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fgiga [73]
3 years ago
6

A news paper article reports the results of an election between candidates the article says that smith received 60% of the votes

and that Murphy received 1/3 of the votes why Is this article not accurate
Mathematics
1 answer:
Leokris [45]3 years ago
4 0
If you want some help here, I can tell you!

60% is 60%, since the best way to find out is by converting the two numbers into percents.

You need to find out a decimal first for the second number in order to convert it to a percent. To find a decimal out of a fraction, you put the numerator in the house (I forgot what it was called) and then you put the denominator out. You divide, and you find that 1/3 is actually a repeating decimal, 0.33333333 and so on. You round to 0.34, and that's 34% since 0.34 is 34 out of 100, basically. 34% plus 60% is NOT 100%, and 100% is a whole, so that's why it isn't accurate.
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The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 30,393 miles, with a standard
Pie

Answer:

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 30393, \sigma = 2876, n = 37, s = \frac{2876}{\sqrt{37}} = 472.81

What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 - 339 = 30054. So

X = 30732

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

By the Central Limit Theorem

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

Z = \frac{30732 - 30393}{472.81}

Z = 0.72

Z = 0.72 has a pvalue of 0.7642.

X = 30054

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

Z = \frac{30054 - 30393}{472.81}

Z = -0.72

Z = -0.72 has a pvalue of 0.2358

0.7642 - 0.2358 = 0.5284

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

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3 years ago
Divide: 6 ÷ 1
fredd [130]

Answer:

One-sixth times StartFraction 2 over 1 EndFraction

5 0
3 years ago
Read 2 more answers
Choose the multiples of 2.
IgorLugansk [536]
B)
2x6= 12
2x7= 14
2x8= 16
2x8= 18
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