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marysya [2.9K]
3 years ago
6

On June 15, Paradise Park purchased merchandise for the racetrack. The invoice was for $4,500, terms 2/10, n/30. On June 20, Par

adise Park returned $200 of merchandise for credit. On June 25, it paid the amount owed. The debit to Purchases on June 15 is
a. $4,214.00.
b. $4,786.00.
c. $4,300.00.
d. $4,500.00.
Mathematics
1 answer:
vaieri [72.5K]3 years ago
4 0
The correct answer is letter D. $4,500.00
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A sample of 100 footballs showed an average air pressure of 13 psi. The standard deviation of the population is known to be .25
Vladimir79 [104]

Answer:

0.025 is the standard error of mean.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 100

Sample mean = 13 psi

Population standard deviation = 0.25 psi

We have to find the standard error of the mean.

Formula:

Standard error =

\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.25}{\sqrt{100}} = \dfrac{0.25}{10} = 0.025

0.025 is the standard error of mean.

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4/3 divided by 7/8 how
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You have to find the commons denominator first then do the same thing to the top then you divide
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In a clinical trial of 268 subjects treated with a​ drug, 11​% of the subjects reported dizziness. The margin of error is plus o
Mariulka [41]

Answer:

\hat p =0.11 represent the proportion estimated of subjects reported with dizziness

n = 268 represent the random sample selected

\hat q = 1-\hat p = 1-0.11= 0.89 represent the proportion of subjects No reported with dizziness

E= 0.04 = 4% represent the margin of error for the confidence interval

\alpha= 1-0.90 =0.1 and this value represent the significance level of the test or the probability of error type I

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

The margin of error is given by:

ME= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

Lower= 0.11-0.04 = 0.07

Upper= 0.11+0.04 = 0.15

And for this case we have the following info:

\hat p =0.11 represent the proportion estimated of subjects reported with dizziness

n = 268 represent the random sample selected

\hat q = 1-\hat p = 1-0.11= 0.89 represent the proportion of subjects No reported with dizziness

E= 0.04 = 4% represent the margin of error for the confidence interval

\alpha= 1-0.90 =0.1 and this value represent the significance level of the test or the probability of error type I

4 0
3 years ago
Pythagorean Theorem
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The answer is 10.6 :)
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Whats the square root of 1/5 minus the square root of 5
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You have to simplify the 1/5 first.
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Then simplify

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