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slava [35]
3 years ago
15

Which of the following SQL statements will display all customers who have not recently placed an order? a. SELECT customer# FROM

customers UNION b. SELECT customer# FROM orders; SELECT customer# FROM orders MINUS c. SELECT customer# FROM customers; SELECT customer# FROM orders INTERSECT d. SELECT customer# FROM customers; SELECT customer# FROM customers MINUS
Computers and Technology
1 answer:
JulsSmile [24]3 years ago
3 0

Answer:

The correct option is option 4 from the options indicated below:

Explanation:

The options are as given below

1-SELECT customer# FROM customers

UNION

SELECT customer# FROM orders;

2-SELECT customer# FROM orders

MINUS

SELECT customer# FROM customers;

3-SELECT customer# FROM orders

INTERSECT

SELECT customer# FROM customers;

4-SELECT customer# FROM customers

MINUS

SELECT customer# FROM orders;

Option 1 is not correct as it will provide the list of all the customers whether they have placed order or not.

Option 2 is not correct as it will provide the list of customers who have placed order but are not in the customers table. This will return the empty records.

Option 3 is not correct as it will provide the list of customers who have placed the order.

Option 4 is correct as it will provide the list of customers from the CUSTOMERS table which are not in the Orders table. thus the list of customers who have not placed the order yet.

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A signal has a wavelength of 1 11m in air. How far can the front of the wave travel during 1000 periods?
olchik [2.2K]

Answer:

A signal has a wavelength of 1 μm in air

Explanation:

looked it up

6 0
2 years ago
Which cpu type would most likely be found in a smartphone?.
dlinn [17]

Answer:

dual-core (two), quad-core (four) and octa-core (eight)

Explanation:

These are the most common according to my research.

If I helped you, can I please have brainliest?

Have a great day/night!

4 0
2 years ago
You load an image file into a numpy array and look at its shape, which is (433, 650). What does this indicate?
Nat2105 [25]

Answer:

option C

Explanation:

The correct answer is option C

the uploaded image shape is ( 433 , 650 )

this shape means that the image is a grayscale image which is 433 pixels high by 650 pixels wide.

a gray scale image is  in white and black color.

433 pixels high by 650 pixel wide means that the image is formed with the combination of 433 vertical dots and 650 horizontal dots.

Resolution of an image can be found out by the  pixels present in the images.

higher the pixel higher is he resolution of the image.

6 0
2 years ago
The problem with the media giving equal air time to those who are __________ about the effects of media violence on violent beha
WINSTONCH [101]

The problem with the media giving equal air time to those who are <u>not worthy</u> about the effects of media violence on violent behaviour is that the public is then misled into thinking that the evidence for such effects is <u>strong and unbiased</u> than it actually is.

<u>Explanation:</u>

Media has an extensive impact on the public. Whatever they air reaches each and every citizen of the specified region through direct and indirect means.

The media should take precautions and make sure whatever they are showing is unbiased and supported by true facts, figures, and objective evidence.

5 0
2 years ago
Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth
telo118 [61]
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.

public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}


So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.

A quick explanation:

We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.

The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.


4 0
3 years ago
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