All categories

3 years ago

10 easy points!!!! What is the x-intercept of the line?

Mathematics

1 answer:

fgiga [73]3 years ago

5 0

Answer:

Step-by-step explanation:

As x increases from -74 to -54 (a 'run' of 20), y decreases from 18 to 12 (a 'rise' of -6. Thus, the slope of this line is m = rise/run = -6/20 = -13/10.

From y = mx + b we get 12 = (-13/10)(12) + b, or (after dividing all terms by 12)

1 = -13/10 + b/12, or

60 = -3(6) + 5b, or

42 = - 5b, or b = -42/5

The line is y = (-13/10)x - 42/5.

At the x-intercept, y = 0. Setting y = 0, we get:

(13/10)X = -42/5, or 13x = -84.

Thus, x = -84/13 = -6.46

and so the x-intercept of the line is (-6.46, 0)

You might be interested in

Note that the right triangles with sides equal to 5, 12, 13 and 9, 12, 13 both have a side equal to 12. using this fact find the

kondor19780726 [428]

There is a mistake 9,12,13 is not a right triangle. but 9,12,15 is. IF you use this triangle instead, then the area of the 13,14,15 triangle is the sum of the area of the other two triangles.

6 0

3 years ago

Match the following items by evaluating the expression for x = -7. x-2 x-1 x0 x1 x2

frosja888 [35]

Answer:

${x}^{ - 2} = \frac{1}{ {x}^{2} } \: \: –> {7}^{ - 2} = \frac{1}{ {7}^{2} } = \frac{ 1}{49} \\ \\ {x}^{ - 1} = \frac{1}{ {x}^{1} } \: –> {7}^{ - 1} = \frac{1}{ {7}^{1} } = \frac{1}{7} \\ \\ {x}^{0} –> {7}^{0} = 1 \\ \\ {x}^{1} –> {7}^{1} = 7 \\ \\ {x}^{2} –> {7}^{2} = 49$

7 0

2 years ago

A scale reports a 15-pound weight as weighing 12.375 pounds. Which of the following is true?

rusak2 [61]

Answer:

Step-by-step explanation:

If it is accurate then it would have gotten the correct weight, but by going down to the thousandths it would be precise

4 0

3 years ago

Read 2 more answers

There is a spinner with 12 equal areas, numbered 1 through 12. If the spinner is spun

Tomtit [17]

Answer:

1/12

Step-by-step explanation:

There is only number between 1 and 12 that is both a multiple of 5 and 2, that being 10. 10 has an equal chance out of all the other numbers that can be spun, and there are 12 numbers, so 10 has a 1/12 chance of being spun.

6 0

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

Firlakuza [10]

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, $0.9664 \times 100 = 96.64$

6 0

3 years ago