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Mekhanik [1.2K]
3 years ago
6

Write an expression for the sequence of operations described below. divide 10 by the sum of n and m Do not simplify any part of

the expression.
Mathematics
2 answers:
nata0808 [166]3 years ago
6 0

Answer:

\frac{10}{n+m}

Step-by-step explanation:

We are asked to divide 10 by the sum of n and m.

The sum of n and m would be n added to m: n+m.

Now, we will divide 10 by n+m as shown below:

\frac{10}{n+m}

Therefore, our required expression would be \frac{10}{n+m}.

lina2011 [118]3 years ago
3 0

Answer:

(m+n)/10

Step-by-step explanation:

divide 10 by the sum of n and m

(m+n)/10

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Write 5 equivalent fractions for each fraction. 1/2, 1/4, 1/8, 1/3, 1/6
agasfer [191]

<u>Equivalent Fractions 1/2</u>

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3 years ago
Segment A'B' is parallel to segment AB.<br>What is the length of segment AB?
Nonamiya [84]

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Step-by-step explanation:

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cross multiply

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7 0
3 years ago
For a given IQ test, an individual is considered a genius if their score falls more than three standard deviations from the mean
ki77a [65]

Answer:

P(Z>3) = 1-P(Z

So then the probability that an individual present and IQ higher than 3 deviation from the mean is 0.00135

And if we find the number of individuals that can be considered as genius we got: 0.00135*1500=2.025

And we can say that the answer is a.2

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu=?,\sigma=?)  

We are interested on this probability

P(X>X+3\mu)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

And we can find the following probablity:

P(Z>3) = 1-P(Z

So then the probability that an individual present and IQ higher than 3 deviation from the mean is 0.00135

And if we find the number of individuals that can be considered as genius we got: 0.00135*1500=2.025

And we can say that the answer is a/2.0

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3 years ago
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