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tangare [24]
3 years ago
5

Determine whether the sequence appears to be an arithmetic sequence. If so, find the common difference and the next three terms

in the sequence. –5, –11, –17, –23, –29, . . .
Mathematics
1 answer:
guapka [62]3 years ago
4 0

an aritmetic sequence increases by the same amount each time (if it decreases, then it increases by a negative number)


the common difference is the number that must be added to a term to get the next term


let's see if it increases by the same amount

-5 to -11 is increase of -6

-11 to -17 is increase of -6

-17 to -23 is increase of -6

-23 to -29 is increase of -6


so it appears to be aritmetic

common difference is -6

-29-6=-35

-35-6=-41

-41-6=-47

the next 3 terms are -35, -41, -47

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Probability of someone that made a reservation not showing up = 50% = 0.5

Probability of someone that made a reservation showing up = 1 - 0.5 = 0.5

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For this to happen, 5 or 6 people have to show up since the limousine can accommodate a maximum of 4 people

Let P(X=x) represent x people showing up

probability that at least one individual with a reservation cannot be accommodated on the trip = P(X = 5) + P(X = 6)

P(X = x) can be evaluated using binomial distribution formula

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 6

x = Number of successes required = 5 or 6

p = probability of success = 0.5

q = probability of failure = 0.5

P(X = 5) = ⁶C₅ (0.5)⁵ (0.5)⁶⁻⁵ = 6(0.5)⁶ = 0.09375

P(X = 6) = ⁶C₆ (0.5)⁶ (0.5)⁶⁻⁶ = 1(0.5)⁶ = 0.015625

P(X=5) + P(X=6) = 0.09375 + 0.015625 = 0.109375

b) If six reservations are made, what is the expected number of available places when the limousine departs?

Probability of one person not showing up after reservation of a seat = 0.5

Expected number of people that do not show up = E(X) = Σ xᵢpᵢ

where xᵢ = each independent person,

pᵢ = probability of each independent person not showing up.

E(X) = 6(1×0.5) = 3

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So, expected number of unoccupied seats = 1

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