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3 years ago

If, when asked for a date of birth, the user enters a future date, this error should be caught by a ________ check.

Computers and Technology

1 answer:

DiKsa [7]3 years ago

6 0

I believe believe it is a reasonableness check if that doesn't work maybe a date check.

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Write the UPDATE command to increase the commission (column name :COMM) by 500 of all the salesman who have achieved sales (Colu

lara31 [8.8K]

Answer:

UPDATE COMPANY

SET COMM=COMM + 500

WHERE SALES> 20000;

Explanation:

The update command is used for alter the record in the database management system in the table .The update command command modify the record on the some condition in the table in the database management system.

Following are syntax for using the update command .

UPDATE table-name

SET column name 1,column name 2 .........

Where condition

In the given question table -name is "COMPANY" the condition is on the sales column and it update the table only when sales is more than 20000 and set the column COMM by 500.

4 0

3 years ago

Oxnard Casualty wants to ensure that their e-mail server has 99.98 percent reliability. They will use several independent server

sattari [20]

Answer:

The smallest number of servers required is 3 servers

Explanation:

Given

$Reliability = 99.98\%$

$Individual Servers = 95\%$

Required

Minimum number of servers needed

Let p represent the probability that a server is reliable and the probability that it wont be reliable be represented with q

Such that

$p = 95\%\\$

It should be noted that probabilities always add up to 1;

So,

$p + q = 1$ '

Subtract p from both sides

$p - p + q = 1 - p$

$q = 1 - p$

Substitute $p = 95\%\\$

$q = 1 - 95\%$

Convert % to fraction

$q = 1 - \frac{95}{100}$

Convert fraction to decimal

$q = 1 - 0.95$

$q = 0.05$

------------------------------------------------------------------------------------------------

To get an expression for one server

The probabilities of 1 servers having 99.98% reliability is as follows;

$p = 99.98\%$

Recall that probabilities always add up to 1;

So,

$p + q = 1$

Subtract q from both sides

$p + q - q = 1 - q$

$p = 1 - q$

So,

$p = 1 - q = 99.98\%$

$1 - q = 99.98\%$

Let the number of servers be represented with n

The above expression becomes

$1 - q^n = 99.98\%$

Convert percent to fraction

$1 - q^n = \frac{9998}{10000}$

Convert fraction to decimal

$1 - q^n = 0.9998$

Add $q^n$ to both sides

$1 - q^n + q^n= 0.9998 + q^n$

$1 = 0.9998 + q^n$

Subtract 0.9998 from both sides

$1 - 0.9998 = 0.9998 - 0.9998 + q^n$

$1 - 0.9998 = q^n$

$0.0002 = q^n$

Recall that $q = 0.05$

So, the expression becomes

$0.0002 = 0.05^n$

Take Log of both sides

$Log(0.0002) = Log(0.05^n)$

From laws of logarithm $Loga^b = bLoga$

So,

$Log(0.0002) = Log(0.05^n)$ becomes

$Log(0.0002) = nLog(0.05)$

Divide both sides by $Log0.05$

$\frac{Log(0.0002)}{Log(0.05)} = \frac{nLog(0.05)}{Log(0.05)}$

$\frac{Log(0.0002)}{Log(0.05)} = n$

$n = \frac{Log(0.0002)}{Log(0.05)}$

$n = \frac{-3.69897000434}{-1.30102999566}$

$n = 2.84310893421$

$n = 3 (Approximated)$

Hence, the smallest number of servers required is 3 servers

6 0

3 years ago

What is the best way to pay for college?

Virty [35]

Answer:

I think it would be Federal student loans if I am sure sorry if you get it wrong.

Explanation:

3 0

3 years ago

Read 2 more answers

What determines the keystroke to open a cmos editor? how can you find this information?

Brums [2.3K]

Determining the keystrokes of opening the cmos editor depends on the ram contained in the parameters in BIOS.It can be a very daunting task however made more accessible by instructions detailed in various ways by others.Research is needed.

3 0

3 years ago

Need help fast this is do a 4

Stells [14]

Answer:

I believe the answer is B.

8 0

3 years ago