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3 years ago

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20. solve the equation by completing the square. round to the nearest hundredth if necessary. x^2+10x=18 a. -11.56, 1.56 b.11.56

, 1.56 c.-11.56, -1.56 d. 11.56, -1.56

1 answer:

larisa86 [58]3 years ago

8 0

The answer was A but i am found

1.557, -11.557

and hope that helps you

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The sum of 14 and 3<br> divided by r

labwork [276]

Answer:

?

Step-by-step explanation:

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3 years ago

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

If you pop 3 bags of popcorn in 15 minutes how many bags did you pop per minute

Harman [31]

1 for every 5 minutes

7 0

3 years ago

Read 2 more answers

2. Mary deposited $8000 in an account that pays 6% annual interest for 5 years. Using A=p(1+r)^t, what is the balance at the end

kap26 [50]

Hi there

1)
A=8,000×(1+0.06)^(5)
A=10,705.8 round your answer to get
A=10706

2)
A=8,000×(1+0.06÷4)^(4×5)
A=10,774.8 round to get
A=10775

3) with depreciation the formula is
V (t)=Vb^t
Where b=(1-r) (r rate of depreciation)
B=1-0.2=0.8

Good luck!

7 0

3 years ago

Read 2 more answers

Please help me with 1-4. I will give an Brainly and a bunch of extra points.

xz_007 [3.2K]

Answer:

$\sf 1) \quad \sf A \cup C = \{a, b, c, d, e\}$

$\sf 2) \quad B \cap A = \{a, b, d\}$

$\sf 3) \quad B' = \{c, f, g, h \}$

$\sf 4) \quad A -(B \cap C') = \{a, c \}$

Step-by-step explanation:

<u>Set Notation</u>

$\begin{array}{|c|c|l|} \cline{1-3} \sf Symbol & \sf N\:\!ame & \sf Meaning \\\cline{1-3} \{ \: \} & \sf Set & \sf A\:collection\:of\:elements\\\cline{1-3} \cup & \sf Union & \sf A \cup B=elements\:in\:A\:or\:B\:(or\:both)}\\\cline{1-3} \cap & \sf Intersection & \sf A \cap B=elements\: in \:both\: A \:and \:B} \\\cline{1-3} \sf ' \:or\: ^c & \sf Complement & \sf A'=elements\: not\: in\: A \\\cline{1-3} \sf - & \sf Difference & \sf A-B=elements \:in \:A \:but\: not\: in \:B}\\\cline{1-3} \end{array}$

Given sets:

U = {a, b, c, d, e, f, g, h}
A = {a, b, c, d}
B = {b, e, a, d}
C = {a, e}

Therefore:

Universal set = {a, b, c, d, e, f, g, h}
B' = not in B = {c, f, g, h}
C' = not in C = {b, c, d, f, g, h}

<u>Question 1</u>

$\begin{aligned}\sf A \cup C & = \sf \{a, b, c, d\} \cup \{a,e\}\\& = \sf \{a, b, c, d, e\}\end{aligned}$

<u>Question 2</u>

$\begin{aligned}\sf B \cap A & = \sf \{b, e, a, d\} \cap \{a, b, c, d\}\\& = \sf \{a, b, d\}\end{aligned}$

<u>Question 3</u>

$\begin{aligned}\sf B' & = \rm U - \sf B\\& = \sf \{a, b, c, d, e, f, g, h \}-\{b, e, a, d\}\\& = \sf \{c, f, g, h \}\end{aligned}$

<u>Question 4</u>

$\begin{aligned}\sf A -(B \cap C') & = \sf \{a, b, c, d \}- \left( \{b, e, a, d \} \cap \{b, c, d, f, g, h \} \right)\\& = \sf \{a, b, c, d \}- \{b, d \} \\& = \sf \{a, c \}\end{aligned}$

Learn more about set notation here:

brainly.com/question/28353607

3 0

2 years ago

Read 2 more answers