Answer:
60.42 grams
Explanation:
So, neon has a molar mass of 20.18 g/mol , which means that every mole of neon has a mass of 20.18 g so that means three of them is 60.42 grams
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One mole methane combusts to form one mole CO2 and 2 moles H2O
To solve this problem, let us say that:
x = volume of 1:2000 drug "i" solution
y = volume of 7% drug "i" solution
Assuming volume additive, then this forms:
x + y = 120 mL
<span>x = 120 – y ---> 1</span>
1:2000 also refers to 0.0005 concentrations and 7% also
refers to 0.07 concentrations. By doing a component balance:
0.0005 x + 0.07 y = 0.035 (120 mL)
0.0005 x + 0.07 y = 4.2
Substituting equation 1 into this derived equation to get an
equation in terms of y:
0.0005 (120 – y) + 0.07 y = 4.2
0.06 – 0.0005 y + 0.07 y = 4.2
0.0695 y = 4.14
y = 59.568 mL = 59.57 mL
From equation 1, x would be:
x = 120 - 59.57
x = 60.43 mL
Answers:
59.57 mL of 1:2000 drug "i" solution
60.43 mL <span>of 7% drug "i" solution</span>
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,
and
The entropy of the training examples is given by :
= 0.9911
b). For the attribute all the associating increments and the probability are :
+ -
T 3 1
F 1 4
Th entropy for is given by :
![$\frac{4}{9}[ -\frac{3}{4}\log\left(\frac{3}{4}\right)-\frac{1}{4}\log\left(\frac{1}{4}\right)]+\frac{5}{9}[ -\frac{1}{5}\log\left(\frac{1}{5}\right)-\frac{4}{5}\log\left(\frac{4}{5}\right)]$](https://tex.z-dn.net/?f=%24%5Cfrac%7B4%7D%7B9%7D%5B%20-%5Cfrac%7B3%7D%7B4%7D%5Clog%5Cleft%28%5Cfrac%7B3%7D%7B4%7D%5Cright%29-%5Cfrac%7B1%7D%7B4%7D%5Clog%5Cleft%28%5Cfrac%7B1%7D%7B4%7D%5Cright%29%5D%2B%5Cfrac%7B5%7D%7B9%7D%5B%20-%5Cfrac%7B1%7D%7B5%7D%5Clog%5Cleft%28%5Cfrac%7B1%7D%7B5%7D%5Cright%29-%5Cfrac%7B4%7D%7B5%7D%5Clog%5Cleft%28%5Cfrac%7B4%7D%7B5%7D%5Cright%29%5D%24)
= 0.7616
Therefore, the information gain for is
0.9911 - 0.7616 = 0.2294
Similarly for the attribute 
the associating counts and the probabilities are :
+ -
T 2 3
F 2 2
Th entropy for is given by :
![$\frac{5}{9}[ -\frac{2}{5}\log\left(\frac{2}{5}\right)-\frac{3}{5}\log\left(\frac{3}{5}\right)]+\frac{4}{9}[ -\frac{2}{4}\log\left(\frac{2}{4}\right)-\frac{2}{4}\log\left(\frac{2}{4}\right)]$](https://tex.z-dn.net/?f=%24%5Cfrac%7B5%7D%7B9%7D%5B%20-%5Cfrac%7B2%7D%7B5%7D%5Clog%5Cleft%28%5Cfrac%7B2%7D%7B5%7D%5Cright%29-%5Cfrac%7B3%7D%7B5%7D%5Clog%5Cleft%28%5Cfrac%7B3%7D%7B5%7D%5Cright%29%5D%2B%5Cfrac%7B4%7D%7B9%7D%5B%20-%5Cfrac%7B2%7D%7B4%7D%5Clog%5Cleft%28%5Cfrac%7B2%7D%7B4%7D%5Cright%29-%5Cfrac%7B2%7D%7B4%7D%5Clog%5Cleft%28%5Cfrac%7B2%7D%7B4%7D%5Cright%29%5D%24)
= 0.9839
Therefore, the information gain for is
0.9911 - 0.9839 = 0.0072
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 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 produces the best split.
Answer:
Another way to describe a reaction that gives out free energy is Spontaneous Reaction
