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JulijaS [17]
3 years ago
12

A movie theater charges $5 for an adult ticket and $2 for a child's ticket. One Saturday, the theater sold 785 tickets for $3280

. How many child's tickets and adult's tickets were sold that Saturday?
Mathematics
2 answers:
Nadya [2.5K]3 years ago
6 0

Answer:

5x + 2y = 3280

x+y = 785

x = 785 - y

5*(785 - y) + 2y = 3280

3925 - 5y + 2y = 3280

3925 - 3280 - 3y = 0

645 = 3y

y=645/3=215

x=785-215 = 570

Adults - 570 tickets, childs - 215 tickets.

570*5 + 215*2 = 3280

570 adult and 215 child and 3,280 in total

if i could get brainliest that would be great

Yuki888 [10]3 years ago
5 0

Answer: 215 child tickets, 570 adult tickets.

Step-by-step explanation:

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Suppose that four microchips in a production run of fifty are defective. A sample of six is to be selected to be checked for def
Irina-Kira [14]

Answer:

a) 15,890,700 different samples can be chosen

b) 6,523,881 will contain at least one defective chip

c) 41.1%

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the microchips are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

(a) How many different samples can be chosen?

Samples of 6 from a set of 50.

Then

C_{50,6} = \frac{50!}{6!(50-6)!} = 15890700

15,890,700 different samples can be chosen

(b) How many samples will contain at least one defective chip?

Either a sample contains no defective chip, or it contains at least one. From a), the sum of them is 15890700.

Then

No defective chips:

Four are defective.

So 50-4 = 46 are not.

This is samples of 6 from a set of 46.

C_{46,6} = \frac{46!}{6!(46-6)!} = 9366819

9,366,819 samples contain no defective chips.

At least one:

9366819 + n = 15890700

n = 6523881

6,523,881 will contain at least one defective chip

(c) What is the probability (as a percent) that a randomly chosen sample of six contains at least one defective chip

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6,523,881/15,890,700 = 0.411

As a percent

41.1%

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