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trapecia [35]
1 year ago
13

You need to trace the route that a cat 6 utp cable takes through the ceiling and walls of your building. which tool should you u

se?
Computers and Technology
1 answer:
Veronika [31]1 year ago
4 0

We should use Tone Generator to to trace the route that a cat 6 utp cable takes through the ceiling and walls of your building. The tone generator can be used to test the speakers and their electrical wiring, as well as to find the frequency of hearing loss and tinnitus.

A sinusoidal signal is produced by a tone generator at the selected frequency. Can be used as a learning tool for physics as well as to test audio equipment like speakers or earbuds. You may also test the frequency range of tablet speakers.

Learn more about tone generator brainly.com/question/28017740

#SPJ4

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Which are benefits of modeling a solution? Choose all that apply. allows you to be sure the problem and solution are understood
labwork [276]

Answer:

A,B,D

A. Allows you to be sure the problem and solution are understood

B. Presents the solution in a manner that is easy to share

C. Is a starting point for all other activities

4 0
4 years ago
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Which tab automatically becomes available after inserting a text box? Drawing Tools Insert Text Box Tools Shape Tools
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Answer:

insert text box

Explanation:

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3 years ago
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A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st
Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

^nC_r = \frac{n!}{(n-r)!r!}

Where n = 15\ and\ r = 4

^{15}C_4 = \frac{15!}{(15-4)!4!}

^{15}C_4 = \frac{15!}{11!4!}

^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}

^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}

^{15}C_4 = \frac{32760}{24}

^{15}C_4 = 1365

<em>Hence, there are 1365 ways </em>

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

p = 3/15

p = 0.2

q = 1 - p

q = 1 - 0.2

q = 0.8

Solving further using binomial;

(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

Probability =  ^nC_3pq^3

Substitute 4 for n, 0.2 for p and 0.8 for q

Probability =  ^4C_3 * 0.2 * 0.8^3

Probability =  \frac{4!}{3!1!} * 0.2 * 0.8^3

Probability = 4 * 0.2 * 0.8^3

Probability = 0.4096

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

Probability =  q^n

Substitute 4 for n and 0.8 for q

Probability =  0.8^4

Probability = 0.4096

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

8 0
3 years ago
What is a coverage map used for
erastova [34]

Answer/Explanation:

A coverage map shows how much land something takes up or reaches to.

4 0
3 years ago
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Two systems are connected by a router. Both systems and the router have transmission rates of 1,000bps. Each link has a propagat
Lisa [10]

Answer:

Two systems are connected by a router. Both systems and the router have transmission rates of 1,000bps. Each link has a propagation delay of 10ms. Also, it takes router 2ms in order to process the packet (e.g. decide where to forward it). Suppose the first system wants to send a 10,000 bit packet to the second system. How long will it take before receiver system receives the entire packet.

Transmission time for first Router = 10,000 bits / 1000 bps = 10 seconds

Receiving time for seond route r= 10,000 bits / 1000 bps = 10 seconds

Propagation delay = 10ms = .01 seconds x 2 for two delays = .02 seconds

First router 2ms to process = .002 seconds

Add all the times together and we get 20.022 seconds which is the same as or 20 seconds and 22 ms

4 0
3 years ago
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