Let's call a child's ticket 
and an adult's ticket 
. From this, we can say:
,
since 116 tickets are sold in total.
Now, we are going to need to find another equation (the problem asks us to solve a systems of equations). This time, we are not going to base the equation on ticket quantity, but rather ticket price. We know that an adult's ticket is $17,000, and a child's ticket is thus
.
Given these values, we can say:
,
since each adult ticket costs 17,000 and each child's ticket costs 12,750, and these costs sum to 1,653,250.
Now, we have two equations:
Let's solve:
- Find on its own, which will allow us to substitute it into the first equation
- Substitute in
for
- Apply the Distributive Property
- Subtract 1972000 from both sides of the equation and multiply both sides by -1
We have now found that 75 child's tickets were sold. Thus,
,
41 adult tickets were sold as well.
In sum, 41 adult tickets were sold along with 75 child tickets.
Answer:50
Step-by-step explanation:
Bc 25+25 is 50.
Answer:
The fourth pair of statement is true.
9∈A, and 9∈B.
Step-by-step explanation:
Given that,
U={x| x is real number}
A={x| x∈ U and x+2>10}
B={x| x∈ U and 2x>10}
If 5∈ A, Then it will be satisfies x+2>10 , but 5+2<10.
Similarly, If 5∈ B, Then it will be satisfies 2x>10 , but 2.5=10.
So, 5∉A, and 5∉B.
If 6∈ A, Then it will be satisfies x+2>10 , but 6+2<10.
Similarly, If 6∈ B, Then it will be satisfies 2x>10 , and 2.6=12>10.
So, 6∉A, and 6∈B.
If 8∈ A, Then it will be satisfies x+2>10 , but 8+2=10.
Similarly, If 8∈ B, Then it will be satisfies 2x>10. 2.8=16>10.
So, 8∉A, and 8∈B.
If 9∈ A, Then it will be satisfies x+2>10 , but 9+2=11>10.
Similarly, If 9∈ B, Then it will be satisfies 2x>10. 2.9=18>10.
So, 9∈A, and 9∈B.
Answer: 10%
Step-by-step explanation:
The difference between his estimated and actual distance is 15 - 13.5 = 1.5
The percentage of error is the difference divided by the actual value.
1.5÷15 = 0.10
Converted to a percentage, it is 10%
Answer:
n = 
, n = 
Step-by-step explanation:
6n² - 5n - 7 = - 8 ( add 8 to both sides )
6n² - 5n + 1 = 0 ← in standard form
Consider the product of the factors of the coefficient of the n² term and the constant term which sum to give the coefficient of the n- term
product = 6 × 1 = 6 and sum = - 5
The factors are - 3 and - 2
Use these factors to split the n- term
6n² - 3n - 2n + 1 = 0 ( factor the first/second and third/fourth terms )
3n(2n - 1) - 1(2n - 1) = 0 ← factor out (2n - 1) from each term
(2n - 1)(3n - 1) = 0 ← in factored form
Equate each factor to zero and solve for n
3n - 1 = 0 ⇒ 3n = 1 ⇒ n =
2n - 1 = 0 ⇒ 2n = 1 ⇒ n =
