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4 years ago

Anthony buys one pound of oranges for $12 and sells it for $14. How can he write the profit in his account book?

1 answer:

5 0

The answer would be, A) +$2 because he spend $12 but earned $14 so when you subtract 14 and 12 you get $2, which would be his profit

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3 years ago

Read 2 more answers

A data mining routine has been applied to a transaction dataset and has classified 88 records as fraudulent (30 correctly so) an

chubhunter [2.5K]

Answer:

The classification matrix is attached below

Part a

The classification error rate for the records those are truly fraudulent is 65.91%.

Part b

The classification error rate for records that are truly non-fraudulent is 96.64%

Step-by-step explanation:

The classification matrix is obtained as shown below:

The transaction dataset has 30 fraudulent correctly classified records out of 88 records, that is, 30 records are correctly predicted given that an instance is negative.

Also, there would be 88 - 30 = 58 non-fraudulent incorrectly classified records, that is, 58 records are incorrectly predicted given that an instance is positive.

The transaction dataset has 920 non-fraudulent correctly classified records out of 952 records, that is, 920 records are correctly predicted given that an instance is positive.

Also, there would be 952 - 920 = 32 fraudulent incorrectly classified records, that is, 32 records incorrectly predicted given that an instance is negative.

That is,

Predicted value

Active value Fraudulent Non-fraudulent

Fraudlent 30 58

non-fraudulent 32 920

The classification matrix is obtained by using the information related to the transaction data, which is classified into fraudulent records and non-fraudulent records.

The error rate is obtained as shown below:

The error rate is obtained by taking the ratio of $\left( {b + c} \right)(b+c)$ and the total number of records.

The classification matrix is, shown above

The total number of records is, $30 + 58 + 32 + 920 = 1,040$

The error rate is,

$\begin{array}{c}\\{\rm{Error}}\,{\rm{rate}} = \frac{{b + c}}{{{\rm{Total}}}}\\\\ = \frac{{58 + 32}}{{1,040}}\\\\ = \frac{{90}}{{1,040}}\\\\ = 0.0865\\\end{array}$

The percentage is $0.0865 \times 100 = 8.65$

(a)

The classification error rate for the records those are truly fraudulent is obtained by taking the rate ratio of b and $\left( {a + b} \right)(a+b)$ .

The classification error rate for the records those are truly fraudulent is obtained as shown below:

The classification matrix is, shown above and in the attachment

The error rate for truly fraudulent is,

$\begin{array}{c}\\FP = \frac{b}{{a + b}}\\\\ = \frac{{58}}{{30 + 58}}\\\\ = \frac{{58}}{{88}}\\\\ = 0.6591\\\end{array}$

The percentage is, $0.6591 \times 100 = 65.91$

(b)

The classification error rate for records that are truly non-fraudulent is obtained by taking the ratio of d and $\left( {c + d} \right)(c+d)$ .

The classification error rate for records that are truly non-fraudulent is obtained as shown below:

The classification matrix is, shown in the attachment

The error rate for truly non-fraudulent is,

$\begin{array}{c}\\TP = \frac{d}{{c + d}}\\\\ = \frac{{920}}{{32 + 920}}\\\\ = \frac{{920}}{{952}}\\\\ = 0.9664\\\end{array}$

The percentage is, $0.9664 \times 100 = 96.64$

8 0

3 years ago

Suppose it has been determined that the sales at Ewing Gallery decline after the end of an advertising campaign, with daily sale

gtnhenbr [62]

Answer:

$1050 is the daily sales 5 days after the end of the campaign.

Step-by-step explanation:

We are given the following in the question:

$S(x)= 2100(2^{-0.2x})$

where S is the daily sales in dollars and x is the number of days after the campaign ends.

We have to find the value of the daily sales 5 days after the end of the campaign.

We put x = 5 in the equation:

$S(5)= 2100(2^{-0.2(5)}) = 2100(2^{-1}) = 1050$

Thus, $1050 is the daily sales 5 days after the end of the campaign.

6 0

3 years ago