Which team behavioral characteristic helps team members to freely express ideas and foster approachability?

How to know your pc does not have bluetooth?

sattari [20]

Assuming you're running Windows, click the start button, and then search for "Device Manager" (or open CMD or PowerShell and type devmgmt.msc). Look at the list of devices. If Bluetooth is there, you have it; if it's not there, you don't.

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 5 cards of type 1, 4 cards of type 2, 2 cards of type 3 and 2 cards of type 4 from a set of 13 cards.

Explanation:

The crucial point 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 combination 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 combinations is $\\ \frac{n!}{(n-k)!k!}$ , where n is the total of elements available and k is the size of a selection of k elements from which we can choose from the total n.

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 Multiplication Principle 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 multiplication of all them.

5 0

4 years ago

Is it possible to build something that doesn't exist yet?

Anastasy [175]

Answer:

yes

Explanation:

how do u think other thing were built like when phones were first made

5 0

4 years ago

Your auto insurance policy has a $200 monthly premium and $700 deductible. what is the maximum amount you will have to pay out o

saul85 [17]

Your monthly premium is the payment you make to your car insurance company that keeps your coverage active.

A car insurance deductible is the amount of money you have to pay toward repairs before your insurance covers the rest.

Therefore $700 is the maximum amount you will have to pay
out of pocket for a car accident before your insurance covers your cost.

4 0

3 years ago

which term describes the layer of software that resides between the virtual operating system and the physical hardware it runs o

sasho [114]

Answer:

hypervisor

Explanation:

becauseis a thin layer of software that resides between the virtual operating system(s) and the hardware. The physical host is the actual hardware that the hypervisor software runs on.

8 0

3 years ago