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2 years ago

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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Luba_88 [7]

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3 years ago

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Norma-Jean [14]

The volume of a cylinder is pi r^ h
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3 years ago

Write the following in terms of sin θ only.<br> cot θ

sveta [45]

$\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\\\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
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cos(\theta)=\pm\sqrt{1-sin^2(\theta)}\\\\
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cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}\implies cot(\theta)=\cfrac{\pm\sqrt{1-sin^2(\theta)}}{sin(\theta)}$

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3 years ago

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

belka [17]

Answer:

With j = number of jars: 2,578 / 6 = j

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.

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3 years ago