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stiv31 [10]
3 years ago
12

Divide the combinations. 60C3 / 15C3

Mathematics
2 answers:
kotykmax [81]3 years ago
5 0
The combination, nCr, which means "combination of n taken r" can be calculated by the equation,
                               nCr = n! / (n - r)!r!
where ! is a factorial sign. For the given,
                     60C3 = 60! / (60 - 3)!3! = 34220
                     15C3 = 15! / (15 - 3)!3! = 455
Their quotient is approximately 75.21.
Mariulka [41]3 years ago
4 0
I hope you know what '!' means
2!=2
3!=6
4!=24
basicalkly times every natural number including and before that number


nCr=\frac{n!}{r!(n-r)!}
so
\frac{60C3}{15C3}=
\frac{ \frac{60!}{3!(60-3)!} }{ \frac{15!}{3!(15-3)!} }=
( \frac{60!}{3!(60-3)!})(\frac{3!(15-3)!}{15!})=
( \frac{60!}{3!(57)!})(\frac{3!(12)!}{15!})=
\frac{(60!)(3!)(12!)}{(3!)(57!)(15!)}=
\frac{(60!)(12!)}{(57!)(15!)}=
\frac{(60!)}{(57!)(15)(14)(13)}=
\frac{(60)(59)(58)}{(15)(14)(13)}=
\frac{(30)(59)(58)}{(15)(7)(13)}=
\frac{(6)(59)(58)}{(3)(7)(13)}=
\frac{6844}{91}
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Natasha2012 [34]
(a)
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
swap the variables: t = 2y - 1
solve for y: t + 1 = 2y
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Answer for (a): g^{-1}(t) =  \frac{t + 1}{2}

(b)
Same steps as part (a) above:
h(t) = 4t + 3
rewrite as: y = 4t + 3
swap the variables: t = 4y + 3
solve for y: y =\frac{t - 3}{4}

Answer for (b): h^{-1}(t) = \frac{t - 3}{4}

(c)
g^{-1} ( h^{-1}(t)) =  g^{-1} (\frac{t - 3}{4})
replace all t's in the g^{-1}(t) equation with \frac{t - 3}{4}
 g^{-1} (\frac{t - 3}{4}) = \frac{ \frac{t-3}{4} + 1}{2}
= \frac{ \frac{t-3}{4} +  \frac{4}{4}}{2} = \frac{ \frac{t - 3 + 4}{4}}{2} = \frac{ \frac{t + 1}{4}}{2} =  \frac{t + 1}{8}
Answer for (c): g^{-1} ( h^{-1}(t)) = \frac{t + 1}{8}

 (d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1

(e)
h(g(t)) = 8t - 1
   y = 8 t - 1
   t = 8y - 1
  t + 1 = 8y
\frac{t + 1}{8} = y
Answer for (e): inverse of h(g(t)) = \frac{t + 1}{8}
 
























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= \[\frac{65}{24}\] liters.

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