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3 years ago

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If userA wants to send a secure message to userB using an asymmetric cryptographic algorithm, which key does userB use to decryp

t the message?

Computers and Technology

1 answer:

ANTONII [103]3 years ago

4 0

Answer:

userB will use his/her private key.

Explanation:

The steps involved in using asymmetric cryptographic algorithm are as follow:

1. Each user generates a pair of keys to be used for the encryption and decryption of messages.

2. Each user places one of the two keys in a public register or other accessible file. This is the public key. The companion key is kept private.

3. If UserA wishes to send a confidential message to UserB, UserA encrypts the message using UserB's public key.

4. When UserB receives the message, she decrypts it using her private key.

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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 $n = 15\ and\ r = 4$

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

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

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

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

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

$^{15}C_4 = 1365$

<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

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Mars2501 [29]

Answer:

1. The home tab

2. Font group

3. The small arrow at the bottom right of the font command group.

4. On "Font" heading (then "All text" subheading)

Explanation:

The following explanation will help Su in accomplishing her task.

1. Go to the home tab;

2. From the home tab, Su will have access to various command groups but in her case, she needs the Font command group.

3. There's a small arrow located at the bottom left of the font command group section; clicking on this arrow will open the font dialogue.

4. On the font dialogue, there are two headings. The "font" heading and the "character spacing" heading.

In her case, she needs the "Font" heading. The font heading also has various subheadings. To access the underline colour option, she needs the "All text" subheading.

See attachment for picture guide.

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