The issue arises because the string you are trying to print is not a string, rather a float value. Item1, item2 and item3 are strong values (if you type some alphabets in it and not just numbers), but itemonecost, itemtwocost, and itemthreecost are explicitly type casted to float. In line 22, 23, and 24 you’re trying to print a float, by adding it with the string. One cannot add numbers to string. Rather you can type cast the itemcost to string while printing.
Add str(itemonecost) instead of itemonecost in print statement. Do this for other float variables too.
However do note that there are multiple ways to correct this issue, and I’ve just pointed one out.
These principles are not explicit to one kind of program and are increasingly broad "best practice" rules that assist designers with composing code that is easier to maintain.
<u>Explanation:</u>
A set of programming guidelines that are executed to play out a particular undertaking according to the prerequisites of the client is known as programming. Every product has some essential standards to follow. In light of all product frameworks have basic quality traits, including accessibility, modifiability, execution, security and wellbeing, testability and ease of use, the key programming thoughts give basic arrangements or strategies to help those characteristics.
It is generally less expensive, over the long haul, to utilize programming designing strategies and methods for programming frameworks instead of simply compose the projects as though it was an individual programming venture.
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec
