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cricket20 [7]
3 years ago
11

A ticket to a movie theatre for a student is $7. The cost for an adult is $2 more than for a student. How much would it cost 13

adult and 11 students for tickets to the movie theatre?
Mathematics
1 answer:
MakcuM [25]3 years ago
3 0

Answer: it would cost a total of $194

Step-by-step explanation:

A ticket to a movie theatre for a student is $7. The cost for an adult is $2 more than for a student.

If student ticket costs $x, adult ticket would cost x + 2

This means that the cost for an adult is

7 + 2 = $9

The cost of 13 adult tickets to the movie theatre would be

13 × 9 = $117.

The cost of 11 student tickets to the movie theatre would be

11 × 7 = $77

Total cost of 13 adult tickets and 11 student tickets would be

117 + 77 = $194

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AURORKA [14]
Rate=\frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(0)}{3-0}=\frac{3-(-4)}{3}=\frac{7}{3}
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3 years ago
What is 215% of 44? 9460 946 94.6 9.46
nirvana33 [79]

Answer:

94.6

Step-by-step explanation:

     100% of 44 = 44

⇒  200% of 44 = 44 + 44 = 88

   15% = 15/100 = 0.15

⇒ 15% of 44 = 0.15 x 44 = 6.6

Therefore, 215% of 44 = 88 + 6.6 = 94.6

7 0
2 years ago
The length of the rectangle is 3 cm and the width is 5 cm what will the area be?
WINSTONCH [101]

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15 cm²

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3 0
3 years ago
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Ludmilka [50]

Answer:

A, x>3

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8 0
3 years ago
at a particular high school, 42% of the students participate in sports and 25% of the students participate in drama. if 53% of t
baherus [9]

The probability that a student participates in both sports and drama is 0.14.

<h3>What is the formula for P(AUB), where A and B are any two events?</h3>

If A and B are any two events, then the probability of the joint event (A\cup B) is given by the following formula: P(A\cup B)=P(A)+P(B)-P(A\cap B)

Given that 42% of the students participate in sports and 25% of the students participate in drama and 53% of the students participate in either sports or drama.

Suppose S denotes that "a student participates in sports" and D denotes that "a student participates in drama".
So, we have P(S)=\frac{42}{100}=0.42, P(D)=\frac{25}{100}=0.25, P(S\cup D)=\frac{53}{100}=0.53.

We want to find the probability that a student participates in both sports and drama i.e., we want to find P(S\cap D).

By the above formula, we obtain:

P(S\cup D)=P(S)+P(D)-P(S\cap D)\\\Longrightarrow P(S\cap D)=P(S)+P(D)-P(S\cup D)\\\Longrightarrow P(S\cap D)=0.42+0.25-0.53\\\therefore P(S\cap D)=0.14=\frac{14}{100}

Therefore, the probability that a student participates in both sports and drama is 0.14.

To learn more about probability, refer: brainly.com/question/24756209

#SPJ9

6 0
2 years ago
Read 2 more answers
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