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:
work done = 48.88 × 
J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 × J
Answer:
fracture will occur as the value is less than E/10 (= 22.5)
Explanation:
If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.
![\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}](https://tex.z-dn.net/?f=%5Csigma_m%20%3D%202%5Csigma_o%20%5B%5Cfrac%7Ba%7D%7B%5Crho_t%7D%5D%5E%7B1%2F2%7D)
= 15 GPa
fracture will occur as the value is less than E/10 = 22.5
C because i know what i’m taking about
