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timama [110]
3 years ago
11

3. What are the identification numbers for ten students chosen at random from a population of 78 students based on the following

string of random digits? Start at the left.
27816 78416 01822 73521 37741 016312 68000 53645 56644 97892 63408 77919 44575
Mathematics
1 answer:
Naya [18.7K]3 years ago
7 0

Answer:

27 67 16 01 27 35 21 37 74 10

Step-by-step explanation:

Starting from the left, you pick two numbers together and skip two numbers if it is greater than 78.

Meaning:

27816 78416 01822 73521 37741 016312 68000 53645 56644 97892 63408 77919 44575

You pick [27] and leave 81 since it is greater than 78, the population number.

Then you pick [67] and leave 84 as it is also greater than 78.

Then you pick [16] and [01] but leave 82, as it is greater than 78.

Then pick [27], [35], [21], [37], [74] and [10] since they are all less than 78.

At this juncture, you have identified 10 numbers: 27 67 16 01 27 35 21 37 74 10

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Intersect with x axis indicates that x value would be Zero

Thus when x = 0 

y would be 
y = -18

Therefor intersection would be at (0,-18)
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3 years ago
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PolarNik [594]

the answer is D-n is greater then or equal to nine.

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3 years ago
A parallelogram is a four-sided figure
stellarik [79]
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4 0
2 years ago
In this exercise, we consider strings made from uppercase letters in the English alphabet and decimal digits.
Yuri [45]

Answer:

  • a) 26^2 36^8
  • b) 21\cdot10\cdot36^7
  • c) 5^3 31^7
  • d) 10\cdot 9\cdot 8 \cdot 7 \cdot 26^6

Step-by-step explanation:

We will use the product rule from combinatorics.

  • a) There are 26 letters in the English alphabet, so there are 26 possible choices for the first character and 26 possible choices for the last one. Each one of the remaining eight characters of the string has 36 choices (letters or digits). By the product rule, there are 26\cdot36\cdot 36\cdots 36\cdot 26=26^2 36^8 strings.
  • b) We have 5 possible choices for the first character, it must be some vowel a,e,i,o,u. The second character can be chosen in 21 ways, selecting some consonant. There are 10 possibilities for the last character because only of the digits are allowed. The other seven characters have no restrictions, so each one can be chosen in 36 ways. By the product rule there are 21\cdot 10\cdot 36^7 strings.    
  • c) The third character has 5 possibilities. Repetition of vowels is allowed, so the sixth and eighth characters have each one 5 possible choices. There are seven characters left. None of them are a vowel, but they are allowed to take any other letter or digit, so each one of them can be chosen in 36-5=31 ways. Therefore there are 5^3 31^7 strings.
  • d) Remember that the binomial coefficient \binom{n}{k} is the number of ways of choosing k elements from a set of n elements. In this case, to count all the possible strings, we first need to count in how many ways we can select the four positions that will have the digits. This can be done in \binom{10}{4} ways, since we are choosing four elements from the set of the ten positions of the string. Now, for the first position, we can choose any digit so it has 10 possibilities. The second position has 9 possibilities, because we can't repeat the digit used on the first position. Similarly, there are 8 choices for the third position and there are 7 choices for the fourth. Now, these are the only digits on the string, so the remaining 6 characters must be letters, then each one of them has 26 possibilities. By the product rule, there are 10\cdot 9\cdot 8 \cdot 7 \cdot 26^6 strings.
3 0
3 years ago
Plssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss help
almond37 [142]

Answer:

5, 6, 4

Step-by-step explanation:

10+(?)=15

10 and 15 have a difference of 5

10+5=15

(?)+9=15

9 and 15 have a difference of 6

6+9=15

11+(?)=15

11 and 15 have a difference of 4

11+4=15

I hope this helps!

3 0
2 years ago
Read 2 more answers
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