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EleoNora [17]
3 years ago
11

Your thrteater company sells tickets to their holiday showing off the nutcracker. Tickets cost $5 each. Tickets for adults cost

$8 each. The total sales on opening night is $720. Write an equation to show the total sales as a combanation of kid tickets and adult tickets sold.
Mathematics
1 answer:
ivann1987 [24]3 years ago
4 0

Answer:

5x + 8y = 720

Step-by-step explanation:

Let the :

Number of kid tickets sold = x

Number of adult tickets sold = y

Hence:

Tickets for kids cost $5 each.

Tickets for adults cost $8 each.

The total sales on opening night is $720.

The equation to show the total sales as a combanation of kid tickets and adult tickets sold is:

$5 × x + $8 × y = $720

5x + 8y = 720

Therefore, the equation to show the total sales as a combanation of kid tickets and adult tickets sold is

5x + 8y = 720

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Answer:

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Step-by-step explanation:

The slope-intercept form of the equation of a line is

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where m = slope, and b = y-intercept.

From the slope-intercept equation y = 5x - 3, we see that the slope of the line is 3.

The point-slope form of the equation of a line is:

y - y1 = m(x - x1)

where m = slope, and (x1, y1) is a point on the line.

We have point (-2, -13), so x1 = -2, and y1 = -13.

We also have slope 5, so m = 5.

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Suppose we have a population whose proportion of items with the desired attribute is p = 0:5. (a) If a sample of size 200 is tak
crimeas [40]

Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28

We can now calculate the probabilities:

P(0.47

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2

We can now calculate the probabilities:

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If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

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