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Deffense [45]
3 years ago
15

20. There are 15 members of the show choir. In how many ways can you arrange 4 members in the front row when order does not matt

er?
Mathematics
2 answers:
sladkih [1.3K]3 years ago
8 0
Say that the front four people's names are John, Kathy, Mia, and Alex, in that order. You can arrange them as follows.

John, Kathy, Alex, Mia
John, Alex, Kathy, Mia
John, Alex, Mia, Kathy
John, Mia, Alex, Kathy
John, Mia, Kathy, Alex
Alex, John, Kathy, Mia
Alex, John, Mia, Kathy
There are many more, but my guess would be 16 combinations. Test for yourself if you don't think thats right!
Aleonysh [2.5K]3 years ago
5 0
The first chair can be any  one of the 15.
   For each of those ...
The second chair can be any one of the remaining 14.
      For each of those ...
The third chair can be any one of the remaining 13.
   For each of those ...
The fourth chair can be any one of the remaining 12.

Number of ways to fill the 4 chairs = (15 x 14 x 13 x 12) = 32,760 .

But ...

Each set of 4 people can be seated in (4 x 3 x 2 x 1) = 24 orders.
So each group of 4 people is represented 24 times among the 32,760.

If the order doesn't matter, you're really asking how many different
groups of 4 people can occupy the front row.

That's        (32,760) / (24) = 1,365 sets of 4 members, in any order.
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As this is already in its simplest form, I believe the only other expression is 4*X +9
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Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3
Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

\frac{dV}{dt} = -kA

Using Euler's method

\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i}    \\

Where initial droplet volume is:

V(0) = \frac{4pi}{3} * r(0)^3 =  \frac{4pi}{3} * 2.5^3 = 65.45 mm^3

Hence, the iterative solution will be as next:

  • i = 1, ti = 0, Vi = 65.45

V'_{i}  = -k *4pi*(\frac{3*65.45}{4pi})^(2/3)  = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88

  • i = 2, ti = 0.5, Vi = 63.88

V'_{i}  = -k *4pi*(\frac{3*63.88}{4pi})^(2/3)  = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33

  • i = 3, ti = 1, Vi = 62.33

V'_{i}  = -k *4pi*(\frac{3*62.33}{4pi})^(2/3)  = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\

The average change of droplet radius with time is:

Δr/Δt = \frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

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