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Inessa [10]
3 years ago
11

You are planning a road trip and there are 15 different cities you would like to visit. Due to time constraints, you can only vi

sit 4 of the cities. If you do not want to repeat any of the cities, how many different ways can you visit 4 of the 15 cities you are interested in
Mathematics
1 answer:
kumpel [21]3 years ago
3 0

Answer: 32760 ways.

Step-by-step explanation:

To solve this, we first determine the number of ways to select 4 countries from 15, we calculate number of ways to visit these 4 places then we multiply.

Selecting number of ways to choose 4 cities from 15 is by using the combination formula, which becomes 15C4

Number of ways to visit these 4 cities = 4!

Hence,total number of ways of visiting these 4 cities = 15C4 * 4! = 32760ways

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Somebody help please!!
dimaraw [331]
The correct answer is 19
3 0
1 year ago
21. x² +9x - 22<br> Factors of cSum of Factors<br> 7<br> 7<br> ?<br> ?<br> 7<br> 7<br> 9<br> 7
Eduardwww [97]

Answer:

\large \begin{array}{| c | c |}\cline{1-2} \sf Factors\:of\:c & \sf Sum\:of\:Factors\\\cline{1-2} 1, -22 & -21 \\\cline{1-2} -1, 22 & 21\\\cline{1-2} 2, -11 & -9 \\\cline{1-2} -2, 11 & 9 \\\cline{1-2}\end{array}

Step-by-step explanation:

Quadratic equation: ax² + bx + c = 0, where a ≠ 0

x² + 9x - 22

  • a = 1
  • b = 9
  • c = -22

Factors of c (-22):

1 × -22     or     -1 × 22

2 × -11      or     -2 × 11

Sum of factors:

1 + (-22) = -21          or          -1 + 22 = 21

2 + (-11) = -9            or           -2 + 11 = 9

Table:

\large \begin{array}{| c | c |}\cline{1-2} \sf Factors\:of\:c & \sf Sum\:of\:Factors\\\cline{1-2} 1, -22 & -21 \\\cline{1-2} -1, 22 & 21\\\cline{1-2} 2, -11 & -9 \\\cline{1-2} -2, 11 & 9 \\\cline{1-2}\end{array}

Hope this helps!

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2 years ago
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olga nikolaevna [1]
4/8 just multiply each number by 4!
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Inga [223]
The answer is B not mutually exclusive
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notsponge [240]
That would be the second choice. Translation of 2 to the left
4 0
2 years ago
Read 2 more answers
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