The most popular input device of a computer is a(n) ___

_____ refers to unsolicited commercial emails, usually sent to a large number of people with little regard to the users interest

frozen [14]

Spam refers to large unsolicited email irregardless of the recipients views on the matter

5 0

3 years ago

What is the big-O performance estimate of the following function?

larisa [96]

Answer:

The time complexity of the code is O(log₇n).

Explanation:

The i is updated by 7*i.On each iteration i is multiplied by 7.So on finding the time complexity of the code given above it will come out to be log base 7.

When we divide the input by 2 the time complexity is log base 2.

So on dividing it by 7 we get the time complexity of log base 7.

8 0

3 years ago

Read 2 more answers

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

2 years ago

With landscape photography, which depth of field do you normally want? A) Small B)Small to Medium C)Large D)With landscapes it d

RideAnS [48]

I am not 100% on this but i think you would want a Large depth of field.

3 0

3 years ago

Read 2 more answers

What is the command to disable any Processes in linix??

stiks02 [169]

Answer: killall[process_name] or kill[PID]

Explanation:

Killall is a tool for disabling running processes on the system. It will disable all programs that matches the name mentioned.

kill disables processes based on process id numbers. it does not disable the process directly. The process recieves a signal where the process will follow instructions which it has to follow if it receives the signal.

7 0

2 years ago

The most popular input device of a computer is a(n) ____.