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anyanavicka [17]
3 years ago
8

HELP PLS! WILL REWARD BRAINLIEST, TYSM !

Mathematics
2 answers:
jonny [76]3 years ago
6 0

Answer:

-0.9b can be combined as well a b/12 and -2b

Helga [31]3 years ago
5 0

Answer:

b and d

Step-by-step explanation:

they both have b

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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
mash [69]

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

X = 21

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

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

Z = 0.56

Z = 0.56 has a p-value of 0.7123

X = 19

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

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%

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

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Norma-Jean [14]
I’m sorry i don’t know but i’m thinking it’s B not sure tho
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3 years ago
How to do property eqautionseqautions
Tanya [424]

Answer:


Step-by-step explanation:


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2 years ago
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