Answer:
- def median(l):
- if(len(l) == 0):
- return 0
- else:
- l.sort()
- if(len(l)%2 == 0):
- index = int(len(l)/2)
- mid = (l[index-1] + l[index]) / 2
- else:
- mid = l[len(l)//2]
- return mid
-
- def mode(l):
- if(len(l)==0):
- return 0
-
- mode = max(set(l), key=l.count)
- return mode
-
- def mean(l):
- if(len(l)==0):
- return 0
- sum = 0
- for x in l:
- sum += x
- mean = sum / len(l)
- return mean
-
- lst = [5, 7, 10, 11, 12, 12, 13, 15, 25, 30, 45, 61]
- print(mean(lst))
- print(median(lst))
- print(mode(lst))
Explanation:
Firstly, we create a median function (Line 1). This function will check if the the length of list is zero and also if it is an even number. If the length is zero (empty list), it return zero (Line 2-3). If it is an even number, it will calculate the median by summing up two middle index values and divide them by two (Line 6-8). Or if the length is an odd, it will simply take the middle index value and return it as output (Line 9-10).
In mode function, after checking the length of list, we use the max function to estimate the maximum count of the item in list (Line 17) and use it as mode.
In mean function, after checking the length of list, we create a sum variable and then use a loop to add the item of list to sum (Line 23-25). After the loop, divide sum by the length of list to get the mean (Line 26).
In the main program, we test the three functions using a sample list and we shall get
20.5
12.5
12
Answer: True.
Explanation: Coolant is flammable and toxic.
Answer:
W = - 184.8 kW
Explanation:
Given data:





We know that work is done as
![W = - [ Q + \dor m[h_2 - h_1]]](https://tex.z-dn.net/?f=W%20%3D%20-%20%5B%20Q%20%2B%20%5Cdor%20m%5Bh_2%20-%20h_1%5D%5D)
for
density of air is 1.22 kg/m^3 and 
for 


W = -[14 + 1.708[400-300]]
W = - 184.8 kW
Answer:
a) 23.89 < -25.84 Ω
b) 31.38 < 25.84 A
c) 0.9323 leading
Explanation:
A) Calculate the load Impedance
current on load side = 0.75 p.u
power factor angle = 25.84
= 0.75 < 25.84°
attached below is the remaining part of the solution
<u>B) Find the input current on the primary side in real units </u>
load current in primary = 31.38 < 25.84 A
<u>C) find the input power factor </u>
power factor = 0.9323 leading
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<em>attached below is the detailed solution </em>