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pshichka [43]
3 years ago
5

Point B ∈ AC . If points K, M, N, and P are the midpoints of respectively AB , AC , MC , and NC , identify and write a statement

about each pair of congruent segments.
Mathematics
2 answers:
matrenka [14]3 years ago
5 0

Answer:

AK and KB

AM and MC

MN and NC

NP and PC

Step-by-step explanation:

* <em>Lets explain how to solve the problem</em>

- B ∈ AC

∴ B is located between A and C

- K is the mid point of AB

∴ K divides AB into two equal parts AK and KB

∴ AK = KB

- M is the mid point of AC

∴ M divides AC into two equal parts AM and MC

∴ AM = MC

- N is the mid point of MC

∴ N divides MC into two equal parts MN and NC

∴ MN = NC

- P is the mid point of NC

∴ P divides NC into two equal parts NP and PC

∴ NP = PC

* The congruent segments are:

AK and KB

AM and MC

MN and NC

NP and PC

Ira Lisetskai [31]3 years ago
4 0

Answer:

AK ≅ KB

AM ≅ MC

MN ≅ NC

NP ≅ PC

Step-by-step explanation:

Given that point K is the midpoint of AB, then by definition of midpoint:

AK ≅ KB

Given that point M is the midpoint of AC, then by definition of midpoint:

AM ≅ MC

Given that point N is the midpoint of MC, then by definition of midpoint:

MN ≅ NC

Given that point P is the midpoint of NC, then by definition of midpoint:

NP ≅ PC

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First, we need to calculate the number of ways we can select the students in all cases, and then, the probability.

First, we'll use the combination formula, to calculate the number of ways we can select the 5 students out of the 21. We use combination, because it does not matter the order that the students are selected.

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With this expression we will calculate first, how many ways we can choose the 5 students out of 21:

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Now let's calculate the number of ways you can get the all 5 students are non majors:

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Now we need to know the number of ways we can get 4 non majors and 1 major:

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Now the way to get 3 non majors and 2 majors, we do the same thing we do to get 4 non majors and 1 major, but changing the numbers. Then the way to get 2 non majors and 3 majors, and finally 1 non major and 4 majors:

3 non majors and 2 majors:

C4 = C4' * C4'' = [14! / 3!(14 - 3)!] * [7! / 2!(7 - 2)!] = 7,644

2 non majors and 3 majors:

C5 = C5' * C5'' = [14! / 2!(14 - 2)!] * [7! / 3!(7 - 3)!] = 3,185

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C6 = C6' * C6'' = [14! / 1!(14 - 1)!] * [7! / 4!(7 - 4)!] = 490

Finally to know the probability of getting 1 out of the 5 to be non major, we have to sum all the previous results, and divide them by the ways we can choose the 5 students (C1):

P = 2,002 + 7,007 + 7,644 + 3,185 + 490 / 20,349

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