Answer:
sentence = "hello wow a stores good"
same_letter_count = 0
sentence_list = sentence.split()
for s in sentence_list:
if s[0] == s[-1]:
same_letter_count += 1
print(same_letter_count)
Explanation:
*The code is in Python.
Initialize the sentence with a string
Initialize the same_letter_count as 0
Split the sentence using split method and set it to the sentence_list
Create a for loop that iterates through the sentence_list. If the first and last of the letters of a string are same, increment the same_letter_count by 1
When the loop is done, print the same_letter_count
Answer: 50
Explanation: let amit's current age = a and armaan's current age be b.
(a-5) = 3* (b-5)
i.e. a= 3b-10 -(i)
10 years later,
(a+10) = 2(b+10)
i.e. a=2b+10 -(ii)
From eqn (i) and (ii),
b=20,
and a=50
Answer:
Visual Basic for Applications runs as an internal programming language in Microsoft Office applications such as Access, Excel, PowerPoint, Publisher, Word, and Visio. VBA allows users to customize beyond what is normally available with MS Office host applications by manipulating graphical-user-interface (GUI) features such as toolbars and menus, dialogue boxes, and forms. You may use VBA to create user-defined functions (UDFs), access Windows application programming interfaces (APIs), and automate specific computer processes and calculations. Macros can automate just about any task—like generating customized charts and reports, and performing word- and data-processing functions. Programmers,like replicating large pieces of code, merging existing program functions, and designing specific languages. VBA can also work in non-Microsoft settings by using a technology called "COM interface," which allows commands to interact across computer boundaries. Many firms have implemented VBA within their own applications, both proprietary and commercial, including AutoCAD, ArcGIS, CATIA, Corel, raw, and SolidWorks.
<em>(Hope this helps/makes sense!)</em>
Answer:
True
Explanation:
Using the t-distribution table, the critical value for a one-tailed test with 6 degrees of freedom and 0.05 significance level is 2.447
Conclusion:
Reject the null hypothesis because the test statistic 2.045 is less than the critical value 2.447
