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Sergio039 [100]
3 years ago
7

The work day has just started and you receive reports that the inventory management server is not accessible on your company's n

etwork. You recall that the new network administration assistant was working on that server last night. Which tool can you use to determine if the network administration assistant left that server's NIC disabled?1. Netnable2. MAC configuration tool3. Server manager4. System window
Computers and Technology
1 answer:
drek231 [11]3 years ago
7 0

Answer:

Option 3 i.e., Server manager is the correct option.

Explanation:

The server manager is the MS windows tool for the purpose to examine and maintain the function of the server and alter the configuration. So, when the user gets the reports of the management server of the inventory or stock which is not usable by the company's server of the following user. Then, he recalls the new admin of the server. The server manager tool should be used by the admin of the server.

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Input 3 positive integers from the terminal, determine if tlrey are valid sides of a triangle. If the sides do form a valid tria
butalik [34]

Answer:

In Python:

side1 = float(input("Side 1: "))

side2 = float(input("Side 2: "))

side3 = float(input("Side 3: "))

if side1>0 and side2>0 and side3>0:

   if side1+side2>=side3 and side2+side3>=side1 and side3+side1>=side2:

       if side1 == side2 == side3:

           print("Equilateral Triangle")

       elif side1 == side2 or side1 == side3 or side2 == side3:

           print("Isosceles Triangle")

       else:

           print("Scalene Triangle")

   else:

       print("Invalid Triangle")

else:

   print("Invalid Triangle")

Explanation:

The next three lines get the input of the sides of the triangle

<em>side1 = float(input("Side 1: ")) </em>

<em>side2 = float(input("Side 2: ")) </em>

<em>side3 = float(input("Side 3: ")) </em>

If all sides are positive

if side1>0 and side2>0 and side3>0:

Check if the sides are valid using the following condition

   if side1+side2>=side3 and side2+side3>=side1 and side3+side1>=side2:

Check if the triangle is equilateral

<em>        if side1 == side2 == side3: </em>

<em>            print("Equilateral Triangle") </em>

Check if the triangle is isosceles

<em>        elif side1 == side2 or side1 == side3 or side2 == side3: </em>

<em>            print("Isosceles Triangle") </em>

Otherwise, it is scalene

<em>        else: </em>

<em>            print("Scalene Triangle") </em>

Print invalid, if the sides do not make a valid triangle

<em>    else: </em>

<em>        print("Invalid Triangle") </em>

Print invalid, if the any of the sides are negative

<em>else: </em>

<em>    print("Invalid Triangle")</em>

4 0
3 years ago
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
Microsoft word's spell checker?
vekshin1
The answer to the given picture is - Microsoft word's spell checker only recognizes very common words. Microsoft word has it's own dictionary which will be compared to the texts on your document. If they see an unfamiliar word they will underline it with red. Anyways you can add your own words and names to the custom dictionary.
8 0
3 years ago
Types in java are divided into two categories. the primitive types are boolean, byte, char, short, int, long, float and double.
Rudik [331]
Types in java are divided into two categories. the primitive types are boolean, byte, char, short, not, long, float, and double. all other types are REFERENCE types
4 0
2 years ago
Technician A says that generic scan tools must be able to read all generic OBD-II codes. Technician B says that all generic scan
Likurg_2 [28]

Answer:

A. Technicians A and B

Explanation:

When we're talking about generic scanners and about all OBD-II codes, in this case, both technician A and B is the correct answer. Because we can scan all OBD-II codes with a generic scan.

But the technician A just says generic tools must be able to read all generic OBD-II codes and technicians B just says generic scan tools must be able to read manufacture OBD-II code, both are correct.

4 0
3 years ago
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