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maria [59]
2 years ago
6

MARKING AS BRAINLIEST! LAST ATTEMPT! ) show ur work

Mathematics
1 answer:
Ivenika [448]2 years ago
5 0

- 3 \sqrt{180 {x}^{5}y }  =  - 3 \sqrt{36 {x}^{4} \times 5xy }  =  - 18  {x}^{2} \sqrt{5xy}

5 \sqrt{108xy {z}^{2} }  = 5 \sqrt{36 {z}^{2} \times 3xy }  = 30z \sqrt{3xy}

3x \sqrt{x {y}^{10} {z}^{7}  }  = 3x \sqrt{ {y}^{10} {z}^{6}  \times xz }  = 3x {y}^{5}  {z}^{3}  \sqrt{xz}

\sqrt{ {x}^{3} }  =  \sqrt{ {x}^{2} \times x }  = x \sqrt{x}

\sqrt{x {y}^{3}z }  = y \sqrt{xyz}

\sqrt{18 {x}^{2}y }  =  \sqrt{9 {x}^{2} \times 2y }  = 3x \sqrt{2y}

\sqrt{32 {x}^{4}  {y}^{2} }  =  \sqrt{16 ({ {x}^{2} })^{2} {y}^{2}  \times 2 }  =  4 {x}^{2} y \sqrt{2}

2 \sqrt{16 {x}^{7} {y}^{3} {z}^{4}   }  = 2 \sqrt{ {4}^{2} {x}^{6} {y}^{2} ( { {z}^{2} )}^{2}  \times xy  }    = 8 {x}^{3} y {z}^{2}  \sqrt{xy}

{x}^{2}  \sqrt{ {x}^{3} {y}^{2}  }  =  {x}^{2}  \sqrt{ {x}^{2} {y}^{2}  \times x }  =  {x}^{3} y \sqrt{x}

Answer: ......

Ok done. Thank to me :>

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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.

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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

X = 21

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Z = \frac{21 - 20.5}{0.9}

Z = 0.56

Z = 0.56 has a p-value of 0.7123

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Z = \frac{19 - 20.5}{0.9}

Z = -1.67

Z = -1.67 has a p-value of 0.0475

0.7123 - 0.0475 = 0.6648

0.6648*100% = 66.48%

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