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kykrilka [37]
2 years ago
7

What are three ratios equivalent to 12:16

Mathematics
1 answer:
zepelin [54]2 years ago
4 0

Answer:

\dfrac{6}{8}, ~ \dfrac 34, ~\dfrac{24}{32}

Step-by-step explanation:

\dfrac{12}{16} = \dfrac{12 \times \tfrac 12}{16 \times \tfrac 12} = \dfrac{6}8\\\\\\\dfrac{12}{16} = \dfrac{12\times \tfrac 14}{16 \times \tfrac 14} = \dfrac 34\\\\\\\dfrac{12}{16} = \dfrac{12 \times 2}{16 \times 2} = \dfrac{24}{32}

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When water is released from the Kinzua Dam it is released gradually. The water level in Kinzua reservoir changes at an average r
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Option D

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Step-by-step explanation:

To easily solve this problem, we apply a rule of three

-916 inch ---------------------------------- 3 hours

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(Lets remember we are talking about average values)

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What is <br><br> 4.2 = c/8<br><br> And it would be a lot helpful
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7 0
2 years ago
Read 2 more answers
Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut
defon

Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

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 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 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.

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125:

This means that n = 125, s = \frac{7320}{\sqrt{125}} = 654.72

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

Z = \frac{39000 - 39725}{654.72}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

0.8665 = 86.65% probability that the sample mean would be at least $39000

4 0
2 years ago
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