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lisov135 [29]
2 years ago
14

Find the sum , write fractions in simple form 11/12 + (-7/12)

Mathematics
2 answers:
mrs_skeptik [129]2 years ago
8 0
The answer is 1/3 hope that helps
dybincka [34]2 years ago
3 0

Answer:

1/3

Step-by-step explanation:

11/12 -7/12 = 11-7/12 =4/12 =1/3

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Write the following in terms of sin θ only.<br> cot θ
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Jon collects 2,578 ounces of honey from his beehive. he will put the honey in jars that hold 6 ounces each. which expression is
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Answer:

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Step-by-step explanation:

8 0
3 years ago
The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit
Alik [6]

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.

8 0
3 years ago
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