In access, all the tables, reports, forms, and queries that are created are stored in a single file called a ___

The network architecture component that is a special LAN with a group of servers that enables electronic data exchange of betwee

Jlenok [28]

Answer:

E-Commerce edge

Explanation:

E-Commerce edge can be defined as a special LAN with a group of servers that enables electronic data exchange between the organization and the external entities with which it does business such as its customers and suppliers.

It facilitates faster, better, more secure and more convenient data exchange between an organization and the business it have transactions with.

4 0

3 years ago

All of the following are typical characteristics of internet predators, except white. male. between the ages of 18 and 35. high

tigry1 [53]

The one that does not belong there is high income.

7 0

3 years ago

Write a recursive method called repeat that accepts a string s and an integer n as parameters and that returns s concatenated to

svp [43]

Answer:

public static String repeat(String text, int repeatCount) {

if(repeatCount < 0) {

throw new IllegalArgumentException("repeat count should be either 0 or a positive value");

}

if(repeatCount == 0) {

return "";

} else {

return text + repeat(text, repeatCount-1);

}

Explanation:

Here repeatCount is an int value.

at first we will check if repeatCount is non negative number and if it is code will throw exception.

If the value is 0 then we will return ""

If the value is >0 then recursive function is called again untill the repeatCount value is 0.

6 0

3 years ago

The running time of Algorithm A is (1/4) n2+ 1300, and the running time of another Algorithm B for solving the same problem is 1

Mnenie [13.5K]

Answer:

Answer is explained below

Explanation:

The running time is measured in terms of complexity classes generally expressed in an upper bound notation called the big-Oh ( "O" ) notation. We need to find the upper bound to the running time of both the algorithms and then we may compare the worst case complexities, it is also important to note that the complexity analysis holds true (and valid) for large input sizes, so, for inputs with smaller sizes, an algorithm with higher complexity class may outperform the one with lower complexity class i.e, efficiency of an algorithm may vary in cases where input sizes are smaller & more efficient algorithm might be outperformed by the lesser efficient algorithms in those cases.

That's the reason why we consider inputs of larger sizes when comparing the complexity classes of the respective algorithms under consideration.

Now coming to our question for algorithm A, we have,

let F(n) = 1/4x² + 1300

So, we can tell the upper bound to the function O(F(x)) = g(x) = x2

Also for algorithm B, we have,

let F(x) = 112x - 8

So, we can tell the upper bound to the function O(F(x)) = g(x) = x

Clearly, algorithmic complexity of algorithm A > algorithmic complexity of algorithm B

Hence we can say that for sufficiently large inputs , algorithm B will be a better choice.

Now to find the exact location of the graph in which algorithmic complexity for algorithm B becomes lesser than

algorithm A.

We need to find the intersection point of the given two equations by solving them:

We have the 2 equations as follows:

y = F(x) = 1/4x² + 1300 __(1)

y = F(X) = 112x - 8 __(2)

Let's put the value of from (2) in (1)

=> 112x - 8 = 1/4x² + 1300

=> 112x - 0.25x² = 1308

=> 0.25x² - 112x + 1308 = 0

Solving, we have

=> x = (112 ± 106) / 0.5

=> x = 436, 12

We can obtain the value for y by putting x in any of the equation:

At x=12 , y= 1336

At x = 436 , y = 48824

So we have two intersections at point (12,1336) & (436, 48824)

So before first intersection, the

Function F(x) = 112x - 8 takes lower value before x=12

& F(x) = 1/4x² + 1300 takes lower value between (12, 436)

& F(x) = 112x - 8 again takes lower value after (436,∞)

Hence,

We should choose Algorithm B for input sizes lesser than 12

& Algorithm A for input sizes between (12,436)

& Algorithm B for input sizes greater than (436,∞)

8 0

3 years ago

The valence electron configurations of several atoms are shown. how many bonds can each atom make without hybridization? 2s^2 2p

Pie

Answer: This element needs 3 electrons to complete its octet state

Why?:
since the electrons in 2s2 are already paired they cannot form a bond pair with electrons from other elements without hybridisation however the 3 electrons in 2p3 are unpaired so can form bond pairs with electrons from other elements without hybridisation.

Hope this helps!

3 0

4 years ago

In access, all the tables, reports, forms, and queries that are created are stored in a single file called a ____