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3 years ago

Which one of these is NOT a successful budgeting strategy?

Computers and Technology

2 answers:

Elina [12.6K]3 years ago

7 0

Answer D isn't a successful budgeting strategy.

Jet001 [13]3 years ago

6 0

I think it's hope it helps

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You are configuring IP settings on a new network. For the external interfaces, you decide to obtain registered IP addresses from

frez [133]

Answer:

131.16.0.0 to 131.16.255.255

Explanation:

I wouldn't consider 131.16.0.0 to 131.16.255.255 because;

the 131.16 range is from the Class B range. This males it not one of the recognized private IP address ranges.

All the other address however, fall within valid private IP address ranges.

Cheers

3 0

3 years ago

Using complete sentences post a detailed response to the following.

lesya [120]

Answer:

an interest group

Explanation:

When designing a video game I believe that the best target group would be an interest group. Focusing on a specific interest group will allow you to design a game, knowing that there are people truly passionate about that interest and who will play the game. Targeting an age group makes things difficult because you have to limit your creativity to that age group and even still there may not be much interest in that age group for your category of game. Therefore, an interest group is the best option and gives you the freedom to unlock your creativity without having to limit your game due to age concerns.

7 0

3 years ago

What method is used in the following line of code to associate clicking the left button with the event handler "clickSingle"?

Olegator [25]

Answer:

Bind

Explanation:

widget.bind('<Button-1>', clickSingle)

7 0

2 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

A citizen of any group both

erastovalidia [21]

Answer:

Rights and Responsibilities

Explanation:

4 0

2 years ago