8.3.5: Max In List Write the function max_int_in_list that takes a list of ints and returns the biggest int in the list. You can assume that the list has at least one int in it. A call to this function would look like: my_list = [5, 2, -5, 10, 23, -21] biggest_int = max_int_in_list(my_list) # biggest_int is now 23 Do not use the built-in function max in your program! Hint: The highest number in the list might be negative! Make sure your function correctly handles that case.
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Subject:
Computers and Technology -
Author:
kendal -
Created:
1 year ago
Answers 3
Answer:
here!!
Explanation:
# Write code here...
def max_int_in_list(my_list):
highest = my_list[4]
for num in my_list:
if num > highest:
highest = num
return highest
my_list = [5, 2, -5, 10, 23, -21]
biggest_int = max_int_in_list(my_list)
print biggest_int
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Author:
troyg747
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Rate an answer:
4
A version of the required program written in python 3 is given below.
my_list = [5, 2, -5, 10, 23, -21]
def max_int_in_list(my_list):
#iniate the function named max_int_in_list which takes in a list argument
maximum = my_list[0]
#set the first value in the list as maximum
for num in my_list:
#iterate through every value in the list
if num > maximum:
#Check if any of the iterated values in the list is greater than the set maximum value
maximum = num
#if so set the Number as the new maximum value
return maximum
#return the maximum value
print(max_int_in_list(my_list))
A SCREENSHOT of the program output is attached.
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Author:
maggieecn5
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8
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Author:
carmenovog
Answer:
A) attached below
B) I₁ = 18.1 A , I₂ = 69.39 A
C) V( magnitude) = 454.5 ∠ 5.04° V , Voltage regulation = ≈ -1.2%
Explanation:
A) Schematic diagram attached below
attached below
B) magnitude of primary and secondary winding currents
I₂ ( secondary current ) = P / √3 * VL * cos∅ ---------- ( 1 )
VL = Line voltage = 208
cos∅ ( power factor ) = 0.8
P = 20 * 10^3 watts
insert values into equation 1
I₂ = 69.39 A
I₁ ( primary current ) = I₂V2 / V1
I₁ = ( 69.39 * 120 ) / 460 = 18.1 A
C ) Calculate the Primary voltage magnitude and the Voltage regulation
V(magnitude ) = Vp + ( I₁ ∠∅ ) Req ( 1 + j2 = 2.24 ∠63.43° )
= 460 + ( 18.1 * cos^-1 (0.8) ) ( 1 + j2 )
= 460 + 40.544 ∠ 100.3°
∴ V( magnitude) = 454.5 ∠ 5.04° V
Voltage regulation
= ((Vmag - V1) / V1 )) * 100
= (( 454.5 - 460 / 460 )) * 100
= -1.195 % ≈ -1.2%
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Author:
ellorypae6
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