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marta [7]
3 years ago
12

Jim wants to buy a car, he’ll probably only need it for a couple of years he has a short commute to work so he won’t put many mi

les on his vehicle what they best transportation option for him? Yes
Computers and Technology
1 answer:
Papessa [141]3 years ago
4 0
Is there more information? or options?
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If I remember correctly, I believe it's <body></body>
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Are the blank space around the edges of the page
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Answer:

The blank space around the edges of a sheet of paper — as with the page of a book — that surrounds the text is called the margin.

8 0
3 years ago
"what is the problem with using challenge handshake authentication protocol (chap) as an authentication protocol?"
Alexxandr [17]

Its use of the message digest 5 (MD5) hash algorithm for security.

CHAP uses a combination of MD5 hashing and a challenge-response mechanism, and authenticates without sending passwords as plaintext over the network. The security of the MD5 hash function is severely compromised.

5 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
3 years ago
Use the following data definitions data myBytes BYTE 10h,20h,30h,40h myWords WORD 3 DUP(?),2000h myString BYTE "ABCDE" What will
Elena-2011 [213]

Answer:

<em>Given Data:</em>

<em>myBytes BYTE 10h, 20h, 30h, 40h </em>

<em>myWords WORD 3 DUP(?), 2000h </em>

<em>myString BYTE "ABCDE"</em>

<em />

Based on the data that we are given we can conclude that:

(a).     a. EAX = 1

         b. EAX = 4

         c. EAX = 4

         d. EAX = 2

         e. EAX = 4

         f. EAX = 8

         g. EAX = 5

8 0
3 years ago
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