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DENIUS [597]
3 years ago
11

Suppose that this routine has an adjustable cutoff (threshold) mechanism by which you can alter the proportion of records classi

fied as fraudulent. Describe how moving the cutoff up or down would affect 2. a. the classification error rate for records that are truly fraudulent (circle best answer) (i)Lowering the cutoff (here, below 0.5) leads to classifying more non-fraudulent records as fraudulent records (more zeros misclassified as 1) (ii)Lowering the cutoff (here, below 0.5) leads to classifying fewer non-fraudulent records as (iii) Lowering the cutoff (here, below 0.5) leads to classifying more fraudulent records as non- (iv) Lowering the cutoff (here, below 0.5) leads to classifying fewer fraudulent records as non- fraudulent records (more zeros misclassified as 1) fraudulent records (more ones misclassified as 0) fraudulent records (more ones misclassified as 0) b. the classification error rate for records that are truly nonfraudulent (circle best answer (i)Increasing the cutoff (here, above 0.5) leads to classifying more non-fraudulent records as fraudulent records (more 1 misclassified as 0) (ii)Increasing the cutoff (here, above 0.5) leads to classifying fewer non-fraudulent records as fraudulent records (more 1 misclassified as 0) (iii)Increasing the cutoff (here, above 0.5) leads to classifying more fraudulent records as non- fraudulent records (more 1 misclassified as 0) (iv) Increasing the cutoff (here, above 0.5) leads to classifying fewer fraudulent records as non- fraudulent records (more 1 misclassified as 0)
Mathematics
1 answer:
VladimirAG [237]3 years ago
8 0

Answer:

a.The classification error rate for records that are truly fraudulent: By increasing the cutoff value, the non-fraudulent records will go down and this increases the error rate.

Also, if the cutoff value is down, the fraudulent records will increase and this lowers the error rate

.b.The classification error rate for records that are truly non-fraudulent: By moving up the cutoff value, the fraudulent records will go down and this decreases the error rate.

Also, if the cutoff value is down, the non-fraudulent records will go down and this increases the error rate.

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Resolver:
Dafna1 [17]
Aquí está las respuestas y el trabajo de cada una, si tienes alguna pregunta me envías un mensaje o un comentario

3 0
2 years ago
A point in the figure is selected at random. Find the probability that the point will be in the part that is NOT shaded.
creativ13 [48]

Answer:

About 20%

Step-by-step explanation:

The probability of a point on the un-shaded region will be given by;

Area of the shaded region/total area

Considering that the circles are of the same size, then the length of the square is 2d

Area of the square = 4d²

Area of the shaded = (4d² -4π(d/2)²

                                = 4d² -πd², but π = 22/7

                                = 6/7d²

Probability = (6/7d² ÷ 4d²) × 100%

                  = (6/28 d²) × 100%

                  = 21.42 %

Therefore; the answer is about 20%

3 0
2 years ago
2. Which expression is equivalent to 3x(-4x² + 5x - 8) - 6(x² - 5x + 7)?
adoni [48]
It is answer choice B. Using the distributive property, -12x^3+15x^2-24x-6x^2+30x-42. Then combine like terms.
5 0
3 years ago
How many times longer is the wavelength of a sound wave with a frequency of 20 waves per second than the wavelength of a sound w
sergejj [24]

The sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec

<h3>Calculating wavelength </h3>

From the question, we are to determine how many times longer is the first sound wave compared to the second sound water

Using the formula,

v = fλ

∴ λ = v/f

Where v is the velocity

f is the frequency

and λ is the wavelength

For the first wave

f = 20 waves/sec

Then,

λ₁ = v/20

For the second wave

f = 16,000 waves/sec

λ₂ = v/16000

Then,

The factor by which the first sound wave is longer than the second sound wave is

λ₁/ λ₂ = (v/20) ÷( v/16000)

= (v/20) × 16000/v)

= 16000/20

= 800

Hence, the sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec

Learn more on Calculating wavelength here: brainly.com/question/16396485

#SPJ1

6 0
1 year ago
Please help me idk this ???
nydimaria [60]

Answer:

64

Step-by-step explanation:

16/100 = x / 400

Cross multiply:-

100x = 16*400

x = 6400 / 100

= 64

7 0
3 years ago
Read 2 more answers
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