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Slav-nsk [51]
3 years ago
6

Solve for x. Round to the nearest tenth, if necessary.

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
8 0

Answer:

x = 63.9

Step-by-step explanation:

We have:

\frac{QR}{sin35} = \frac{78}{sin90}

=> QR × sin90 = 78 × sin35

=> QR × sin90 = 44.7

=> QR = 44.7

But, we also have: x^{2} + QR^{2} = 78^{2} <=> x^{2} + 44.7^{2} = 78^{2} <=> x^{2} = 6084 - 1998.09

=> x^{2} = 4085.91

=> x = \sqrt{4085.91} = 63.9211.. = 63.9

<3 Have a nice day!!

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Lengths of full-term babies in the US are Normally distributed with a mean length of 20.5 inches and a standard deviation of 0.9
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Answer:

66.48% of full-term babies are between 19 and 21 inches long at birth

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.

Mean length of 20.5 inches and a standard deviation of 0.90 inches.

This means that \mu = 20.5, \sigma = 0.9

What percentage of full-term babies are between 19 and 21 inches long at birth?

The proportion is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19. Then

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Z = 0.56

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0.7123 - 0.0475 = 0.6648

0.6648*100% = 66.48%

66.48% of full-term babies are between 19 and 21 inches long at birth

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