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9 months ago

Simplify the expression. Rewrite using only positive exponents.

Mathematics

1 answer:

Afina-wow [57]9 months ago

8 0

Answer:

The answer is D

Step-by-step explanation:

$((c^6d^5e^4)*(c^-3de^-2))^2\\$

Reorder the terms

$= ((e^4c^6d^5)*(e^-2c^-3d))^2\\$

Calculate

$= (e^4c^6d^5e^-2c^-3d)^2$

Use the properties of exponents

$= (e^2c^3d^6)^2$

Therefore, the answer is c^6 d^12 e^4 or D

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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:

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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.

In this question:

Mean $\mu$ , standard deviation $\sigma$

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b. If the length of pregnancy is in the lowest a%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

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