Answer: Answer below.
Explanation:
I'm not fully sure myself, so don't agree with me fully.
I believe what she may have done wrong is tell the technician about a "program." A program doesn't have to do anything with physical hardware.
Answer:
sectors
Explanation:
hard disk contain tracks which are further divided into sectors, for storage purposes.
Answer:
c. Full Sandbox
d. Partial Sandbox
e. Developer Pro
Explanation:
Universal Containers wants to use a sandbox with real data in it. Which Sandboxes three would you recommend?
a. Test Sandbox
c. Test Sandbox
c. Full Sandbox
d. Partial Sandbox
e. Developer Pro
From the question, we are informed that Universal Containers wants to use a sandbox with real data in it. In this case I will recommend Full Sandbox, Partial Sandbox and Developer Pro.
Sandboxes. In cybersecurity sandbox can be explained as security mechanism that is utilized to separate running programs, and this is an effort utilized to curb system failure as well as software vulnerabilities to disperse.
Sandboxes are crucial when executing suspicious code, it helps to do this so that the host device is is not put to risk of harm. Since, Containers wants to use a sandbox with real data in it then the three types of sandboxes can be use.
✓Full sandboxes allows performance testing as well as staging it can be regarded as copy of production org, and these can be objects attachment and others
✓Partial Copy Sandbox can be allows copying of configuration and part of one's data, in order to allow new configuration testing with one's real data.
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 )
