When pointed over text within a paragraph, the cursor takes the shape of a/an

____________ occurs when a provider does not support data export or when a provider's service is unavailable through others.

elixir [45]

Answer:

The correct answer to the following question will be Vendor Lock-In.

Explanation:

Vendor Lock-In: It is also known as Customer Lock-In. The Vendor Lock-In makes the costumer depends on services and products on the vendor. The costumers are not able to use another vendor without changing costs as it creates barriers.

Some ways to avoid Vendor Lock-In, these are as follows:

Design your application portable.
Keep watching vendor contracts.
Arrange both entry and exit with your vendor.

8 0

3 years ago

How do professionals address their problems?

8090 [49]

Answer:

b, with grace and maturity

6 0

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 $n = 15\ and\ r = 4$

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

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

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

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

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

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

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)

From the given parameters; 3 out of 15 is detective;

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;

Probability that at least one is defective + Probability that at none is defective = 1

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

Why do amusement parks continue to build new roller coasters?

kykrilka [37]

They do it because they want people to experience new rides, instead of the same old boring rides they’ve been on several times.

5 0

4 years ago

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Since the 1960s, there have been five major generations, or cycles, of computing. Which represents those generations in order?

sergiy2304 [10]

Answer:a) Mainframes, miniframes, PCs, desktop Internet computing, mobile computing

Explanation: The hierarchy of the computers or computing are in the generation is:-

Mainframe:-designed for the execution of huge data processing , storage and execution.
Miniframe:-has similar functioning as the mainframe but on smaller scale and version.
Personal computer(PC):-It is designed for the individual purpose and according the need of the user.
Mobile computing:-computing that can be done on the mobile phone similar to the computing on the personal computer
Internet computing:-the computing of the mobile and computer system with the facility of the internet connectivity.

7 0

3 years ago