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Verizon [17]
3 years ago
6

A stadium has 50 comma 00050,000 seats. Seats sell for ​$3030 in Section​ A, ​$2424 in Section​ B, and ​$1818 in Section C. The

number of seats in Section A equals the total number of seats in Sections B and C. Suppose the stadium takes in ​$1 comma 288 comma 8001,288,800 from each​ sold-out event. How many seats does each section​ hold?
Mathematics
1 answer:
Mariulka [41]3 years ago
5 0

A= B+C

As given, A+B+C = 50000 seats

A+(B+C) = 50000

2A = 50000

A= 25000 seats

Now section A sells seats at $30 each

So, 25000\times30=750000

Now, B+C = 25000

B = 25000-C

B sells at $24 and C sells at $18

So, 24B+18C = 538800

Now as section A incurs an amount of $750000 from booking, so B and C will incur 1288800-750000 = 538800 where $1288800 is the total booking amount.

24B+18C = 538800

24(25000-C)+18C=538800

600000-24C+18C=538800

6C = 61200

C = 10200

Now, B = 25000-C

B = 25000 - 10200 = 14800

Hence, section A has 25000 seats

Section B has 14800 seats

Section C has 10200 seats




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2.6795

Step-by-step explanation:

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Mr Thompson please help with my new HW
sesenic [268]

Answers:

  1. C) x = plus/minus 11
  2. B) No real solutions
  3. C) Two solutions
  4. A) One solution
  5. The value <u>  18  </u> goes in the first blank. The value <u>  17  </u> goes in the second blank.

========================================================

Explanations:

  1. Note how (11)^2 = (11)*(11) = 121 and also (-11)^2 = (-11)*(-11) = 121. The two negatives multiply to a positive. So that's why the solution is x = plus/minus 11. The plus minus breaks down into the two equations x = 11 or x = -11.
  2. There are no real solutions here because the left hand side can never be negative, no matter what real number you pick for x. As mentioned in problem 1, squaring -11 leads to a positive number 121. The same idea applies here as well.
  3. The two solutions are x = 0 and x = -2. We set each factor equal to zero through the zero product property. Then we solve each equation for x. The x+2 = 0 leads to x = -2.
  4. We use the zero product property here as well. We have a repeated factor, so we're only solving one equation and that is x-3 = 0 which leads to x = 3. The only root is x = 3.
  5. Apply the FOIL rule on (x+1)(x+17) to end up with x^2+17x+1x+17 which simplifies fully to x^2+18x+17. The middle x coefficient is 18, while the constant term is 17.
3 0
3 years ago
Sherry is a waitress at a country club restaurant. She receives an automatic 15% tip which is added to the member's check for th
cricket20 [7]

Answer:

hii there

the correct answer is option ( D ) $143.50

hope it helps

have a nice day

7 0
2 years ago
There are premium tickets costing $99 each and rest are regualr tickets costing $25 each. A pile of 120 tickets have a total val
polet [3.4K]

Answer:

There are 38 premium tickets and 82 regular tickets in a pile.

Step-by-step explanation:

Given,

Total number of tickets = 120

Total amount = $5812

Solution,

Let the number of premium tickets be x.

And the number of regular tickets be y.

Total number of tickets is the sum of total number of premium tickets and  total number of regular tickets.

So the equation can be written as;

x+y=120\ \ \ \ \ equation\ 1

Again,total amount  is the sum of total number of premium tickets multiplied by cost of each premium ticket and  total number of regular tickets multiplied by cost of each regular ticket.

So the equation can be written as;

99x+25y=5812\ \ \ \ equation\ 2

Now We will multiply equation 1 by 25 we get;

x+y=120\\25(x+y)=120\times25\\25x+25y= 3000 \ \ \ \ equation\ 3

Now Subtracting equation 3 from equation 2 we get;

(99x+25y)-(25x+25y)=5812-3000\\\\99x+25y-25x-25y=2812\\\\74x=2812\\\\x=\frac{2812}{74}=38

We will now substitute the value of x in equation 1 we get;

x+y=120\\38+y=120\\y=120-38\\y=82

Hence There are 38 premium tickets and 82 regular tickets in a pile.

3 0
3 years ago
Suppose you have a binomial distribution with n = 27 and p = 0.6. find p(8 ≤ x ≤ 11).
krek1111 [17]
The answer appears to be 3.335%

4 0
3 years ago
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