Let the boxes be Box 1, Box 2, Box 3.
consider the 3 white balls. They can be all of them in one box:
(3, 0, 0) (3 in Box 1, 0 in box 2 and 0 in box 0)
(0, 3, 0)
(0, 0, 3)
We can have 2 in one box, and 1 in one of the remaining boxes:
(2, 0, 1)
(2, 1, 0)
(0, 2, 1)
(1, 2, 0)
(0, 1, 2)
(1, 0, 2)
and there is only one way: (1, 1, 1) to place one white ball in each box
In total there are: 3+6+1=10 ways to place the white balls. Similarly there are 10 ways to place the black ones.
Since every placement of the white balls can be combined with any placement of the black balls, there are 10*10=100 ways to place the 3white balls and the 3 black bals in the boxes.
Answer: 100
Answer:
Step-by-step explanation:
it is a long time I have not applied Ito's lemma
I would say the following
for 
f'(x)=2x
f''(x)=2
so using Ito's lemma we can write that
so it comes
Answer:
P(A) = 3/20
Step-by-step explanation:
P(A)=P(blue)P(head)=(3/10)(1/2)=3/20
as there are 10 cards in total, out of which 3 are blue so the probability to get the blue card is, P(blue) = 3/10. and the probability of getting a head when a coin is tossed is P(head) = 1/2.
So in total
P(A) = P(blue)*p(head) = (3/10)*(1/2) = 3/20 = 0.15
Answer:
Below
Step-by-step explanation:
All you need to do to find angle x is to subtract 48 from 90 :)
So, 90 - 48 = 42
Angle x = 42 degrees
Hope this helps!
Use the formula for area of a triangle:
Area = 1/2 x base x height.
Fill in what you know:
40 = 1/2 x 20 x height
Multiply both sides by 2:
80 = 20 x height
Divide both sides by 20
Height = 4 m
