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borishaifa [10]
3 years ago
12

Can u help me pls they are exponents with multiplication and division

Mathematics
1 answer:
Ronch [10]3 years ago
3 0
When there is a multiplication involving exponents, we need to evaluate the numbers and each variables separately. In this case, we have 5 and 4 as the numbers, and w and h for the variables.

When we multiply 5 and 4 we get:
5 \times 4 = 20

For w, we have {w}^{6}, {w}^{4}, w. While we are multiplying the w's, we need to add the exponents together to get the final answer: {w}^{6} \times {w}^{4} \times w = {w}^{11}

For h's, there are {h}^{2}, {h}^{5}, {h}^{4}. When we apply the same thing to h's, we get {h}^{2} \times {h}^{5} \times {h}^{4} = {h}^{11}

For the final step, we need to write all of these together: 20 \times {w}^{11} \times {h}^{11} \: \: or \: \: 20 {w}^{11} {h}^{11} \: \: or \: \: 20 {(wh)}^{11}
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Step-by-step explanation:

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The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit
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Answer:

a) 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) 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.

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.

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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mariarad [96]

Answer:

4 week ago question free epoint yasss

Step-by-step explanation:

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Answer:

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