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vitfil [10]
4 years ago
15

Use the row of numbers shown below to generate 12 random numbers between 01 and 99.

Mathematics
1 answer:
sattari [20]4 years ago
6 0

Answer:

Below is the complete step-by-step explanation.

Step-by-step explanation:

Given the row of numbers

25060 12315 86244 97348 36173 32710 80033 16160

<u>Taking 25060 as seed</u>

Total random numbers to be generates = 12

Range of random number generators

  • Minimum value = 01
  • Maximum Value = 99

Seed the number (25060) into your calculator. 25060 ➡️ rand

So, the operation randInt(1,99,12) will generate the following 12 random numbers between 01 and 99.

85 90 62 45 26 97 40 55 67 17 10 32

<u>Taking 12315 as seed</u>

Total random numbers to be generates = 12

Range of random number generators

  • Minimum value = 01
  • Maximum Value = 99

Seed the number (12315) into your calculator. 12315 ➡️ rand

So, the operation randInt(1,99,12) will generate the following 12 random numbers between 01 and 99.

33 20 26 68 73 09 46 14 82 74 17 04

<u>Taking 86244 as seed</u>

Total random numbers to be generates = 12

Range of random number generators

  • Minimum value = 01
  • Maximum Value = 99

Seed the number (86244) into your calculator. 86244 ➡️ rand

So, the operation randInt(1,99,12) will generate the following 12 random numbers between 01 and 99.

54 93 02 50 37 01 86 51 38 28 23 36

<u>Taking 97348 as seed</u>

Total random numbers to be generates = 12

Range of random number generators

  • Minimum value = 01
  • Maximum Value = 99

Seed the number (97348) into your calculator. 97348 ➡️ rand

So, the operation randInt(1,99,12) will generate the following 12 random numbers between 01 and 99.

04 79 01 86 51 02 50 37 38 28 23 36

<u>Taking 36173 as seed</u>

Total random numbers to be generates = 12

Range of random number generators

  • Minimum value = 01
  • Maximum Value = 99

Seed the number (36173) into your calculator. 36173 ➡️ rand

So, the operation randInt(1,99,12) will generate the following 12 random numbers between 01 and 99.

13 03 28 17 14 31 98 47 40 48 68 19

<u>Taking 32710 as seed</u>

Total random numbers to be generates = 12

Range of random number generators

  • Minimum value = 01
  • Maximum Value = 99

Seed the number (32710) into your calculator. 32710 ➡️ rand

So, the operation randInt(1,99,12) will generate the following 12 random numbers between 01 and 99.

22 80 42 03 27 89 98 46 14 31 72 56

<u>Taking 80033 as seed</u>

Total random numbers to be generates = 12

Range of random number generators

  • Minimum value = 01
  • Maximum Value = 99

Seed the number (80033) into your calculator. 80033 ➡️ rand

So, the operation randInt(1,99,12) will generate the following 12 random numbers between 01 and 99.

29 86 51 01 23 64 37 36 35 72 14 59

<u>Taking 16160 as seed</u>

Total random numbers to be generates = 12

Range of random number generators

  • Minimum value = 01
  • Maximum Value = 99

Seed the number (16160) into your calculator. 16160 ➡️ rand

So, the operation randInt(1,99,12) will generate the following 12 random numbers between 01 and 99.

93 54 72 58 30 14 48 87 99 95 83 29

Keywords: random number generator

Learn more about random number generator from brainly.com/question/1601128

#learnwithBrainly

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Answer:

Angle of A = 90 degree-62 degree =  28 degree.

Step-by-step explanation:

tan (62) = opposite / adjacent = 10 / a ---> a = 10/tan (62) = 10/ 1.88 = 5.319

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3 years ago
An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed
masya89 [10]

Answer:

P(X\leq 1) = 0.331

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

P(X = x) = \frac{y^xe^{-y}}{x!}

Where y represents λ (average)

y = 2.3

<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>

<em />

Using probability notations;

P(X\leq 1) = P(X=0) + P(X =1)

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}

P(X = 0) = \frac{1 * e^{-2.3}}{1}

P(X = 0) = e^{-2.3}

P(X = 0) = 0.10025884372

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}

P(X = 1) = \frac{2.3 * e^{-2.3}}{1}

P(X = 1) =2.3 * e^{-2.3}

P(X = 1) = 2.3 * 0.10025884372

P(X = 1) = 0.23059534055

P(X\leq 1) = P(X=0) + P(X =1)

P(X\leq 1) = 0.10025884372 + 0.23059534055

P(X\leq 1) = 0.33085418427

P(X\leq 1) = 0.331

Hence, the required probability is 0.331

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