Answer:
16.2 cents
Explanation:
Given that a homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time.
Where Base monthly charge of $10.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 300 kWh per month at 6 cents/kWh.
For the first 100 kWh:
16 cent × 100 = 1600 cents = 16 dollars
Since 1 dollar = 100 cents
For the remaining energy:
260 - 100 = 160 kwh
10 cents × 160 = 1600 cents = 16 dollars
The total cost = 10 + 16 + 16 = 42 dollars
Note that the base monthly of 10 dollars is added.
The cost of 260 kWh of energy consumption in July is 42 dollars
To determine the average cost per kWh for the month of July, divide the total cost by the total energy consumed.
That is, 42 / 260 = 0.1615 dollars
Convert it to cents by multiplying the result by 100.
0.1615 × 100 = 16.15 cents
Approximately 16.2 cents
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
“Thinking about pleasant things to pass the time” would not promote safety in the shop because it would be taking the focus away from important tasks, which in turn decreases safety.
Answer:
material remove in 3 min is 16790.4 mm³/s
Explanation:
given data
length L = 80 cm = 800 mm
width W = 30 cm
height H = 15 cm
make grove length = 80 cm
width = 8 cm
depth = 10 cm
mill toll diameter = 4 mm
axial cutting depth = 20 mm
to find out
How much material removed in 3 minutes
solution
first we find time taken for length of advance that is
time = 
here advance is given as 0.001166 mts / sec
so time = 
time = 686.106 seconds
now we find material remove rate that is
remove rate = mill toll rate × axial cutting depth × advance
remove rate = 4 × 20×0.001166 ×1000
remove rate = 93.28 mm³/s
so
material remove in 3 minute = 3 × 60 = 180 sec
so material remove in 3 min = 180 × 93.28
material remove in 3 min is 16790.4 mm³/s
