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Julli [10]
3 years ago
14

In how many ways can people be lined up to get on a bus if a particular person always boards the bus first or last?

Mathematics
1 answer:
Ronch [10]3 years ago
6 0

Incomplete question. I Assumed the number of people is 6. In other words, In how many ways can 6 people be lined up to get on a bus if a particular person always boards the bus first or last?

Answer:

<u>6</u>

Step-by-step explanation:

Note, this question involves the use of the permutation formula; which helps <u>determine the arrangement or rearrangement of a set of objects in an ordered way.</u> Since we are told a particular person (ie 1 person) from among the group always boards first or last, it forms our r.

P(n,r)= \frac{n!}{(n-r)!} where n = number of people; r = the difference taken at a time.

P(n,r) = \frac{6!}{(6-1)!} = \frac{720}{120} = 6

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Nookie1986 [14]

Step-by-step explanation:

\lim_{n \to \infty} \sum\limits_{k=1}^{n}f(x_{k}) \Delta x = \int\limits^a_b {f(x)} \, dx \\where\ \Delta x = \frac{b-a}{n} \ and\ x_{k}=a+\Delta x \times k

In this case we have:

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So the integral is:

∫₁⁴ √x dx

To evaluate the integral, we write the radical as an exponent.

∫₁⁴ x^½ dx

= ⅔ x^³/₂ + C |₁⁴

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If ∫₁⁴ f(x) dx = e⁴ − e, then:

∫₁⁴ (2f(x) − 1) dx

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1/(2k) = 1/4

2k = 4

k = 2

8 0
3 years ago
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