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Tanzania [10]
3 years ago
10

Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 8x and y = 2x + 2 intersect are the so

lutions of the equation 8x = 2x + 2. (4 points)
Part B Make tables to find the solution to 8x = 2x + 2. Take the integer values of x between -3 and 3.

Part C: How can you solve the equation 8x = 2x + 2 graphically?

Mathematics
2 answers:
son4ous [18]3 years ago
3 0

Answer:

(A)

As per the given condition.

You have 2 equations for y.

i,e y =8x and y= 2x+2

then, they will intersect at some point where y is the same for both equations.

That is why in equation y=8x you exchange y with other equation you got which is y=2x+2 once you do this you will have

8x = 2x+2  and the solution of which will satisfy both equation.

(B)

8x = 2x + 2

to find the solutions take the integer values of x between -3 and 3.

x = -3 , then

8(-3) = 2(-3) +2

-24 = -6+2

-12 = -4    False.

similarly, for x = -2

8(-2) = 2(-2)+2

-16 = -2   False

x = -1

8(-1) = 2(-1)+2

-8= 0   False

x = 0

8(0) = 2(0)+2

0= 2   False

x = 1

8(1) = 2(1)+2

8= 4   False

x = 2

8(2) = 2(2)+2

16 = 6   False

x = 3

8(3) = 2(3)+2

24 = 8   False

there is no solution to 8x = 2x +2 for the integers values of x between -3 and 3.

(C)

The equations cab be solved graphically by plotting the two given functions on a coordinate plane and identifying the point of intersection of the two graphs.

The point of intersection are the values of the variables which satisfy both equations at a particular point.

you can see the graph as shown below , the point of intersection at x =0.333 and value of y = 2.667









Salsk061 [2.6K]3 years ago
3 0
A) At the intersection, the values of of x and y of the equations are the same. Because of this, we can equate their y values to get an equation in the form of only x.

B)
8(-3) = 2(-3) + 2 → -24 =/= -4
8(-2) = 2(-2) + 2 → -16 =/= -2
8(-1) = 2(-1) + 2 → -8 =/= 0
8(0) = 2(0) + 2 → 0 =/= 2
8(1) = 2(1) + 2 → 8 =/= 4
8(2) = 2(2) + 2 → 16 =/= 6
8(3) = 2(3) + 2 → 24 =/= 14
The solution lies between x = 0 and x = 1.

We can graph both equations and find the value of x and y at which they intersect.
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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

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