How many ways are there for a horse race with four horses to finish if ties are possible?

1 answer:

6 0

1. 4 horses tie for 1st place
2. 3 horses tie for 1st place and 1 comes in 2nd place
3. 2 horses tie for 1st place and 2 tie for 2nd place
4. 2 horses tie for 1st place, 1 horse comes in 2nd and 1 comes in 3rd
5. 1 horse comes in 1st and 3 horses tie for 2nd place
6. 1 horse comes in 1st, 2 tie for 2nd place, and 1 comes in 3rd place
7. 1 horse comes in 1st, 1 comes in 2nd, and 2 tie for 3rd place.
8. 1 horse comes in 1st, 1 comes in 2nd, 1 comes in 3rd, and 1 comes in 4th.

75.

Hope this helps.

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DochEvi [55]

Answer:

Please check the explanation

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:\left(x-5\right)^3-1$

Given that the output = -3

i.e. y = -3

now substituting the value y=-3 and solve for x to determine the input 'x'

$\:\:y=\:\left(x-5\right)^3-1$

$-3\:=\:\left(x-5\right)^3-1\:\:\:$

switch sides

$\left(x-5\right)^3-1=-3$

Add 1 to both sides

$\left(x-5\right)^3-1+1=-3+1$

$\left(x-5\right)^3=-2$

$\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

Thus, the input values are:

$x=-\sqrt[3]{2}+5,\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}-i\frac{\sqrt[3]{2}\sqrt{3}}{2},\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}+i\frac{\sqrt[3]{2}\sqrt{3}}{2}$