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Nikolay [14]
3 years ago
6

The ballpark made a total of $15,000 from ticket sales at Wednesday's game. The ballpark charges $20 for each adult ticket and $

10 for each child's ticket. They sold 3 times as many children's tickets as adult tickets. Write a system of equations that can be used to determine the number of adult and child tickets sold. How many adult and child tickets were sold?
Mathematics
1 answer:
IRISSAK [1]3 years ago
8 0
<h3>Answer:</h3>

equations

  • 20a +10c = 15000
  • c = 3a

solution

  • 300 adult
  • 900 children's
<h3>Step-by-step explanation:</h3>

Let "a" and "c" represent the numbers of adult and children's tickets sold, respectively. The problem statement tells us two relationships between these values:

... 20a +10c = 15000 . . . . . . total revenue from ticket sales

... c = 3a . . . . . . . . . . . . . . . . relationship between numbers of tickets sold

Using the expression for c, we can substitute into the first equation to get ...

... 20a +10(3a) = 15000

... 50a = 15000

... a = 15000/50 = 300 . . . . . adult tickets sold

... c = 3·300 = 900 . . . . . children's tickets sold

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Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

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The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that \mu = 38.72, \sigma = 3.17

Sample of 10:

This means that n = 10, s = \frac{3.17}{\sqrt{10}}

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

Z = \frac{X - \mu}{\sigma}

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\mu = 266, \sigma = 16

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This is the p-value of Z when X = 260. So

Z = \frac{X - \mu}{\sigma}

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0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

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Now n = 20, so:

Z = \frac{X - \mu}{s}

Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}

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Z = -1.68 has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

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Now n = 50, so:

Z = \frac{X - \mu}{s}

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Z = -2.65 has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that n = 15. This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

Z = \frac{X - \mu}{s}

Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}

Z = 2.42

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X = 256

Z = \frac{X - \mu}{s}

Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}

Z = -2.42

Z = -2.42 has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

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