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:
by providing an alternative pathway with a lower activation energy
Explanation:
Cataysts are the substances that neither consumed nor produced during a chemical reaction, but increase the rate of the reaction.
So, catalysts remain chemically unchnaged after the completion of the reaction.
For a reaction to occur, reactants molecule must collide and successful collisions lead to product formation.
A certain minimum amount of energy is required for successful collision. This minimum amounr of energy is called activation energy.
One way to increase speed of reaction is to provide energy to the reactants and other way to increase speed is to provide alternate pathway with lower activation energy.
Catalyst increases a rate of reaction by providing an alternate pathway with lower activation energy.
Therefore, the correct option is option b 'by providing an alternative pathway with a lower activation energy'
Answer:
Most viscous to least viscous: 
Explanation:
For hydrocarbons, viscosity increases with increasing molar mass. Because increasing molar mass signifies increase in number of electrons in molecules.
We know that in non-polar hydrocarbons, only van der waal intermolecular force exists. Van der waal force is proportional to number of electrons in a molecule.
Therefore with increasing molar mass, van der waal force increases. hence molecules gets more tightly bind with each other resulting increase in viscosity.
Here molar mass order : 
Therefore viscosity order : 
Answer:
Curbing global carbon dioxide emissions has been a challenge, primarily because they are being driven higher by countries with low per capita emissions.
Explanation:
"just trust me bro"