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Answer:
Here is the Python program:
small_container = int(input("Enter the number of small containers you recycled?"))
large_container = int(input("Enter the number of large containers you recycled?"))
refund = (small_container * 0.10) + (large_container * 0.25)
print("The total refund for returning the containers is $" + "{0:.2f}".format(float(refund)))
Explanation:
The program first prompts the user to enter the number of small containers. The input value is stored in an integer type variable small_container. The input is basically an integer value.
The program then prompts the user to enter the number of large containers. The input value is stored in an integer type variable large_container. The input is basically an integer value.
refund = (small_container * 0.10) + (large_container * 0.25) This statement computers the refund that will be recieved for returning the small and larger containers. The small containers holding one litre or less have a $0.10 deposit so the number of small containers is multiplied by 0.10. The large containers holding more than one litre have a $0.25 deposit so the number of large containers is multiplied by 0.25. Now both of these calculated deposits of containers of each side are added to return the refund that will be received for returning these containers. This whole computation is stored in refund variable.
print("The total refund for returning the containers is $" + "{0:.2f}".format(float(refund))) This print statement displays the refund in the format given in the question. The output includes a $ sign and displays exactly two decimal places by using {0:.2f} where .2f means 2 decimal places after the decimal point. Then the output is represented in floating point number using. format(float) is used to specify the output type as float to display a floating point refund value up to 2 decimal places.
Answer:
a.) k² - 3k + 3
b.) 1/(1 - k)²
c.)
Explanation:
a.) A packet can make 1,2 or 3 hops
probability of 1 hop = k ...(1)
probability of 2 hops = k(1-k) ...(2)
probability of 3 hops = (1-k)²...(3)
Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)
= (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)
= k + 2k - 2k² + 3(1 + k² - 2k)
∴mean number of hops = k² - 3k + 3
b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3
if k = 0 then number of hops is 3
if k = 1 then number of hops is (1 - 3 + 3) = 1
multiple transmissions can be needed if K is between 0 and 1
The probability of successful transmissions through the entire path is (1 - k)²
for one transmission, the probility of success is (1 - k)²
for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)
for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on
∴ for transmitting a single packet, it makes:
∞ n-1
T = ∑ n(1 - k)²(1 - (1 - k)²)
n-1
= 1/(1 - k)²
c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)
from (a) above, mean number of hops when transmitting a packet = k² - 3k + 3
from (b) above, mean number of transmissions by a packet = 1/(1 - k)²
substituting: mean number of required packet =