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ehidna [41]
3 years ago
9

How many ways are there to put 6 balls in 3 boxes if the balls are distinguishable but the boxes are not?

Mathematics
1 answer:
FromTheMoon [43]3 years ago
5 0

This is essentially asking how many different ways to partition 6 into 3 segments.


I am assuming "no ball in a box" is not allowed.


6 can be partitioned as

(4,1,1), (3,2,1), and (2,2,2)


So, calculate each partition, we get

(6 choose 4) + (6 choose 3)*(3 choose 2) + (6 choose 2) * (4 choose 2)

= 15 + 20*3 + 15*6

=165

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Answer:

dh/dt =  0.4 m/min

Step-by-step explanation:

The volume of the cone is:

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The volume of the cone as a function of h will be:

V(h)  =  (1/3)* (h/3)²*h

V(h) =  (1/27)*h³

The increasing rate of the volume is equal to the rate of sand added the:

D(V)/dt   = (1/27)*3*h²*dh/dt

D(v) / dt  =  10 m³/min

h =  15 m      and   dh/dt is the rate of increasing of the height

By substitution

10 m³/min  = ( 1/9)* 225 * dh/dt  (m²)

dh/dt =  90 / 225   m/min

dh/dt =  0.4 m/min

3 0
3 years ago
The difference between twice a number, x, and a smaller number, y, is 3. The sum of twice the number and the smaller
azamat

Answer:

y=2x+3 and y=-2x-3

Step-by-step explanation:

First situation:

2x-y=3

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Second situation:

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7 0
2 years ago
1. Kylie wrote the expression below. (9 + k) x 5 What is the value of Kylie's expression for k= 4? A 65 B 41 C 29 D 28​
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