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Nata [24]
3 years ago
11

No question here...............

Chemistry
1 answer:
Lady bird [3.3K]3 years ago
4 0

no answer here ......

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A sample of gas occupies 17 mL at –112°C. What volume does the sample occupy at 70°C?
Snowcat [4.5K]

final answer is

V

2

=

4

×

10

1

 

mL

8 0
3 years ago
Groups on the Periodic Tables have the same number of valence electrons which affects
Anastasy [175]

Answer:

The vertical columns (group) of the periodic table are arranged such that all it's elements have the same number of valence electrons. All elements within a certain group thus group share similar properties.

8 0
3 years ago
What is the entropy of this collection of training examples with respect to the positive class B. What are the information gains
Alenkinab [10]

The data set is missing in the question. The data set is given in the attachment.

Solution :

a). In the table, there are four positive examples and give number of negative examples.

Therefore,

$P(+) = \frac{4}{9}$   and

$P(-) = \frac{5}{9}$

The entropy of the training examples is given by :

$ -\frac{4}{9}\log_2\left(\frac{4}{9}\right)-\frac{5}{9}\log_2\left(\frac{5}{9}\right)$

= 0.9911

b). For the attribute all the associating increments and the probability are :

  $a_1$   +   -

  T   3    1

  F    1    4

Th entropy for   $a_1$  is given by :

$\frac{4}{9}[ -\frac{3}{4}\log\left(\frac{3}{4}\right)-\frac{1}{4}\log\left(\frac{1}{4}\right)]+\frac{5}{9}[ -\frac{1}{5}\log\left(\frac{1}{5}\right)-\frac{4}{5}\log\left(\frac{4}{5}\right)]$

= 0.7616

Therefore, the information gain for $a_1$  is

  0.9911 - 0.7616 = 0.2294

Similarly for the attribute $a_2$  the associating counts and the probabilities are :

  $a_2$  +   -

  T   2    3

  F   2    2

Th entropy for   $a_2$ is given by :

$\frac{5}{9}[ -\frac{2}{5}\log\left(\frac{2}{5}\right)-\frac{3}{5}\log\left(\frac{3}{5}\right)]+\frac{4}{9}[ -\frac{2}{4}\log\left(\frac{2}{4}\right)-\frac{2}{4}\log\left(\frac{2}{4}\right)]$

= 0.9839

Therefore, the information gain for $a_2$ is

  0.9911 - 0.9839 = 0.0072

$a_3$     Class label      split point       entropy        Info gain

1.0         +                        2.0            0.8484        0.1427

3.0        -                         3.5            0.9885        0.0026

4.0        +                        4.5            0.9183         0.0728

5.0        -

5.0        -                        5.5            0.9839        0.0072

6.0        +                       6.5             0.9728       0.0183

7.0        +

7.0        -                        7.5             0.8889       0.1022

The best split for $a_3$  observed at split point which is equal to 2.

c). From the table mention in part (b) of the information gain, we can say that $a_1$ produces the best split.

4 0
3 years ago
If a person weighs 87lbs and has to be given a drug with a dosage of 50 milligrams per kilogram body weight, how many grams of t
Katarina [22]

Answer:

1 957,5

Explanation:

87libs=39.5kg that equals 50=1.957.5

7 0
3 years ago
A gas cylinder contains exactly 1 mole of oxygen gas (O2). How many molecules of oxygen are in the cylinder?
Aloiza [94]

The number of molecules of oxygen in the cylinder is 6.02×1023 molecules

Data obtained from the question

Number of mole of oxygen = 1 mole

Number of molecule of oxygen =?

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×1023 molecules. This implies that 1 mole of oxygen contains 6.02×1023 molecules.

Since the cylinder contains 1 mole of oxygen, then the number of molecules of oxygen present in the cylinder is 6.02×1023 molecules.

8 0
2 years ago
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