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arlik [135]
3 years ago
15

In a horse​ race, how many different finishes among the first 3 places are possible if 7 horses are​ running? (Exclude​ ties)

Mathematics
1 answer:
kodGreya [7K]3 years ago
3 0

Answer:

7 x 6 x 5 = 210

Step-by-step explanation:

its a combination's problem.

7 horses are eligible for first place.

6 of those are eligible for 2nd given that one of the 7 finished first.

similarly 5 of the remaining are eligible for 3rd place given 2 horses have taken the first 2 spots.

therefore 7 x 6 x 5 possible combinations = 210 combinations.

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Step-by-step explanation:

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8 0
3 years ago
Mia used 3 different base 10 blocks to model a 3 digit number what is the number?
spin [16.1K]

Answer:

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5 0
3 years ago
Read 2 more answers
To the nearest tenth, what is the perimeter of the triangle with vertices at (−2, 3), (3, 6), and (2, −2)?
katovenus [111]

9514 1404 393

Answer:

  20.3

Step-by-step explanation:

The distance formula can be used to find the side lengths.

  d = √((x2 -x1)^2 +(y2 -y1)^2)

For the first two points, ...

  d = √((3 -(-2))^2 +(6 -3)^2) = √(5^2 +3^2) = √34 ≈ 5.83

For the next two points, ...

  d = √((2 -3)^2 +(-2-6)^2) = √(1 +64) = √65 ≈ 8.06

For the last and first points, ...

  d = √((-2-2)^2 +(3-(-2)^2) = √(16 +25) = √41 ≈ 6.40

Then the sum of the side lengths is ...

  5.83 +8.06 +6.40 = 20.29 ≈ 20.3

The perimeter of the triangle is about 20.3 units.

7 0
3 years ago
it takes someone 15 minutes to prepare 3 1/4 cups of juice how many hours would it take to prepare 32 1/2 cups of juice​
wariber [46]

\bf \begin{array}{ccll} minutes&\stackrel{juice}{cups}\\ \cline{1-2} 15&3\frac{1}{4}\\\\ x&32\frac{1}{2} \end{array}\implies \cfrac{15}{x}=\cfrac{3\frac{1}{4}}{32\frac{1}{2}}\implies \cfrac{15}{x}=\cfrac{\frac{3\cdot 4+1}{4}}{\frac{32\cdot 2+1}{2}}\implies \cfrac{15}{x}=\cfrac{\frac{13}{4}}{\frac{65}{2}}

\bf \cfrac{15}{x}=\cfrac{~~\begin{matrix} 13 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{\underset{2}{~~\begin{matrix} 4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\cdot \cfrac{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{\underset{5}{~~\begin{matrix} 65 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}\implies \cfrac{15}{x}=\cfrac{1}{10}\implies 150=x\leftarrow \begin{array}{llll} \textit{150 minutes or}\\\\ \textit{2 hours and a half} \end{array}

4 0
3 years ago
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