.576

e. When responding that "My spouse" is better at scoring deals, what is the likelihood that the claim originated from a husband?

We aim to compute: P ("Husband" / "My spouse")

Applying the conditional probability formula:

  • P("Husband" / "My spouse") = P("Husband" ∩ "My spouse")/P("My spouse")

  • P("Husband" / "My spouse") = 0.126/0.236

  • P("Husband" / "My spouse") = 0.534

f. When the response indicates "We are equal," what likelihood exists that this response is from a husband? What is the chance that it hails from a wife?

What is the likelihood that this response came from a husband?

  • P("Husband" / "We are equal") = P("Husband" ∩ "We are equal") / P ("We are equal")

  • P("Husband" / "We are equal") = 0.101 / 0.502 = 0.201

What is the chance the response originated from a wife:

  • P("Wife") / "We are equal") = P("Wife" ∩ "We are equal") / P("We are equal")

  • P("Wife") / "We are equal") = 0.101 / 0.498 = 0.208
6 0
3 months ago
for the level 3 course, examination hours cost twice as much as workshop hours and workshop hours cost twice as much as lecture
Leona [12618]

Answer:

The hourly rate for lectures is $7.33

Step-by-step explanation:

* Let's break down how to tackle the problem.

- For the level 3 course, examination hours are priced at double that of workshop hours.

- Workshop hours cost twice the rate of lecture hours.

- The total includes examination, workshop, and lecture hours.

- Examination lasts 3 hours, workshops 24 hours, and lectures 12 hours.

* Let’s denote the cost of lecture hours as $x per hour.

∴ The lectures cost $x per hour.

∵ Workshop charge is twice that of lectures

∴ Workshop hours cost 2(x) = 2x per hour.

∵ Examination fees are double that of workshop hours

∵ The workshop cost is 2x

∴ Examination fees are 2(2x) = 4x per hour.

- Combining costs for level 3 gives us the total of lecture, workshop, and examination hours.

∵ 12 hours for lectures

∵ 24 hours for workshops

∵ 3 hours for examinations

∵ Thus the total cost for level 3 = 12(x) + 24(2x) + 3(4x).

∴ Total cost for level 3 = 12x + 48x + 12x.

∵ Therefore, total cost = $528.

∴ 12x + 48x + 12x = 528.

∴ 72x = 528; hence we divide both sides by 72.

∴ x = 7.33.

∵ x represents the cost of lecture hours per hour.

∴ Therefore, the hourly price for lectures is $7.33.

6 0
3 months ago
Which steps can be used to solve StartFraction 6 Over 7 EndFraction x + one-half = StartFraction 7 Over 8 EndFraction for x? Che
lawyer [12517]

Response:

First, subtract one-half from each side of the equation.

Next, divide both sides by 6/7.

Then, multiply each side by 7/6.

Detailed explanation:

I simply took the question on ed.

8 0
3 months ago
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