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patriot [66]
2 years ago
12

At a carnival, food tickets cost $2 each and ride tickets cost $3 each. A total of $1,240 was collected at the carnival. The num

ber of food tickets sold was 10 less than twice the number of ride tickets sold.
The system of equations represents x, the number of food tickets sold, and y, the number of ride tickets sold.

2x + 3y = 1240

x = 2y – 10

How many of each type of ticket were sold?

180 food tickets and 293 ride tickets
180 food tickets and 350 ride tickets
293 food tickets and 180 ride tickets
350 food tickets and 180 ride tickets
Mathematics
1 answer:
maria [59]2 years ago
7 0

Answer:

180 food tickets and 350 ride tickets

Step-by-step explanation:

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Question has missing details

One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction. Even when a process is running correctly, there is a small probability that a particular point will signal a problem with the process. Suppose that this probability is 0.05. What is the probability that at least one of 10 successive points indicates a problem when in fact the process is operating correctly?

Answer:

The probability that at least one of 10 successive points indicates a problem when in fact the process is operating is 0.4013

Step-by-step explanation:

Given

Let P = Probability that a point signals an error incorrectly = 0.05

Let Q = Probability that a point signals an error correctly

P + Q = 1 ---- Make Q the subject of formula

Q = 1 - P where P = 0.05

So, Q = 1 - P becomes

Q = 1 - 0.05

Q= 0.95

Solving for the probability that at least one of 10 successive points indicates a problem when in fact the process is operating.

If two events (P and Q) are independent

Then

P(P n Q) = P(P) * P(Q)

From De Morgan law;

P(P u Q) = 1 - P(P' n Q')

Where P(P u Q) represent the probability that at least one of 10 successive points

P(P' n Q') is calculated as follows;

P(P' n Q') = 0.95^10

P(P' n Q') = 0.59873693923837890625

So,

P(P u Q) = 1 - P(P' n Q') becomes

P(P u Q) = 1 - 0.59873693923837890625

P(P u Q) = 0.40126306076162109375

P(P u Q) = 0.4013 ----- Approximated

The probability that at least one of 10 successive points indicates a problem when in fact the process is operating is 0.4013

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