Answer:
The correct Answer is 0.0571
Explanation:
53% of U.S. households have a PCs.
So, P(Having personal computer) = p = 0.53
Sample size(n) = 250
np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10
So, we can just estimate binomial distribution to normal distribution
Mean of proportion(p) = 0.53
Standard error of proportion(SE) = 
= 
= 0.0316
For x = 120, sample proportion(p) = 
= 
= 0.48
So, likelihood that fewer than 120 have a PC
= P(x < 120)
= P( p^ < 0.48 )
= P(z < 
) (z=
)
= P(z < -1.58)
= 0.0571 ( From normal table )
Answer:
The answer is (a).Presentation Layer
Explanation:
This layer is located at the 6th level of the OSI model, responsible for delivering and formatting the information to the upper layer for further processing. This service is needed because different types of computer uses different data representation. The presentation layer handles issues related to data presentation and transport, including encryption,translation and compression.
<h2><u>Given</u> :-</h2>
Create a HTML code for writing a web page for your school time-table.
<h2><u>Answer</u> :-</h2>
<html>
<head>
<title> Time Table </title>
</head>
<hrcolor = "red">
<hrcolor = "blue">
<body>
<table><tr = "7"><md = "7"><Period 1_></table>
</body>
</html>
Answered by - ItzMaster
Question 1: Best way is option 1, Worst way is option 4.
Question 2: Best way is option 4, Worst way is option 3
