Answer: The correct number of balls is (b) 4.
Step-by-step explanation: Given that a single winner is to be chosen in a random draw designed for 210 participants. Also, there is an equal probability of winning for each participant.
We are using 10 balls, numbered through 0 to 9. We are to find the number of balls which needs to be picked up, regardless of order, so that each of the 210 participants can be assigned a unique set of numbers.
Let 'r' represents the number of balls to be picked up.
Since we are choosing from 10 balls, so we must have
The value of 'r' can be any one of 0, 1, 2, . . , 10.
Now,
if r = 1, then
If r = 2, then
If r = 3, then
If r = 4, then
Therefore, we need to pick 4 balls so that each participant can be assigned a unique set of numbers.
Thus, (b) is the correct option.
Answer:
Step-by-step explanation:
Given data
Total units = 250
Current occupants = 223
Rent per unit = 892 slips of Gold-Pressed latinum
Current rent = 892 x 223 =198,916 slips of Gold-Pressed latinum
After increase in the rent, then the rent function becomes
Let us conside 'y' is increased in amount of rent
Then occupants left will be [223 - y]
Rent = [892 + 2y][223 - y] = R[y]
To maximize rent =
Since 'y' comes in negative, the owner must decrease his rent to maximixe profit.
Since there are only 250 units available;
![y=-250+223=-27\\\\maximum \,profit =[892+2(-27)][223+27]\\=838 * 250\\=838\,for\,250\,units](https://tex.z-dn.net/?f=y%3D-250%2B223%3D-27%5C%5C%5C%5Cmaximum%20%5C%2Cprofit%20%3D%5B892%2B2%28-27%29%5D%5B223%2B27%5D%5C%5C%3D838%20%2A%20250%5C%5C%3D838%5C%2Cfor%5C%2C250%5C%2Cunits)
Optimal rent - 838 slips of Gold-Pressed latinum
Answer:
$154,750 ~ APEX
Step-by-step explanation:
so basically ummmmmm well I guessed and got it right so hope this helps
Answer:
Two distinct concentric circles: 0 max solutions
Two distinct parabolas: 4 max solutions
A line and a circle: 2 max solution
A parabola and a circle: 4 max solutions
Step-by-step explanation:
<u>Two distinct concentric circles:</u>
The maximum number of solutions (intersections points) 2 distinct circles can have is 0. You can see an example of it in the first picture attached. The two circles are shown side to side for clarity, but when they will be concentric, they will have same center and they will be superimposed. So there can be ZERO max solutions for that.
<u>Two distinct parabolas:</u>
The maximum solutions (intersection points) 2 distinct parabolas can have is 4. This is shown in the second picture attached. <em>This occurs when two parabolas and in perpendicular orientation to each other. </em>
<u>A line and a circle:</u>
The maximum solutions (intersection points) a line and a circle can have is 2. See an example in the third picture attached.
<u>A parabola and a circle:</u>
The maximum solutions (intersection points) a parabola and a circle can have is 4. If the parabola is <em>compressed enough than the diameter of the circle</em>, there can be max 4 intersection points. See the fourth picture attached as an example.
Answer:
On the first hour she gathered 8 mushroons, on the second 15 mushroons and on the third she gathered 14 mushroons.
Step-by-step explanation:
Helen gathred a total of 37 mushroons in three hours. On the first hour she gathered "x" mushroons, on the second hour she gathered "x + 7" mushroons and on the third hour she gathered "x + 6". Therefore the sum of mushrrons she gathered on each hour should be equal to the total she gathered as shown below:
37 = x + (x+ 7) + (x + 6)
37 = 3*x + 13
3*x + 13 = 37
3*x = 37 - 13
3*x = 24
x = 24 / 3
x = 8
On the first hour she gathered 8 mushroons, on the second 15 mushroons and on the third she gathered 14 mushroons.
