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Andreyy89
3 years ago
5

Tickets for a concert were \$5$5dollar sign, 5 for each child and \$8$8dollar sign, 8 for each adult. At one of the concerts, ea

ch adult brought 444 children with them, and 101010 children attended without an adult. The total ticket sales were \$1{,}730$1,730dollar sign, 1, comma, 730. Which of the following systems of equations can be solved to determine the number of children, ccc, and adults, aaa, who attended the concert?
Mathematics
1 answer:
OlgaM077 [116]3 years ago
4 0

Answer:

Step-by-step explanation:

Let c represent the number of children that attended the concert.

Let a represent the number of adults that attended the concert.

At one of the concerts, each adult brought 4 children with them, and 10 children attended without an adult. This means that the number of children that attended is

c = 4a + 10

Tickets for a concert were $5 for each child and $8 for each adult. The total ticket sales were $1730. It means that

5c + 8a = 1730

Therefore, the systems of equations that can be solved to determine the number of children, c, and adults that attended the concert is

c = 4a + 10

5c + 8a = 1730

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(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

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Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

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If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

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