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Mnenie [13.5K]
2 years ago
12

How to represent 5 rational numbers on line. the numbers should be of form p/q where p or q should be a taxicab number. the frac

tion can be then reduced to its standard form. write about the taxicab numbers in brief.
Mathematics
1 answer:
Rudiy272 years ago
7 0

The five rational numbers are 1/1729, 1/2, 3/2, 2/1, 1729/1 and the numbers should be of form p/q where p or q should be a taxicab number.

According to the question,

To represent 5 rational numbers on line, the numbers should be of form p/q where p or q should be a taxicab number.

A taxicab number is a smallest integer can be expressed as a sum of two positive integer cube.

The most famous taxicab number is 1729. Thus,

1729 = 1³ + 12³

1729 = 9³+ 10³

Now, 2 = 1³ + 1³ so, 2 is the first taxicab number. Therefore, five rational numbers are  1/1729, 1/2, 3/2, 2/1, 1729/1.

In 1/1729 -> 1729 is taxicab number

In 1/2 -> 2 is taxicab number

In 3/2 -> 2 is taxicab number

In 1/1729 -> 2 is taxicab number

In 1729/1  -> 1729 is taxicab number

Hence, five rational numbers are 1/1729, 1/2, 3/2, 2/1, 1729/1 and the numbers should be of form p/q where p or q should be a taxicab number.

Learn more about rational numbers here

brainly.com/question/9983341

#SPJ4

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Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem, we have that:

10 student are sampled, so n = 10

34% of college students say they use credit cards because of the rewards program, so \pi = 0.34

(a) exactly​ two

This is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than​ two

This is P(X > 2).

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

P(X \leq 2) + P(X > 2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157

P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838

P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573

P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320

P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

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