Answer:
Please see explanation for the answer. The code is written in python and is as given below:
Step-by-step explanation:
The solution is obtained on the Python with the following code
import matplotlib.pyplot as plotter
import numpy as npy
x_s = npy.linspace(-5,5,100) #Defining a linear sample space with boundaries as -5 to 5 and 100 as number of samples.
def sigmo(z):return 1/(1 + npy.exp(-z)) #Defining sigmoid function for the f(x).
plotter.plot(x_s, sigmo(x_s))
plotter.plot([-5,5],[.5,.5])
plotter.xlabel("z")
plotter.ylabel("sigmoid(z)")
plotter.show()
Step-by-step explanation:
the answer is in the image above
Answer:
<h2>3. Infinitely many</h2>
Step-by-step explanation:
Answer:
the answers is -4
Step-by-step explanation:
-1+3 I think this the answer
Since we are already given the amount of jumps from the first trial, and how much it should be increased by on each succeeding trial, we can already solve for the amount of jumps from the first through tenth trials. Starting from 5 and adding 3 each time, we get: 5 8 (11) 14 17 20 23 26 29 32, with 11 being the third trial.
Having been provided 2 different sigma notations, which I assume are choices to the question, we can substitute the initial value to see if it does match the result of the 3rd trial which we obtained by manual adding.
Let us try it below:
Sigma notation 1:
10
<span> Σ (2i + 3)
</span>i = 3
@ i = 3
2(3) + 3
12
The first sigma notation does not have the same result, so we move on to the next.
10
<span> Σ (3i + 2)
</span><span>i = 3
</span>
When i = 3; <span>3(3) + 2 = 11. (OK)
</span>
Since the 3rd trial is a match, we test it with the other values for the 4th through 10th trials.
When i = 4; <span>3(4) + 2 = 14. (OK)
</span>When i = 5; <span>3(5) + 2 = 17. (OK)
</span>When i = 6; <span>3(6) + 2 = 20. (OK)
</span>When i = 7; 3(7) + 2 = 23. (OK)
When i = 8; <span>3(8) + 2 = 26. (OK)
</span>When i = 9; <span>3(9) + 2 = 29. (OK)
</span>When i = 10; <span>3(10) + 2 = 32. (OK)
Adding the results from her 3rd through 10th trials: </span><span>11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 = 172.
</span>
Therefore, the total jumps she had made from her third to tenth trips is 172.
