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CaHeK987 [17]
1 year ago
15

I need the right answer!thanks in advance^^ ​

Mathematics
1 answer:
LiRa [457]1 year ago
3 0

The questions are illustrations of combination and permutation.

<h3>The different arrangement of LOVE</h3>

From the question, we understand that the letter V must come first.

So, we have:

  • First letter = 1; i.e. V
  • The second letter can be any of the remaining 3
  • The third letter can be any of the remaining 2
  • The last letter can be the remaining 1

So, we have:

Arrangement = 1 * 3 * 2 * 1

Arrangement = 6

Hence, there are 6 different arrangement of the letters.

<h3>Ways to pick two vegetables</h3>

We have:

  • Vegetables, n = 5
  • To pick, r = 2

The number of different combination is:

Combination = ^5C_2

Evaluate

Combination = 10

Hence, there are 10 different ways

<h3>Number of 4-digit code</h3>

The numbers are given as {1,4,5,7,8}

So, we have:

  • The first digit can be any of the 5 digits
  • The second digit can be any of the remaining 4 digits
  • The third digit can be any of the remaining 3 digits
  • The fourth digit can be any of the remaining 2 digits

So, we have:

Digits = 5 * 4 * 3 * 2

Digits = 120

Hence, there are 120 different codes of 4-digit

<h3>Arrangement of books in a shelf</h3>

We have:

  • Books, n = 4

The number of arrangement is:

Arrangement = 4!

Evaluate

Arrangement = 24

Hence, there are 24 different arrangements

<h3>Outcomes in tossing a coin</h3>

A coin has 2 faces

When four coins are tossed, the number of outcomes is:

Outcome = 2 * 2 * 2* 2

Evaluate

Outcome = 16

Hence, there are 16 different outcomes

<h3>Ways of sitting in a row</h3>

The number of people is given as:

People = 5

So, we have:

  • The first person can sit on any of the 5 seats
  • The second person can sit on any of the 4 seats
  • The third person can sit on any of the 3 seats
  • The fourth person can sit on any of the 2 seats
  • The last person would sit on the remaining seat1

So, we have:

Ways = 5 * 4 * 3 * 2 * 1

Ways = 120

Hence, there are 120 different ways

Read more about combination and permutation at:

brainly.com/question/11732255

#SPJ1

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Para decorar una pared se disponen de tiras de papel azules de 15 cm, verdes de 20 cm, y rojas de 25 cm. En la pared se quiere a
den301095 [7]

Answer:

a) Smallest line that can be made with each color = 300 cm

b) Total strips should be used = 47 strips

c) Total strips used of blue color = 20

  Total strips used of green color = 15

  Total strips used of red color = 12

Step-by-step explanation:

Given - To decorate a wall, there are 15 cm blue, 20 cm green, and 25 cm red strips of paper. On the wall you want to build three lines of the same size, one of each color and without cutting any strip.

To find - a) How long is the smallest line that can be made with each color?

              b) How many strips should be used?

              c) How many of each color?

Proof -

a)

For the smallest line that can be made with each color, we just have to find the lcm (least common multiple) of the 3 srtips.

Firstly,

Decompose the 3 strips to its prime factors , we get

15 = 3×5

20 = 2²×5

25 = 5²

So,

The Lcm(15, 20, 25) = 3×2²×5² = 3×4×25 = 300

∴ we get

Smallest line that can be made with each color = 300 cm

b)

Now,

Total strips used = 300 cm

Strips used by 15 cm blue = \frac{300}{15} = 20 strips

Strips used by 20 cm green = \frac{300}{20} = 15 strips

Strips used by 25 cm red = \frac{300}{25} = 12 strips

So,

Total strips should be used = 20 + 15 + 12 = 47 strips

c)

Total strips used of blue color = 20

Total strips used of green color = 15

Total strips used of red color = 12

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B is in between 0 and -1, so it has to be -1/3
C is below -1 and so its the improper fraction

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