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OLga [1]
3 years ago
11

A band sold 10,598 tickets to their concert at the stadium. If the stadium can only hold 15,000 people, how many more tickets ca

n be sold?
Mathematics
1 answer:
anzhelika [568]3 years ago
6 0

Answer:

4402

Step-by-step explanation:

15,000-10,598=4,402

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Step-by-step explanation:

2 1/3

=2*3+1/3

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3 years ago
Simplify 5/6<br><br> IM KINDA CONFUSED UMM PLEASE HELP
gulaghasi [49]
The answer is 5/6. It can’t be simplified any further.
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Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly samp
aleksklad [387]

Answer:

95.44% probability the resulting sample proportion is within .04 of the true proportion.

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes 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 \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the sampling distribution of the sample proportion in sample of size n, the mean is \mu = p and the standard deviation is s = \sqrt{\frac{p(1-p)}{n}}

In this question:

p = 0.2, n = 400

So

\mu = 0.2, s = \sqrt{\frac{0.2*0.8}{400}} = 0.02

How likely is the resulting sample proportion to be within .04 of the true proportion (i.e., between .16 and .24)?

This is the pvalue of Z when X = 0.24 subtracted by the pvalue of Z when X = 0.16.

X = 0.24

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

By the Central Limit Theorem

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

Z = \frac{0.24 - 0.2}{0.02}

Z = 2

Z = 2 has a pvalue of 0.9772.

X = 0.16

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

Z = \frac{0.16 - 0.2}{0.02}

Z = -2

Z = -2 has a pvalue of 0.0228.

0.9772 - 0.0228 = 0.9544

95.44% probability the resulting sample proportion is within .04 of the true proportion.

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3 years ago
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On a pleasant Monday, the Dow-Jones Industrial average closed at 11,315. On Wednesday, it closed down twice as many points as it
kirza4 [7]
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3 years ago
Jacob paid $987 for a laptop which was 60% off the original price. What was the
Ray Of Light [21]

Answer:

$2467.50

Step-by-step explanation:

987 = ( 1 - 0.6 )x

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x = 2467.5

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