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Anuta_ua [19.1K]
3 years ago
14

1) Last Tuesday, Regal Cinemas sold a total of 85 movie tickets. Ticket sales totaled

Mathematics
1 answer:
irakobra [83]3 years ago
7 0

In this question, some of the data is missing so its a complete question and the solution can be defined as follows:

\bold{\text{m = matinee tickets}} \\\\\bold{\text{s = student tickets}}\\\\\bold{\text{r = regular tickets}}

By given question, calculating equations are:

\to \bold{m + s + r = 8500}............................................(a)\\\\\to \bold{5m + 6s + 8.50r = 64600}............................(b)\\\\\to \bold{s = 2m}...................(c)

Putting the equation (c) value in the equation (a) to calculate value:

m + 2m + r = 8500\\\\3m + r = 8500\\\\r = 8500 - 3m........................................(d)

Putting equation (c) and (d) value into equation (b):

5m + 6(2m) + 8.50(8500 - 3m) = 64600\\\\5m + 12m + 72250 - 25.50m = 64600\\\\-8.50m = 64600 - 72250 \\\\-8.50m = -7650\\\\m = \frac{-7650}{-8.50}\\\\m = \frac{7650}{8.50}\\\\m = 900

Putting m value into equation (c) and equation (d) to calculate its value:

Calculating the s value:

\to s = 2(900) \\\\ \to s = 1800

Calculating the r value:

\to r = 8500 - 3(900) \\\\\to r = 8500 - 2700\\\\\to r = 5800\\\\

So, the final value of "m, s, and r" are "900, 1800, and 5800".

Learn more:

brainly.com/question/2929577

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Answer:

a) \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b) \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( i-1 )st and the ith failures. Then, the T_{i} are independent with \mathrm{T}_{\mathrm{i}} being exponential with rate \frac{(101-i)}{200} .

Therefore,

a) E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]

=\sum_{i=1}^{5} \frac{200}{101-i}

\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b)

The variance is given by, \mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]

\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

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