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:
they are not its just that people haven't changed their profile picture. i haven't yet lol
Explanation:
Answer:
JTextField
Explanation:
JTextField is a Swing control which can be used for user input or display of output. It corresponds to a textfield and can be included with other controls such as JLabel, JButton,JList etc. on a comprehensive user interface which is application dependent. JTextField supports a read-write mode where the user can enter a value and it also supports a read-only mode which can be used to display non-editable output to the user.