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3 years ago

Is 0.27 greater than 0.3 ?

Mathematics

1 answer:

Zinaida [17]3 years ago

7 0

Answer:

Step-by-step explanation:

here's a little trick, add a zero to make both numbers the same.

it's doesn't change the value.

so .27 and .3 become .27 and .30

and .30 is greater

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What is the answer to the question?

Naily [24]

It is 1/3, add up the total number of kids, which is 30, the number of kids 6-8 is 10. 10/30 is equivalent to 1/3

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3 years ago

Read 2 more answers

Find two unit vectors orthogonal to a=⟨2,−2,−3⟩a=⟨2,−2,−3⟩ and b=⟨3,2,2⟩b=⟨3,2,2⟩ enter your answer so that the first non-zero c

Ronch [10]

We need the cross product of the two vectors.
a x b
=<2,-2,-3> x <3,2,2>
=
i j k
2 -2 -3
3 2 2
=<-4+6, -(4+9), 4+6>
=<2,-13,10>
The second vector is obtained by reversing the direction, namely <-2,13,-10>
Thus the two vectors are <2,-13,10> and <-2,13,-10>.

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3 years ago

Could someone help me with this im so confused

Arlecino [84]

Ok so basically the number by the X is your slope (m) and the other number is your y-intercept (b) so for example the first question your slope (m) 2/3 and your y-intercept (b) -2

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3 years ago

Write a function describing the relationship of the given variables.

zhuklara [117]

Answer:

The function is A = 10√r

Step-by-step explanation:

* Lets explain the meaning of direct variation

- The direct variation is a mathematical relationship between two

variables that can be expressed by an equation in which one

variable is equal to a constant times the other

- If Y is in direct variation with x (y ∝ x), then y = kx, where k is the

constant of variation

* Now lets solve the problem

# A is varies directly with the square root of r

- Change the statement above to a mathematical relation

∴ A ∝ √r

- Chang the relation to a function by using a constant k

∴ A = k√r

- To find the value of the constant of variation k substitute A and r

by the given values

∵ r = 16 when A = 40

∵ A = k√r

∴ 40 = k√16 ⇒ simplify the square root

∴ 40 = 4k ⇒ divide both sides by 4 to find the value of k

∴ 10 = k

- The value of the constant of variation is 10

∴ The function describing the relationship of A and r is A = 10√r

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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

Is 0.27 greater than 0.3 ?​

Is 0.27 greater than 0.3 ?