Answer:
All i know is The Adoption Of The Bill Of Rights was in 1791
Explanation:
Q=hc A (Ts-Ta)
q = heat transfered per unit time
A= heat transfer area of the surface
hc= convective heat transfer coefficient of the process
Ts= temperature of the surface
Ta= temperature of the air
Answer:
A₃B₈ + 5C₂ –> 3AC₂ + 4B₂C
Explanation:
A₃B₈ + C₂ –> AC₂ + B₂C
The equation can be balance as illustrated below:
A₃B₈ + C₂ –> AC₂ + B₂C
There are 3 atoms of A on the left side and 1 atom on the right side. It can be balance by writing 3 before AC₂ as shown below:
A₃B₈ + C₂ –> 3AC₂ + B₂C
There are 8 atoms of B on the left side and 2 atoms on the right side. It can be balance by writing 4 before B₂C as shown below:
A₃B₈ + C₂ –> 3AC₂ + 4B₂C
There are 2 atoms of C on the left side and a total of 10 atoms on the right side. It can be balance by writing 5 before C₂ as shown below:
A₃B₈ + 5C₂ –> 3AC₂ + 4B₂C
Now, the equation is balanced.
ethane is the compound of the answer you are asking help
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.
