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GREYUIT [131]
3 years ago
6

Drag steps in the given order to evaluate this expression. (4)(−3)−5−3(−6)

Mathematics
1 answer:
never [62]3 years ago
3 0

Answer:

1

Step-by-step explanation:

using PEDMAS(parentheses, exponents, division, multiply, addition, subtraction) to solve the problem

First we multiply (4)(-3)

(4)(−3)−5−3(−6)

Then multiply -3(-6)

(-12)-5-3(-6)

=-12-5+18

subtract 12-5

-12-5+18

=-17+18

Adding -17+18 gives us 1.

Therefore,  (4)(−3)−5−3(−6) is 1.

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Sveta_85 [38]

Answer:

a) Z = -2.88

b) Z = -0.96

c) 40 weeks gestation babies

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.

a. Find the standardized score (z-score), relative to all U.S. births, for a baby with a birth length of 45 cm.

Here, we use \mu = 52.2, \sigma = 2.5. So

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

Z = \frac{45 - 52.2}{2.5}

Z = -2.88

b. Find the standardized score of a birth length of 45 cm. for babies born one month early, using 47.4 as the mean.

Here, we use \mu = 47.4, \sigma = 2.5. So

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

Z = \frac{45 - 47.4}{2.5}

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c. For which group is a birth length of 45 cm more common?

For each group, the probability is 1 subtracted by the pvalue of Z.

Z = -2.88 has a lower pvalue than Z = -0.96, so for Z = -2.88 the probability 1 - pvalue of Z will be greater. This means that for 40 weeks gestation babies a birth length of 45 cm is more common.

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3 years ago
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Answer:

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