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Basile [38]
3 years ago
13

A stadium has 50,000 seats. Seats in section A cost $30, seats in section B cost $24, and seats in section C cost $18. the numbe

r of seats in section A is equal to the total number of seats in section B and C. Suppose the stadium takes in $1,287,600 from each sold out event. How many seats does each section have?
Mathematics
1 answer:
svetoff [14.1K]3 years ago
7 0

Answer:

Section A = 25,000 seats

Section B = 14,600 seats

Section C = 10,400 seats

Step-by-step explanation:

Total Seats = 50,000

Seats in Section A cost = $30

Seats in Section B cost = $24

Seats in Section C cost = $18

Total sales from the event = $1,287,600

No. of Seats in section A = No. seats in Section B + No. seats in Section C

A = B + C

or, 2A = 50,000

A = 25,000 seats @ $30/seat = $750,000

B + C = 25,000

24B + 18C = 537,600

24B + 18(25,000 - B) = 537,600

24B + 450,000 - 18B = 537,600

6B = 87600

B = 14,600

C = 10,400

Hence;

A = 25,000 seats

B = 14,600 seats

C = 10,400 seats

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Answer:

\bar X = \frac{\sum x_i f_i}{n} = \frac{1220750}{500}=2441.5

s^2= \frac{3408029125 -\frac{(1220750)^2}{500}}{500-1} =856849.7

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Step-by-step explanation:

For this case we can create the following table

Interval      Frequency (f)    Midpoint(xi)       xi *f      xi^2* f

0-499              9                       249.5            2245.5   560252.3

500-999         13                      749.5            9743.5    7302753

1000-1499      33                     1249.5          41233.5   51521258.25

1500-1999      115                    1749.5          201193.5  361986278.8

2000-2499     125                   2249.5         281187.5   632531281.3

2500-2999      81                    2749.5          222709.5 612339770.3

3000-3499      47                    3249.5         152726.5   496284761.8

3500-3999      45                    3749.5         168727.5    632643761.3

4000-4499      22                    4249.5         93489        397281505.5

4500-4999      10                     4749.5         47495        225577502.5

Total                500                                      1220750      3408029125

\sum f_i = 500 , \sum x_i f_i = 1220750, \sum x^2_i f_i = 3408029125

For this case we can calculate the sample mean with this formula:

\bar X = \frac{\sum x_i f_i}{n} = \frac{1220750}{500}=2441.5

And for the sample variance we can use the following formula:

s^2= \frac{\sum x^2_i f_i - \frac{(\sum x_i f_i)^2}{n}}{n-1}

And if we replace we got:

s^2= \frac{3408029125 -\frac{(1220750)^2}{500}}{500-1} =856849.7

And the deviation is just the square root of the sample variance and for this case is:

s= \sqrt{856849.7}=925.662

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