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faltersainse [42]

4 years ago

5

Which of the following best explains why some people invest their saving in the stock market and others put their saving in bank

accounts

Computers and Technology

2 answers:

Tanya [424]4 years ago

7 0

The answer to this question is B. Some people feel that the stock market is too risky for them.

dusya [7]4 years ago

5 0

Answer:

B a P e x proven

Explanation:

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Select which is true for for loop

uranmaximum [27]

Answer:

i dont understand what you mean and what you are asking in the qestion

Explanation:

7 0

3 years ago

Compare pseudocode and flowcharts for designing computer programs, and discuss the advantages and disadvantages of each.

Zarrin [17]

Answer:

<h2 /><h2>Pseudocode</h2>

<h3>Advantages of Pseudocode:</h3>

Easier to translate into a high level programming language
No pressure of syntax (grammar) of the coding
Partially resembles standard English so it's easier for programmers to understand

<h3>Disadvantages of Pseudocode:</h3>

Can't immediately spot errors in coding such as logic errors
Runtime errors don't exist in Pseudocode

<h2>Flowcharts</h2><h3 /><h3>Advantages of Flowcharts</h3>

Focuses on the logic of the program
Displays the manipulation (and flow) of data easily

<h3>Disadvantages of Flowcharts</h3>

Not ideal for big programs (only for subprograms)
Shapes may not be clear or obvious to what they are in terms of their functions

8 0

4 years ago

Describe how computer are used in ticket counter?

m_a_m_a [10]

Answer:

These systems are commonly used in facilities such as public libraries to ensure equitable use of limited numbers of computers. Bookings may be done over the internet or within the library itself using a separate computer set up as a booking terminal.

6 0

3 years ago

Fromwhich object do you ask for DatabaseMetaData?ConnectionResultSetDriverManagerDriver

icang [17]

Answer:

Connection

Explanation:

We can get DatabaseMetaData from the java.sql.Connection object.

The relevant API call is Connection.getMetaData().

This returns an object of the type DatabaseMetaData.

Some of the methods available as part of the DatabaseMetaData object are:

getDatabaseMajorVersion()
getDatabaseMinorVersion()
getDriverName()
getDriverVersion()
isReadOnly()
getCatalogs()
getTables()
supportsBatchUpdate()
supportsUnion()

8 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

4 years ago