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Sliva [168]
3 years ago
12

Is a processor is be the same speed as the chips on the motherboard??​

Computers and Technology
2 answers:
Andrei [34K]3 years ago
7 0

Explanation:

A computer's speed and processing power aren't attributable to a single component. It takes a number of pieces of hardware working together to determine your computer's overall performance. The key is how well, and how quickly, all the important components communicate with each other to perform actions.

garik1379 [7]3 years ago
3 0

Answer:

Explanation:

Two of the basic components in any computer are the motherboard and the central processing unit. Both perform processes vital to running the computer's operating system and programs -- the motherboard serves as a base connecting all of the computer's components, while the CPU performs the actual data processing and computing

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Mad Libs are activities that have a person provide various words, which are then used to complete a short story in unexpected (a
allsm [11]

Answer:

The code is below. The output is "Eric went to Chipotle to buy 12 different types of Cars"

Explanation:

import java.util.scanner;

public class labprogram {

    public static void main (string [ ] args) {

    scanner scnr = new scanner (system.in) ;

    string firstname;

    string genericlocation;

    int wholenumber;

    string pluralNoun;

   firstName = scnr.next();

   genericLocation = scnr.next();

   wholeNumner = scnr.nextInt();

   pluralNoun = scnr.nextLine();

   system.output.println(firstname + " went to " + genericlocation + " to buy " + wholenumber + " different types of " + pluralnoun + " . ");

}

 }

6 0
3 years ago
. How does Word indicate a possible instances of inconsistent formatting?: *
REY [17]

Answer:b)With a wavy blue underline

Explanation:Wavy blue lines in the Word document is for representing that the Format consistency checker is in working/on mode .It can identify the inconsistent format instances for the correction while the user is writing the text.

Other options are incorrect because wavy red line is for incorrect spelling of any word and wavy green line signifies the grammatical mistake .The function is available for the format inconsistency as wavy blue line in Word .Thus, the correct option is option(b).

5 0
3 years ago
In cell i5, enter a formula to calculate the total admission fees collected for the 2018 altamonte springs job fair using a mixe
nika2105 [10]

The formula is to calculate the total admission fees collected is = F5*B14

1. Click on cell I5

2. Type equal sign =

3. Type F5 or click on F5 cell

3. Type multiplication sign *

4. Type B14 or click on B14 cell

4 0
4 years ago
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Ben is seeking a control objective framework that is widely accepted around the world and focuses specifically on information se
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Answer:

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3 years ago
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A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co
tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k  ...(1)

probability of 2 hops = k(1-k)  ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

                                                       = (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

                                                       = k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops                = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

     ∞                             n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

    n-1

   = 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet =  k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet =  k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

6 0
3 years ago
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