Ask question

Ask question

All categories

4 years ago

13

What would happen if a pc that contains a power supply that does not automatically adjust for input voltage is set to 230 volts

and attaches to an outlet in the united states?

Computers and Technology

1 answer:

frutty [35]4 years ago

3 0

It would be underpowered which results in poor performance or won't turn on

You might be interested in

Which subexpression will be solved first in the given statement?

Crazy boy [7]

Answer:

If you dont buy terraria than im going to make mrbeast buy it for you

Explanation:

reeeeeeeeeeeeeeeeeeeee

4 0

4 years ago

Select the correct text in the passage.

aliina [53]

Answer: When was admitted, hospital authorities recorded his medical history. Then, placed in an ICU where his vital signs were constantly monitored.

Explanation: With the help of computers, medical histories are often kept in the computer for future reference. Machines are connected to computers to record vital signs.

4 0

3 years ago

A person gets 13 cards of a deck. Let us call for simplicity the types of cards by 1,2,3,4. In How many ways can we choose 13 ca

natima [27]

Answer:

There are 5,598,527,220 ways to choose <em>5</em> cards of type 1, <em>4 </em>cards<em> </em>of type 2, <em>2</em> cards of type 3 and <em>2</em> cards of type 4 from a set of 13 cards.

Explanation:

The <em>crucial point</em> of this problem is to understand the possible ways of choosing any type of card from the 13-card deck.

This is a problem of <em>combination</em> since the order of choosing them does not matter here, that is, the important fact is the number of cards of type 1, 2, 3 or 4 we can get, no matter the order that they appear after choosing them.

So, the question for each type of card that we need to answer here is, how many ways are there of choosing 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 are of type 4 from the deck of 13 cards?

The mathematical formula for <em>combinations</em> is $\\ \frac{n!}{(n-k)!k!}$ , where <em>n</em> is the total of elements available and <em>k </em>is the size of a selection of <em>k</em> elements from which we can choose from the total <em>n</em>.

Then,

Choosing 5 cards of type 1 from a 13-card deck:

$\frac{n!}{(n-k)!k!} = \frac{13!}{(13-5)!5!} = \frac{13*12*11*10*9*8!}{8!*5!} = \frac{13*12*11*10*9}{5*4*3*2*1} = 1,287$ , since $\\ \frac{8!}{8!} = 1$ .

Choosing 4 cards of type 2 from a 13-card deck:

$\\ \frac{n!}{(n-k)!k!} = \frac{13!}{(13-4)!4!} = \frac{13*12*11*10*9!}{9!4!} = \frac{13*12*11*10}{4!}= 715$ , since $\\ \frac{9!}{9!} = 1$ .

Choosing 2 cards of type 3 from a 13-card deck:

$\\ \frac{n!}{(n-k)!k!} =\frac{13!}{(13-2)!2!} = \frac{13*12*11!}{11!2!} = \frac{13*12}{2!} = 78$ , since $\\ \frac{11!}{11!}=1$ .

Choosing 2 cards of type 4 from a 13-card deck:

It is the same answer of the previous result, since

$\\ \frac{n!}{(n-k)!k!} = \frac{13!}{(13-2)!2!} = 78$ .

We still need to make use of the <em>Multiplication Principle</em> to get the final result, that is, the ways of having 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 cards of type 4 is the multiplication of each case already obtained.

So, the answer about how many ways can we choose 13 cards so that there are 5 of type 1, there are 4 of type 2, there are 2 of type 3 and there are 2 of type 4 is:

1287 * 715 * 78 * 78 = 5,598,527,220 ways of doing that (or almost 6 thousand million ways).

In other words, there are 1287 ways of choosing 5 cards of type 1 from a set of 13 cards, 715 ways of choosing 4 cards of type 2 from a set of 13 cards and 78 ways of choosing 2 cards of type 3 and 2 cards of type 4, respectively, but having all these events at once is the <em>multiplication</em> of all them.

5 0

3 years ago

Which tool is a GUI used to build PowerShell scripts?

enyata [817]

A will be your answer!!! I’m pretty sure

8 0

4 years ago

All of the following can cause a fatal execution-time error except: Group of answer choices Dereferencing a pointer that has not

7nadin3 [17]

Answer: Dereferencing a variable that is not a pointer

Explanation:

The execution time also refered to as the CPU time pertaining to a given task is the time that is used by the system to execute a task.

Some of the reasons for a fatal execution-time error include:

• Dereferencing a pointer that has not been initialized properly

• Dereferencing a null pointer

• Dereferencing a pointer that has not been assigned to point to a specific address.

It should be noted that dereferencing a variable that is not a pointer doesn't cause a fatal execution-time error.

6 0

3 years ago