Answer:
It is not possible.
Explanation:
In this example, we need to accommodate 473 computers for six clients that are 473 IP addresses.
For this request just we have /22 IPv4 address blocks, this mean
22 red bits 11111111111111111111110000000000 <--- 10 host bits
We must increase red bits to 25, we need these 3 bits to create 6 sub red, in this case, 2^3 = 8 sub red.
Why did we ask 3 bits? Because if we ask only 2, 2^2 = 4, and we need 6 sub red.
25 red bits 11111111111111111111111110000000 7 host bits
In this case, we need more than 260 computers, but just we have 7 bits, this means.
2^7 = 128 and just one customer needs 260, for that is impossible.
Answer:
b
Explanation:
C and D have equivalent iterations
C: D:
99 990
90 900
81 810
72 720
63 630
54 540
45 450
36 360
27 270
18 180
9 90
Answer:
a is the correct answer
Explanation:
correct me if I'm wrong hope it's help thanks
Answer:
Here’s one!
Given [math]R[/math], the radius of the circle.
Let [math]N,D\leftarrow 0[/math]
Repeat until [math]D[/math] is large enough (about 1,000,000)
[math]x,y\leftarrow U[0,1][/math]
If [math]x^2 + y^2\le 1[/math] then [math]N\leftarrow N+1[/math]
[math]D\leftarrow D+1[/math]
[math]P\leftarrow\frac{8NR}{D}[/math]
Return [math]P[/math]
[math]U[0,1][/math] is a uniform random number in the range [math][0,1][/math].
Explanation:
It really depends what device you're on.
For example, my wifes laptop shows a lock screen, then it shows a sign in screen.
My laptop shows ONLY the sign in screen.
My old laptop shows a sign in screen with the date and time. So I dunno man