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strojnjashka [21]

3 years ago

15

____ contain instructions for the OS for hardware devices, such as the keyboard, mouse, and video card, and are stored in the sy

stemroot\Windows\System32\Drivers folder.

Computers and Technology

1 answer:

AnnyKZ [126]3 years ago

3 0

The Motherboard contains instructions for the OS for hardware devices, such as the keyboard, mouse, and video card, and are stored in the systemroot\Windows\System32\Drivers folder.

Learned this in middle school, your welcome ;)

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Masja [62]

I'm not too experienced with code including external memory, but i believe there was an error with the code mixing that either mixed lines of code up or got rid of code to save files onto your device. Try looking up code for the device, see if that helps at all.

6 0

2 years ago

The 3rd generation programming language that most students learned when most computers used MS DOS was ___ . It remains a safe p

Diano4ka-milaya [45]

Answer:

Basic

Explanation:

The 3rd generation programming language that most students learned when most computers used MS DOS was basic. It remains a safe programming language for using 3rd party code and provides a coding standard that integrates easily with other programming languages.

7 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

2 years ago

Java Homework:(The Person, Student, Employee, Faculty, and Staff classes) Design a class named Person and its two subclasses nam

miss Akunina [59]

Answer: Wait, so what is the question exactly..?

8 0

3 years ago

In some video games, the player cannot obtain the reward without doing what with something that they already have?

Crank

Answer:

risking

Explanation:

8 0

2 years ago