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vichka [17]
3 years ago
5

A school club is comparing prices at two movie theaters. Theater 1 charges $7.50 for a movie ticket and $5.75 per item at the co

ncession stand it would cost a total of $187.75 for their group. Theater 2 charges $22.50 a ticket and $7.50 per item for a total of $397.50. Using this information solve for how many tickets and how many concession items they plan to purchase how many movies do they plan to purchase? and how many snack bins did they get?? I will give you brainlist !!!!!
Mathematics
1 answer:
salantis [7]3 years ago
7 0

Step-by-step explanation:

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A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as
Kisachek [45]

Answer:

(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.

Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

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

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

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)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0
3 years ago
Help what’s the rule
solong [7]
Doubling or quadruple
5 0
3 years ago
Evaluate this expression <br><br> <img src="https://tex.z-dn.net/?f=8%5E%7B-2%7D" id="TexFormula1" title="8^{-2}" alt="8^{-2}" a
shepuryov [24]

Answer:

0.015625

Step-by-step explanation:

Rule: a^{-n} = \dfrac{1}{a^n}

8^{-2} = \dfrac{1}{8^2} = \dfrac{1}{64} = 0.015625

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Write 2.8 as a mixed number in a simplest form
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It would be 2 4/5............i hope this helps you
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Ignore the highlighted ones please thxs
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I believe it’s D. 2/3
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