A theater has a seating capacity of 750 and charges $5 for children, $7 for students, and $9 for adults. At a certain screening with full attendance, there were half as many adults as children and students combined. The receipts totaled $4950. How many children attended the show

Answers

Answer 1
Answer:

Let 'c' be the number of children, 'a' be the number of adults and 's' be the number of students in the theater.

Since, there are total 750 people in the theater.

So, c + s + a = 750  (Equation 1)

Now, it is given that there were half as many adults as children and students combined.

So, a= (c+s)/(2)

So, c+s = 2a

Putting the value of "c+s" in equation 1.

c+s+a=750

2a+a = 750

3a= 750

a = 250

Now, it is given that the receipts totaled $4950 and it charges $5 for children, $7 for students, and $9 for adults.

So, ( 5 * c) +( 7 * s )+ (9 * a) = \$4950 (equation 2)

Since, c+s = 2a

c+s = 500

So, s = 500-c

Substituting the values of 's' and 'a' in equation 2, we get

5c+7s+9a= 4950

5c+ 7(500-c)+(9 * 250) = 4950

5c+3500-7c+2250=4950

-2c= -800

So, c = 400

Therefore, there were 400 children at the theater.


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If f(1) =8 and f(n) =1/2f(n-1) + 2 then what is the value of f(4)?I can't get it! PLEASE HELP!!!!!

Answers

If you would like to know the value of f(4), you can calculate this using the following steps:

f(n) = 1/2 * f(n-1) + 2
f(1) = 8

f(2) = 1/2 * f(2-1) + 2 = 1/2 * f(1) + 2 = 1/2 * 8 + 2 = 4 + 2 = 6
f(3) = 1/2 * f(3-1) + 2 = 1/2 * f(2) + 2 = 1/2 * 6 + 2 = 3 + 2 = 5
f(4) = 1/2 * f(4-1) + 2 = 1/2 * f(3) + 2 = 1/2 * 5 + 2 = 5/2 + 2 = 5/2 + 4/2 = 9/2 = 4 1/2

The correct result would be 4 1/2.

1/3*h(-8) h(x)= x^2 - 5x +7

Answers

Answer: 37

Step-by-step explanation:

    First, we need to find what h(-8) is. We can do this by substituting and simplifying the given function.

Given:

  h(x) = x² - 5x +7

Substitute:

  h(-8) = (-8)² - 5(-8) +7

Square and multiply:

  h(-8) = 64 + 40 +7

Add:

  h(-8) = 111

    Next, we can find one-third of this. To do this, we will multiply.

Multiply:

  (1/3) * 111 = 37

Let \( A \) and \( B \) be events with \( P(A)=1 / 3 \) and \( P(B)=2 / 3 \). Let \( N \) be the number of events from among \( A \) and \( B \) that occur. (For instance, if \( A \) occurs but \( B \) does not, then \( N=1 \).) Suppose that the occurrence of events in \( A \) and \( B \) are independent. Find, for example, the probability that no events occur.

Answers

Answer:

( 2/9 )

Step-by-step explanation:

Okay, so since the events A and B are independent, the probability that both of them do not occur is the product of their individual probabilities of not occurring.

The probability of event A not occurring, denoted as ( P(A') ), is equal to 1 minus the probability of A occurring. Similarly, the probability of event B not occurring, denoted as ( P(B') ), is equal to 1 minus the probability of B occurring.

So, ( P(A') = 1 - P(A) = 1 - 1/3 = 2/3 ) and ( P(B') = 1 - P(B) = 1 - 2/3 = 1/3 ).

Since A and B are independent, the probability that both A and B do not occur is simply the product of their individual probabilities of not occurring:

[ P(N = 0) = P(A' cap B') = P(A')P(B') = (2/3) times (1/3) = 2/9 ]

So, the probability that no events occur is ( 2/9 ).

Hope this helps! :)

What are the domain and range of the relation {(–5, 5), (–3, 2), (0, 3), (3, 2)}?

Answers

domain is th set of number you input
range is the set of number you get from inputing the domain
normally domain is x and range is y
(x,y)

domain is all 1st number
range is all 2nd numbers

doman={-5,-3,0,3}
range={5,2,3} (if it repeats, don't list)

adult tickets cost $10 and children's tickets to a show cost $5. if the number of children's tickets was twice the number of adult tickets, how many of each type of tickets was sold if $800 was collcted in tickets sales? ( make sure you write a legend stating what your variable represents)

Answers

x=amount of children tickes sold
y=amount of adult tickets sold

2 times of children as adult
x=2y

total cost is 800
5x+10y=800
divide both sides by 5
x+2y=160
x=2y
2y+2y=160
4y=160
divie by 4
y=40

x=2y
x=2(40)
x=80


80 children tickets
40 adult tickets

When embedding fraction addition problems in familiar contexts, the lowest common denominator for parts of a dozen is

Answers

Answer:

The lowest common denominator for parts of a dozen is 12.

Step-by-step explanation: