Answer:I love Python, very useful
Explanation:python is very easy and user friendly!
Answer:
Option B and Option D
are correct answers.
Explanation:
To view the invoice data on the related account records in sales force when invoices are stored in SAP we must:
- Create an External Object connected to an invoice table in SAP
OR
- Connect to an O-Data Publisher Service for SAP databases.
SAP can be defined as Systems, Applications and Products. SAP is basically a software which has a backbone of SAP ERP (most advance Enterprise Resource Planing). SAP software helps to manage many business areas by providing powerful tools that include financial and logistic areas.
<h2 /><h2>I hope it will help you!</h2>
Answer:
def analyze_text(sentence):
count = 0
e_count = 0
for s in sentence:
s = s.lower()
if s.isalpha():
count += 1
if s == "e":
e_count += 1
return "The text contains " + str(count) + " alphabetic characters, of which " + str(e_count) + " (" + str(e_count*100/count) + "%) are ‘e’."
Explanation:
Create a function called analyze_text takes a string, sentence
Initialize the count and e_count variables as 0
Create a for loop that iterates through the sentence
Inside the loop, convert all letters to lowercase using lower() function. Check each character. If a character is a letter, increment the count by 1. If a character is "e", increment the e_count by 1.
Return the count and e_count in required format
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
Answer:
Answers in explanation. Try to ask one question at a time, it is easier for people to answer single questions and you will get answers faster.
Explanation:
15. A
16. D
17. B + C
18. A+B+D
19. B+C+D
**20. is NOT Planned personal leave for documentation developers and proof readers. The other 4 answers are correct
21. Presentation notes + Outline
22.B
23.D (im not entirely sure about this one)
