Answer:
{14,23,654,0,9}
Step-by-step explanation:
The first three contain negative numbers and decimal points therefor the last one {14,23,654,0,9} is correct.
Here we must see in how many different ways we can select 2 students from the 3 clubs, such that the students <em>do not belong to the same club. </em>We will see that there are 110 different ways in which 2 students from different clubs can be selected.
So there are 3 clubs:
- Club A, with 10 students.
- Club B, with 4 students.
- Club C, with 5 students.
The possible combinations of 2 students from different clubs are
- Club A with club B
- Club A with club C
- Club B with club C.
The number of combinations for each of these is given by the product between the number of students in the club, so we get:
- Club A with club B: 10*4 = 40
- Club A with club C: 10*5 = 50
- Club B with club C. 4*5 = 20
For a total of 40 + 50 + 20 = 110 different combinations.
This means that there are 110 different ways in which 2 students from different clubs can be selected.
If you want to learn more about combination and selections, you can read:
brainly.com/question/251701
if you take the radius of the first and second sphere you find that they are doubled there fore the volume changed at a rate of 2 squared because 2 is what the radius was multiplied by going from the second sphere to the first so you would square what the sphere was multiplied by there fore making your answer: the volume is compares with 2 squared
Answer:
f(x) = -1/x+5 +2
Step-by-step explanation:
