A star appears under the <br> menu of the word processing program.

Imagine that you are preparing a multimedia presentation. What are the four things you need to consider when getting started?

Marrrta [24]

Explanation:

Consider the Content

Create an Outline

Develop Your Presentation

PRACTICE!!

8 0

3 years ago

Read 2 more answers

ACTIVITY NO. 5 (DAY 5) FACT OR BLUFF. Write Fact if the statement is correct and write Bluff if it is incorrect.

KengaRu [80]

All of the above are Facts.

preventive maintenance does indeed help with the lifespan etc.

If you are feeling tired or fatigued when operating machinery, stop and rest for fear of mistake. So , two is also correct

4if you dress inappropriately it may cause danger from getting caught in machinery or catching a flame depending on the worksite. so four is correct. Newer models are most likely more updated and have different operations then the last machine. so yes read the manual before using it. so yet again 5 is also correct.

hopefully this helped

3 0

3 years ago

What key do I use to start my presentation

Len [333]

Answer:

The answer to this question is given below in this explanation section.

Explanation:

Knowing how o start a presentation is crucial if you fail to capture the audience attention right of the bat your entire presentation will flop.Few listeners will stick with you to the end and retain what you have told.

That is mildly unpleasant when you are doing an in house presentation in front of your colleagues.But can become utterly embrassing when you are presenting in front of larger audience presentation to prospective customers.

Let you have most of your presentation slides and polished up you dont check and quick and effective power point presentation design tips.Many ideas are broken into bits sized statements for your slides and complemented with visuals.The best way is to appeal and invoke certain emotions with your audience curiosity,surprise fear or good old amusements.

8 0

3 years ago

Suppose we have two String objects and treat the characters in each string from beginning to end in the following way: With one

Simora [160]

Answer:

If all the character pairs match after processing both strings, one string in stack and the other in queue, then this means one string is the reverse of the other.

Explanation:

Lets take an example of two strings abc and cba which are reverse of each other.

string1 = abc

string2 = cba

Now push the characters of string1 in stack. Stack is a LIFO (last in first out) data structure which means the character pushed in the last in stack is popped first.

Push abc each character on a stack in the following order.

c

b

a

Now add each character of string2 in queue. Queue is a FIFO (first in first out) data structure which means the character inserted first is removed first.

Insert cba each character on a stack in the following order.

a b c

First c is added to queue then b and then a.

Now lets pop one character from the stack and remove one character from queue and compare each pair of characters of both the strings to each other.

First from stack c is popped as per LIFO and c is removed from queue as per FIFO. Then these two characters are compared. They both match

c=c. Next b is popped from stack and b is removed from queue and these characters match too. At the end a is popped from the stack and a is removed from queue and they both are compared. They too match which shows that string1 and string2 which are reverse of each other are matched.

6 0

3 years ago

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

A star appears under the menu of the word processing program.