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Gwar [14]
3 years ago
5

Enter an estimate. Round each mixed number to the nearest half in your estimate.

Mathematics
1 answer:
zhenek [66]3 years ago
3 0
I got 9/14 ( sorry if I’m wrong )
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Find the quotient. h^ 6 ÷h
Arada [10]
h^6:h=h^{6-1}=\boxed{h^5}
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Look at pic 10 pts will mark brainilest tell me if u need multiple choice answers
stealth61 [152]

Answer:

Step-by-step explanation:

1/2 b*h *2

1/2 * 14*24*2= 336

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The Arc Electronic Company had an income of 59 million dollars last year. Suppose the mean income of firms in the same industry
Zolol [24]

Answer:

0.0228 = 2.28% probability that a randomly selected firm will earn more than Arc did last year

Step-by-step explanation:

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.

Suppose the mean income of firms in the same industry as Arc for a year is 45 million dollars with a standard deviation of 7 million dollars

This means that \mu = 45, \sigma = 7

What is the probability that a randomly selected firm will earn more than Arc did last year?

Arc earned 59 million, so this is 1 subtracted by the pvalue of Z when X = 59.

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

Z = \frac{59 - 45}{7}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that a randomly selected firm will earn more than Arc did last year

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3 years ago
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pishuonlain [190]
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Flura [38]

Answer:

1 1/8 or 9/8

have a great day :)

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3 years ago
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