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Alexxandr [17]
3 years ago
13

Which of these categories of computer software can add thousands of dollars to the cost of a computer system?

Computers and Technology
1 answer:
Radda [10]3 years ago
6 0
Specialized software. hope this helps
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From what I know about landscapers and from the description that makes the most sense.

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Which feature of virtualisation helps increase the IT productivity of a business
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because

8 0
3 years ago
How do you do this question?
elena55 [62]

Answer:

for (scores.Entry<String, Integer> entry : scores.entrySet()) {

          int value=0; String key="Rambo";

          String k = entry.getKey();

           int v = entry.getValue();

           if (v>value)

           {

               value =v;

           }

           else

           {

               value=value;

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       }

       for( scores.Entry<String, Integer> entry : scores.entrySet())

      {

        if( entry.getValue()==value)            

        System.out.println("Key = " + entry.getKey() + ", Value = " + entry.getValue());

      }

Explanation:

The above scores are an object of Map type.  And this is a parameter of findUopStudent function. So scores is a Map, and entry is an item. And we get its key and value. Now we compare each value using senteniel method and find the maximum value. Now we iterate through the scores using for the look and check the value of each entry against the maximum value found, and we print out the entry with maximum value. And that is the top student, which is required.  

4 0
2 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
2 years ago
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